Maths in Focus - Margaret Grove - ch7

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Linear Functions TERMINOLOGY Collinear points: Two or more points that lie on the same straight line

Interval: A section of a straight line including the end points

Concurrent lines: Two or more lines that intersect at a single point

Midpoint: A point lying exactly halfway between two points

Gradient: The slope of a line measured by comparing the vertical rise over the horizontal run. The symbol for gradient is m

Perpendicular distance: The shortest distance between a point and a line. The distance will be at right angles to the line

Chapter 7 Linear Functions

INTRODUCTION IN CHAPTER 5, YOU STUDIED functions and their graphs. This chapter looks at the linear function, or straight-line graph, in more detail. Here you will study the gradient and equation of a straight line, the intersection of two or more lines, parallel and perpendicular lines, the midpoint, distance and the perpendicular distance from a point to a line.

DID YOU KNOW? Pierre de Fermat (1601–65) was a lawyer who dabbled in mathematics. He was a contemporary of Descartes, and showed the relationship between an equation in the form Dx = By, where D and B are constants, and a straight-line graph. Both de Fermat and Descartes only used positive values of x, but de Fermat used the x-axis and y-axis as perpendicular lines as we do today. De Fermat’s notes Introduction to Loci, Method of Finding Maxima and Minima and Varia opera mathematica were only published after his death. This means that in his lifetime de Fermat was not considered a great mathematician. However, now he is said to have contributed as much as Descartes towards the discovery of coordinate geometry. De Fermat also made a great contribution in his discovery of differential calculus.

Class Assignment Find as many examples as you can of straight-line graphs in newspapers and magazines.

Distance The distance between two points (or the length of the interval between two points) is easy to find when the points form a vertical or horizontal line.

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Maths In Focus Mathematics Preliminary Course

EXAMPLES Find the distance between 1. ^ -1, 4 h and ^ -1, -2 h

Solution

Counting along the y-axis, the distance is 6 units. 2. ^ 3, 2 h and ^ -4, 2 h

Solution

Counting along the x-axis, the distance is 7 units.

When the two points are not lined up horizontally or vertically, we use Pythagoras’ theorem to find the distance.


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EXAMPLE Find the distance between points ^ 3, -1 h and ^ -2, 5 h.

Solution

BC = 5 and AC = 6 By Pythagoras’ theorem,

You studied Pythagoras’ theorem in Chapter 4.

c =a +b AB 2 = 5 2 + 6 2 = 25 + 36 = 61 2

2

2

` AB = 61 Z 7.81

DID YOU KNOW? Pythagoras made many discoveries about music as well as about mathematics. He found that changing the length of a vibrating string causes the tone of the music to change. For example, when a string is halved, the tone is one octave higher.

The distance between two points _ x 1, y 1 i and _ x 2, y 2 i is given by d=

2 2 _ x2 - x1 i + _ y2 - y1 i

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Proof

If points A and B were changed around, the formula would be d =

(x 1 - x 2 ) + (y 1 - y 2 ) , 2

2

which would give the same answer.

Let A = _ x 1, y 1 i and B = _ x 2, y 2 i Length AC = x 2 - x 1 and length BC = y 2 - y 1 By Pythagoras’ theorem AB 2 = AC 2 + BC 2 d 2 = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 `d=

2 2 _ x2 - x1 i + _ y2 - y1 i

EXAMPLES 1. Find the distance between the points ^ 1, 3 h and ^ -3, 0 h.

Solution Let ^ 1, 3 h be _ x 1, y 1 i and ^ -3, 0 h be _ x 2, y 2 i d=

2 2 _ x2 - x1 i + _ y2 - y1 i

= ] -3 - 1 g2 + ] 0 - 3 g2 = ] -4 g2 + ] -3 g2 = 16 + 9 = 25 =5 So the distance is 5 units. 2. Find the exact length of AB given that A = ^ -2, -4 h and B = ^ -1, 5 h .

Solution Let ^ -2, -4 h be _ x 1, y 1 i and ^ -1, 5 h be _ x 2, y 2 i d=

You would still get 82 if you used (- 2, - 4) as (x 2 , y 2 ) and (-1, 5) as (x 1 , y 1 ).

2 2 _ x2 - x1 i + _ y2 - y1 i

=

6 -1 - ^ -2 h @ 2 + 6 5 - ^ -4 h @ 2

= = =

12 + 92 1 + 81 82


7.1 Exercises 1.

Find the distance between points (a) ^ 0, 2 h and ^ 3, 6 h (b) ^ -2, 3 h and ^ 4, -5 h (c) ^ 2, -5 h and ^ -3, 7 h

2.

Find the exact length of the interval between points (a) ^ 2, 3 h and ^ -1, 1 h (b) ^ -5, 1 h and ^ 3, 0 h (c) ^ - 2, -3 h and ^ - 4, 6 h (d) ^ -1, 3 h and ^ -7, 7 h

3.

4.

Find the distance, correct to 2 decimal places, between points (a) ^ 1, -4 h and ^ 5, 5 h (b) ^ 0, 4 h and ^ 3, -2 h (c) ^ 8, -1 h and ^ -7, 6 h Find the perimeter of D ABC with vertices A ^ 3, 1 h, B ^ -1, 1 h and C ^ -1, -2 h .

5.

Prove that the triangle with vertices ^ 3, 4 h, ^ -2, 7 h and ^ 6, -1 h is isosceles.

6.

Show that AB = BC, where A = ^ -2, 5 h, B = ^ 4, -2 h and C = ^ -3, -8 h .

7.

Show that points ^ 3, -4 h and ^ 8,1 h are equidistant from point ^ 7, -3 h .

8.

A circle with centre at the origin O passes through the point _ 2 , 7 i . Find the radius of the circle, and hence its equation.

9.

Prove that the points X _ 2 , -3 i, Y _ -1, 10 i and Z _ - 6 , 5 i all lie on a circle with centre at the origin. Find its equation.

10. If the distance between ^ a, -1 h and ^ 3, 4 h is 5, find the value of a. 11. If the distance between ^ 3, -2 h and ^ 4, a h is 7 , find the exact value of a.

12. Prove that A ^ 1, 4 h, B ^ 1, 2 h and C _ 1 + 3 , 3 i are the vertices of an equilateral triangle. 13. If the distance between ^ a, 3 h and ^ 4, 2 h is 37 , find the values of a. 14. The points M ^ -1, -2 h, N (3, 0), P ^ 4, 6 h and Q ^ 0, 4 h form a quadrilateral. Prove that MQ = NP and QP = MN. What type of quadrilateral is MNPQ? 15. Show that the diagonals of a square with vertices A ^ -2, 4 h, B ^ 5, 4 h, C ^ 5, -3 h and D ^ -2, -3 h are equal. 16. (a) Show that the triangle with vertices A ^ 0, 6 h, B ^ 2, 0 h and C ^ -2, 0 h is isosceles. (b) Show that perpendicular OA, where O is the origin, bisects BC. 17. Find the exact length of the diameter of a circle with centre ^ -3, 4 h if the circle passes through the point ^ 7, 5 h . 18. Find the exact length of the radius of the circle with centre (1, 3) if the circle passes through the point ^ -5, -2 h . 19. Show that the triangle with vertices A ^ -2, 1 h, B ^ 3, 3 h and C ^ 7, -7 h is right angled. 20. Show that the points X ^ 3, -3 h, Y ^ 7, 4 h and Z ^ - 4, 1 h form the vertices of an isosceles right-angled triangle.

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Midpoint The midpoint is the point halfway between two other points.

The midpoint of two points _ x 1, y 1 i and _ x 2, y 2 i is given by M=e

x1 + x2 y1 + y2 o , 2 2

Proof

Can you see why these triangles are similar?

Find the midpoint of points A _ x 1, y 1 i and B _ x 2, y 2 i. Let M = ^ x, y h Then D APQ <; D ABR AQ AP = AR AB x - x1 1 ` x -x = 2 2 1 2 _ x - x1 i = x2 - x1 2x - 2x 1 = x 2 - x 1 2x = x 1 + x 2 x1 + x2 ` x= 2 y1 + y2 Similarly, y = 2 `

EXAMPLES 1. Find the midpoint of ^ -1, 4 h and ^ 5, 2 h.

Solution x=

x1 + x2 2


-1 + 5 2 4 = 2 =2 y1 + y2 y= 2 4+2 = 2 6 = 2 =3 So M = (2, 3) . =

2. Find the values of a and b if ^ 2, -3 h is the midpoint between ^ -7, -8 h and ^ a, b h.

Solution x=

x1 + x2

2 -7 + a 2= 2 4 = -7 + a 11 = a y1 + y2 y= 2 -8 + b -3 = 2 -6 = -8 + b 2=b So a = 11 and b = 2.

Note that the x-coordinate of the midpoint is the average of x 1 and x 2 . The same applies to the y-coordinate.

PROBLEM A timekeeper worked out the average time for 8 finalists in a race. The average was 30.55, but the timekeeper lost one of the finalist’s times. The other 7 times were 30.3, 31.1, 30.9, 30.7, 29.9, 31.0 and 30.3. Can you find out the missing time?

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7.2 Exercises 1.

2.

Find the values of a and b if (a) ^ 4, 1 h is the midpoint of ^ a, b h and ^ -1, 5 h (b) ^ -1, 0 h is the midpoint of ^ a, b h and ^ 3, -6 h (c) ^ a, 2 h is the midpoint of (3, b h and ^ -5, 6 h (d) ^ -2, 1 h is the midpoint of ^ a, 4 h and ^ -3, b h (e) ^ 3, b h is the midpoint of ^ a, 2 h and ^ 0, 0 h

3.

Prove that the origin is the midpoint of ^ 3, -4 h and ^ -3, 4 h .

4.

Show that P = Q where P is the midpoint of ^ -2, 3 h and ^ 6, -5 h and Q is the midpoint of ^ -7, -5 h and ^ 11, 3 h .

5.

The locus is the path that P (x, y) follows.

Find the midpoint of (a) ^ 0, 2 h and ^ 4, 6 h (b) ^ -2, 3 h and ^ 4, -5 h (c) ^ 2, -5 h and ^ -6, 7 h (d) ^ 2, 3 h and ^ -8, 1 h (e) ^ -5, 2 h and ^ 3, 0 h (f) ^ -2, -2 h and ^ -4, 6 h (g) ^ 1, -4 h and ^ 5, 5 h (h) ^ 0, 4 h and ^ 3, -2 h (i) ^ 8, -1 h and ^ -7, 6 h (j) ^ 3, 7 h and ^ -3, 4 h

6.

Find the point that divides the interval between ^ 3, -2 h and ^ 5, 8 h in the ratio of 1:1. Show that the line x = 3 is the perpendicular bisector of the interval between the points ^ -1, 2 h and ^ 7, 2 h .

7.

The points A ^ -1, 2 h, B ^ 1, 5 h, C ^ 6, 5 h and D ^ 4, 2 h form a parallelogram. Find the midpoints of the diagonals AC and BD. What property of a parallelogram does this show?

8.

The points A ^ 3, 5 h, B ^ 9, -3 h, C ^ 5, -6 h and D ^ -1, 2 h form a quadrilateral. Prove that the diagonals are equal and bisect one another. What type of quadrilateral is ABCD?

9.

A circle with centre ^ -2, 5 h has one end of a diameter at ^ 4, -3 h . Find the coordinates of the other end of the diameter.

10. A triangle has vertices at A ^ -1, 3 h, B ^ 0, 4 h and C ^ 2, -2 h . (a) Find the midpoints X, Y and Z of sides AB, AC and BC respectively. 1 (b) Show that XY = BC, 2 1 1 XZ = AC and YZ = AB. 2 2 11. Point P ^ x, y h moves so that the midpoint between P and the origin is always a point on the circle x 2 + y 2 = 1. Find the equation of the locus of P. 12. Find the equation of the locus of the point P ^ x, y h that is the midpoint between all points on the circle x 2 + y 2 = 4 and the origin.

Gradient The gradient of a straight line measures its slope. The gradient compares the vertical rise with the horizontal run.


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rise Gradient = run

On the number plane, this is a measure of the rate of change of y with respect to x.

The rate of change of y with respect to x is a very important measure of their relationship. In later chapters you will use the gradient for many purposes, including sketching curves, finding the velocity and acceleration of objects, and finding maximum and minimum values of formulae.

EXAMPLES Find the gradient of each interval. 1. You will study the gradient at different points on a curve in the next chapter.

Solution rise Gradient = run 2 = 3 CONTINUED

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2.

Solution In this case, x is - 3 (the run is measured towards the left). rise Gradient = run 2 = -3 2 =3

Positive gradient leans to the right.

Negative gradient leans to the left.

Gradient given 2 points The gradient of the line between _ x 1, y 1 i and _ x 2, y 2 i is given by y2 - y1 m= x -x 2 1

Proof


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BC = y 2 - y 1 and AC = x 2 - x 1 rise Gradient = run y2 - y1 = x -x 2 1

This formula could also be y1 - y2 written m = x1 - x2

EXAMPLES 1. Find the gradient of the line between points ^ 2, 3 h and ^ -3, 4 h .

Solution y2 - y1 Gradient: m = x - x 2 1 4-3 = -3 - 2 1 = -5 1 =5 2. Prove that points ^ 2, 3 h, ^ -2, -5 h and ^ 0, -1 h are collinear.

Solution To prove points are collinear, we show that they have the same gradient (slope).

CONTINUED

Collinear points lie on the same line, so they have the same gradients.

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Gradient of the interval between ^ -2, -5 h and ^ 0, -1 h : y2 - y1 m= x -x 2 1 -1 - ] -5 g = 0 - ] -2 g -1 + 5 = 2 4 = 2 =2 Gradient of the interval between ^ 0, -1 h and ^ 2, 3 h : y2 - y1 m= x -x 2 1 3 - ] -1 g = 2-0 3+1 = 2 4 = 2 =2 Since the gradient of both intervals is the same, the points are collinear.

Gradient given the angle at the x-axis The gradient of a straight line is given by m = tan i where i is the angle the line makes with the x-axis in the positive direction

Proof rise m = run opposite = adjacent = tan i


For an acute angle tan i 2 0.

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For an obtuse angle tan i 1 0.

Class Discussion 1. Which angles give a positive gradient? 2. Which angles give a negative gradient? Why? 3. What is the gradient of a horizontal line? What angle does it make with the x-axis? 4. What angle does a vertical line make with the x-axis? Can you find its gradient?

EXAMPLES 1. Find the gradient of the line that makes an angle of 135c with the x-axis in the positive direction.

Solution

m = tan i = tan 135c = -1

2. Find the angle, in degrees and minutes, that a straight line makes with the x-axis in the positive direction if its gradient is 0.5.

Solution m = tan i ` tan i = 0.5 i = 26c34l

Can you see why the gradient is negative?

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7.3 Exercises 1.

2.

3.

Find the gradient of the line between (a) ^ 3, 2 h and ^ 1, -2 h (b) ^ 0, 2 h and ^ 3, 6 h (c) ^ -2, 3 h and ^ 4, -5 h (d) ^ 2, -5 h and ^ -3, 7 h (e) ^ 2, 3 h and ^ -1, 1 h (f) ^ - 5, 1 h and ^ 3, 0 h (g) ^ -2, -3 h and ^ -4, 6 h (h) ^ -1, 3 h and ^ -7, 7 h (i) ^ 1, -4 h and ^ 5, 5 h (j) ^ 0, 4 h and ^ 3, -2 h If the gradient of _ 8, y 1 i and ^ -1, 3 h is 2, find the value of y 1 . The gradient of ^ 2, -1 h and ^ x, 0 h is –5. Find the value of x.

4.

The gradient of a line is –1 and the line passes through the points ^ 4, 2 h and ^ x, -3 h . Find the value of x. 5. (a) Show that the gradient of the line through ^ -2, 1 h and ^ 3, 4 h is equal to the gradient of the line between the points ^ 2, -1 h and ^ 7,2 h . (b) Draw the two lines on the number plane. What can you say about the lines? 6.

7.

8.

Show that the points A ^ -1, 2 h, B ^ 1, 5 h, C ^ 6, 5 h and D ^ 4, 2 h form a parallelogram. Find the gradients of all sides. The points A ^ 3, 5 h, B ^ 9, -3 h, C ^ 5, -6 h and D ^ -1, 2 h form a rectangle. Find the gradients of all the sides and the diagonals. Find the gradients of the diagonals of the square with vertices A ^ -2, 1 h, B ^ 3, 1 h, C ^ 3, 6 h and D ^ -2, 6 h .

9.

A triangle has vertices A ^ 3, 1 h, B ^ -1, -4 h and C ^ -11, 4 h . (a) By finding the lengths of all sides, prove that it is a rightangled triangle. (b) Find the gradients of sides AB and BC.

10. (a) Find the midpoints F and G of sides AB and AC where ABC is a triangle with vertices A ^ 0, 3 h, B ^ 2, -7 h and C ^ 8, -2 h . (b) Find the gradients of FG and BC. 11. The gradient of the line between a moving point P ^ x, y h and the point A ^ 5, 3 h is equal to the gradient of line PB where B has coordinates ^ 2, -1 h . Find the equation of the locus of P. 12. Prove that the points ^ 3, -1 h, ^ 5, 5 h and ^ 2, -4 h are collinear. 13. Find the gradient of the straight line that makes an angle of 45c with the x-axis in the positive direction. 14. Find the gradient, to 2 significant figures, of the straight line that makes an angle of 42c51l with the x-axis. 15. Find the gradient of the line that makes an angle of 87c14l with the x-axis, to 2 significant figures. 16. Find the angle, in degrees and minutes, that a line with gradient 1.2 makes with the x-axis. 17. What angle, in degrees and minutes does the line with gradient –3 make with the x-axis in the positive direction?


18. Find the exact gradient of the line that makes an angle with the x-axis in the positive direction of (a) 60c (b) 30c (c) 120c. 19. Show that the line passing through ^ 4, -2 h and ^ 7, -5 h

makes an angle of 135c with the x-axis in the positive direction. 20. Find the exact value of x with rational denominator if the line passing through ^ x, 3 h and ^ 2, 1 h makes an angle of 60c with the x-axis.

Gradient given an equation In Chapter 5 you explored and graphed linear functions. You may have noticed a relationship between the graph and the gradient and y-intercept of a straight line.

Investigation 1. (i) Draw the graph of each linear function. (ii) By selecting two points on the line, find its gradient. (a) y = x (b) y = 2x (c) y = 3x (d) y = - x (e) y = - 2x Can you find a pattern for the gradient of each line? Can you predict what the gradient of y = 5x and y = - 9x would be? 2. (i) Draw the graph of each linear function. (ii) Find the y-intercept. (a) y = x (b) y = x + 1 (c) y = x + 2 (d) y = x - 2 (e) y = x - 3 Can you find a pattern for the y-intercept of each line? Can you predict what the y-intercept of y = x + 11 and y = x - 6 would be?

y = mx + b has m = gradient b = y-intercept

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EXAMPLES 1. Find the gradient and y-intercept of the linear function y = 7x - 5.

Solution The equation is in the form y = mx + b where m = 7 and b = - 5. Gradient = 7 y-intercept = - 5 2. Find the gradient of the straight line with equation 2x + 3y - 6 = 0.

Solution First, we change the equation into the form y = mx + b. 2x + 3y - 6 = 0 2x + 3y - 6 + 6 = 0 + 6 2x + 3y = 6 2x - 2x + 3y = 6 - 2x 3y = 6 - 2x = - 2x + 6 3y - 2x + 6 = 3 3 - 2x 6 y= + 3 3 2 = - x +2 3 2 m=3 2 So the gradient is - . 3

There is a general formula for finding the gradient of a straight line.

The gradient of the line ax + by + c = 0 is given by m=-

Proof ax + by + c = 0 by = - ax - c ax c y=b b a ` m=b

a b


EXAMPLE Find the gradient of 3x - y = 2.

Solution 3x - y = 2 3x - y - 2 = 0 a = 3, b = - 1 a m=b 3 =-1 =3 ` gradient is 3

7.4 Exercises 1.

Find (i) the gradient and (ii) the y-intercept of each linear function. (a) y = 3x + 5 (b) f ] x g = 2x + 1 (c) y = 6x - 7 (d) y = - x (e) y = - 4x + 3 (f) y = x - 2 (g) f ] x g = 6 - 2x (h) y = 1 - x (i) y = 9x (j) y = 5x - 2

2.

Find (i) the gradient and (ii) the y-intercept of each linear function. (a) 2x + y - 3 = 0 (b) 5x + y + 6 = 0 (c) 6x - y - 1 = 0 (d) x - y + 4 = 0 (e) 4x + 2y - 1 = 0 (f) 6x - 2y + 3 = 0 (g) x + 3y + 6 = 0 (h) 4x + 5y - 10 = 0 (i) 7x - 2y - 1 = 0 (j) 5x - 3y + 2 = 0

3.

Find the gradient of the straight line. (a) y = 4x (b) y = - 2x - 1 (c) y = 2 (d) 2x + y - 5 = 0 (e) x + y + 1 = 0 (f) 3x + y = 8 (g) 2x - y + 5 = 0 (h) x + 4y - 12 = 0 (i) 3x - 2y + 4 = 0 (j) 5x - 4y = 15 2 (k) y = x + 3 3 x (l) y = 2 x (m) y = - 1 5 2x (n) y = +5 7 3x -2 (o) y = 5 x 1 (p) 2y = - + 7 3 y (q) 3x - = 8 5 x y (r) + =1 2 3 2x (s) - 4y - 3 = 0 3 x 2y + +7=0 (t) 4 3

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Equation of a Straight Line There are several different ways to write the equation of a straight line.

General form ax + by + c = 0

Gradient form y = mx + b where m = gradient and b = y-intercept

Intercept form

x y a+b =1 where a and b are the x-intercept and y-intercept respectively

Proof b m = - a, b = b `

b y = -ax + b y

`

b y

x = -a + 1

x a+b =1

Point-gradient formula There are two formulae for finding the equation of a straight line. One of these uses a point and the gradient of the line.

The equation of a straight line is given by y - y1 = m _ x - x1 i This is a very useful formula as it is used in many topics in this course.

where _ x 1, y 1 i lies on the line with gradient m


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Proof Given point _ x 1, y 1 i on the line with gradient m Let P = ^ x, y h Then line AP has gradient y2 - y1 m= x -x 2 1 y - y1 ` m= x-x 1 m _ x - x1 i = y - y1

Two-point formula The equation of a straight line is given by y - y1 y2 - y1 = x - x1 x2 - x1

This formula is optional as you can use the point–gradient formula for any question.

where _ x 1, y 1 i and _ x 2, y 2 i are points on the line

Proof

The gradient is the same anywhere along a straight line.

Let P = ^ x, y h D APQ <; D ABR PQ BR So = AR AQ y - y1 y2 - y1 i.e. x - x = x - x 1 2 1 The two-point formula is not essential. The right-hand side of it is the gradient of the line. Replacing this by m gives the point–gradient formula.

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EXAMPLES 1. Find the equation of the straight line with gradient -4 and passing through the point ^ -2, 3 h .

Solution m = -4, x 1 = -2 and y 1 = 3 Equation: y - y 1 = m (x - x 1) y - 3 = - 4 [x - (-2)] = - 4 (x + 2) = - 4x - 8 ` y = - 4x - 5 or 4x + y + 5 = 0

(gradient form) (general form)

2. Find the equation of the straight line that passes through the points ^ 2, -3 h and ^ -4, -7 h .

Solution By two-point formula: y - y1 y2 - y1 = x - x1 x2 - x1 y - ] -7 g -3 - ] -7 g = x - ] -4 g 2 - ] -4 g y+7 -3 + 7 = x+4 2+4 y+7 2 = x+4 3 3 ^ y + 7 h = 2 ]x + 4 g 3y + 21 = 2x + 8 -2x + 3y + 13 = 0 or 2x - 3y - 13 = 0 By point-gradient method: y2 - y1 m= x -x 2 1 -3 - ] -7 g = 2 - ] -4 g -3 + 7 = 2+4 2 = 3 Use one of the points, say ^ -4, -7 h . 2 m = , x 1 = -4 and y 1 = -7 3 Equation:

y - y 1 = m ( x - x 1) y - (-7) =

2 6 x - ( - 4) @ 3


2 ( x + 4) 3 = 2 ]x + 4 g = 2x + 8 =0 =0

y+7= 3^ y + 7h 3y + 21 ` -2x + 3y + 13 or 2x - 3y - 13

3. Find the equation of the line with x-intercept 3 and y-intercept 2.

Solution x y Intercept form is a + = 1, where a and b are the x-intercept and b y-intercept respectively. x y ` + =1 3 2 2x + 3y = 6 ` 2x + 3y - 6 = 0 Again, the point-gradient formula can be used. The x-intercept and y-intercept are the points ^ 3, 0 h and ^ 0, 2 h .

7.5 Exercises 1.

Find the equation of the straight line (a) with gradient 4 and y-intercept -1 (b) with gradient -3 and passing through ^ 0, 4 h (c) passing through the origin with gradient 5 (d) with gradient 4 and x-intercept -5 (e) with x-intercept 1 and y-intercept 3 (f) with x-intercept 3, y-intercept -4 (g) with y-intercept -1 and making an angle of 45c with the x-axis in the positive direction (h) with y-intercept 5 and making an angle of 45c with the x-axis in the positive direction.

2.

Find the equation of the straight line that makes an angle of 135c with the x-axis and passes through the point ^ 2, 6 h .

3.

Find the equation of the straight line passing through (a) ^ 2, 5 h and ^ -1, 1 h (b) ^ 0, 1 h and ^ -4, -2 h (c) ^ - 2, 1 h and ^ 3, 5 h (d) ^ 3, 4 h and ^ -1, 7 h (e) ^ -4, -1 h and ^ - 2, 0 h .

4.

What is the equation of the line with x-intercept 2 and passing through ^ 3, -4 h ?

5.

Find the equation of the line (a) parallel to the x-axis and passing through ^ 2, 3 h (b) parallel to the y-axis and passing through ^ -1, 2 h .

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6.

A straight line passing through the origin has a gradient of - 2. Find its equation.

7.

A straight line has x-intercept 4 and passes through ^ 0, -3 h . Find its equation.

8.

Find the equation of the straight line with gradient -2 that passes through the midpoint of ^ 5, -2 h and ^ -3, 4 h .

9.

What is the equation of the straight line through the point ^ -4, 5 h and the midpoint of ^ 1, 2 h and ^ -9, 4 h ?

10. What is the equation of the straight line through the midpoint of ^ 0, 1 h and ^ -6, 5 h and the midpoint of ^ 2, 3 h and ^ 8, -3 h ?

Parallel and Perpendicular Lines Parallel lines

Class Investigation Sketch the following straight lines on the same number plane. 1. y = 2x 2. y = 2x + 1 3. y = 2x - 3 4. y = 2x + 5 What do you notice about these lines?

If two lines are parallel, then they have the same gradient. That is, m1 = m2

Two lines that are parallel have equations ax + by + c 1 = 0 and ax + by + c 2 = 0


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Proof a b a ax + by + c 2 = 0 has gradient m 2 = b Since m 1 = m 2, the two lines are parallel. ax + by + c 1 = 0 has gradient m 1 = -

EXAMPLES 1. Prove that the straight lines 5x - 2y - 1 = 0 and 5x - 2y + 7 = 0 are parallel.

Solution 5x - 2y - 1 = 0 5x - 1 = 2y 5 1 x- =y 2 2 5 ` m1 = 2 5x - 2y + 7 = 0 5x + 7 = 2y 5 7 x+ =y 2 2 5 ` m2 = 2 5 m1 = m2 = 2 ` the lines are parallel. 2. Find the equation of a straight line parallel to the line 2x - y - 3 = 0 and passing through ^ 1, -5 h .

Solution 2x - y - 3 = 0 2x - 3 = y ` m1 = 2 For parallel lines m 1 = m 2 ` m2 = 2 Equation:

y - y 1 = m (x - x 1) y - (-5) = 2 (x - 1) y + 5 = 2x - 2 0 = 2x - y - 7

Notice that the equations are both in the form 5x - 2y + k = 0.

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DID YOU KNOW? Parallel lines are usually thought of as lines that never meet. However, there is a whole branch of geometry based on the theory that parallel lines meet at infinity. This is called affine geometry. In this geometry there are no perpendicular lines.

Perpendicular lines

Class Investigation Sketch the following pairs of straight lines on the same number plane. 1. (a) 3x - 4y + 12 = 0 2. (a) 2x + y + 4 = 0

(b) 4x + 3y - 8 = 0 (b) x - 2y + 2 = 0

What do you notice about these pairs of lines?

Gradients of perpendicular lines are negative reciprocals of each other.

If two lines with gradients m 1 and m 2 respectively are perpendicular, then m 1 m 2 = -1 1 i.e. m 2 = - m 1

Proof

Let line AB have gradient m 1 = tan a . Let line CD have gradient m 2 = tan b. EB EC +CBE = 180c - a EC tan ] 180c - a g = EB EB ` cot ] 180c - a g = EC tan b =

^ straight angle h


`

So or

tan b = cot ] 180c - a g = - cot a 1 =tan a 1 m2 = - m 1 m 1 m 2 = -1

Perpendicular lines have equations in the form ax + by + c 1 = 0 and bx - ay + c 2 = 0

Proof a b b bx - ay + c 2 = 0 has gradient m 2 = - - a b =a a b m1 m2 = - # a b = -1 ax + by + c 1 = 0 has gradient m 1 = -

Since m 1 m 2 = -1, the two lines are perpendicular.

EXAMPLES 1. Show that the lines 3x + y - 11 = 0 and x - 3y + 1 = 0 are perpendicular.

Solution 3x + y - 11 = 0 y = -3x + 11 m 1 = -3 ` x - 3y + 1 = 0 x + 1 = 3y 1 1 x+ =y 3 3 1 ` m2 = 3 1 m 1 m 2 = - 3# 3 = -1

Notice that the equations are in the form 3x + y + c 1 = 0 and x - 3y + c 2 = 0.

` the lines are perpendicular.

CONTINUED

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2. Find the equation of the straight line through ^ 2, 3 h perpendicular to the line that passes through ^ -1, 7 h and ^ 3, 3 h .

Solution Line through ^ -1, 7 h and ^ 3, 3 h: y2 - y1 m= x -x 2 1 7-3 m1 = -1 - 3 4 = -4 = -1 For perpendicular lines, m 1 m 2 = - 1 i.e. -1m 2 = - 1 m2 = 1 Equation through ^ 2, 3 h: y - y 1 = m (x - x 1) y - 3 = 1 (x - 2 ) =x-2 0=x-y+1

7.6 Exercises 1.

Find the gradient of the straight line (a) parallel to the line 3x + y - 4 = 0 (b) perpendicular to the line 3x + y - 4 = 0 (c) parallel to the line joining ^ 3, 5 h and ^ -1, 2 h (d) perpendicular to the line with x-intercept 3 and y-intercept 2 (e) perpendicular to the line making an angle of 135c with the x-axis in the positive direction (f) perpendicular to the line 6x - 5y - 4 = 0 (g) parallel to the line making an angle of 30c with the x-axis (h) parallel to the line x - 3y - 7 = 0

(i) perpendicular to the line making an angle of 120c with the x-axis in the positive direction (j) perpendicular to the line passing through ^ 4, -2 h and ^ 3, 3 h . 2.

Find the equation of each straight line (a) passing through ^ 2, 3 h and parallel to the line y = x + 6 (b) through ^ -1, 5 h and parallel to the line x - 3y - 7 = 0 (c) with x-intercept 5 and parallel to the line y = 4 - x (d) through ^ 3, -4 h and perpendicular to the line y = 2x (e) through ^ -2, 1 h and perpendicular to the line 2x + y + 3 = 0


(f) through ^ 7, -2 h and perpendicular to the line 3x - y - 5 = 0 (g) through ^ -3, -1 h and perpendicular to the line 4x - 3y + 2 = 0 . 3.

Show that the straight lines y = 3x - 2 and 6x - 2y - 9 = 0 are parallel.

4.

Show that lines x + 5y = 0 and y = 5x + 3 are perpendicular.

5.

Show that lines 6x - 5y + 1 = 0 and 6x - 5y - 3 = 0 are parallel.

6.

Show that lines 7x + 3y + 2 = 0 and 3x - 7y = 0 are perpendicular.

7.

If the lines 3x - 2y + 5 = 0 and y = kx - 1 are perpendicular, find the value of k.

8.

Show that the line joining ^ 3, -1 h and ^ 2, -5 h is parallel to the line 8x - 2y - 3 = 0.

9.

Show that the points A ^ -3, -2 h, B ^ -1, 4 h, C ^ 7, -1 h, and D ^ 5, -7 h are the vertices of a parallelogram.

10. The points A ^ -2, 0 h, B ^ 1, 4 h, C ^ 6, 4 h and D ^ 3, 0 h form a rhombus. Show that the diagonals are perpendicular.

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11. Find the equation of the straight line (a) passing through the origin and parallel to the line x+y+3=0 (b) through ^ 3, 7 h and parallel to the line 5x - y - 2 = 0 (c) through ^ 0, - 2 h and perpendicular to the line x - 2y = 9 (d) perpendicular to the line 3x + 2y - 1 = 0 and passing through the point ^ -2, 4 h . 12. Find the equation of the straight line passing through ^ 6, -3 h that is perpendicular to the line joining ^ 2, -1 h and ^ -5, -7 h . 13. Find the equation of the line through ^ 2, 1 h that is parallel to the line that makes an angle of 135c with the x-axis in the positive direction. 14. Find the equation of the perpendicular bisector of the line passing through ^ 6, -3 h and ^ -2, 1 h . 15. Find the equation of the straight line parallel to the line 2x - 3y - 1 = 0 and through the midpoint of ^ 1, 3 h and ^ -1, 9 h .

Intersection of Lines Two straight lines intersect at a single point ^ x, y h . The point satisfies the equations of both lines. We find this point by solving simultaneous equations.

You may need to revise simultaneous equations from Chapter 3.

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Concurrent lines meet at a single point. To show that lines are concurrent, solve two simultaneous equations to find the point of intersection. Then substitute this point of intersection into the third and subsequent lines to show that these lines also pass through the point.

EXAMPLES 1. Find the point of intersection between lines 2x - 3y - 3 = 0 and 5x - 2y - 13 = 0.

Solution Solve simultaneous equations: 2x - 3y - 3 = 0 5x - 2y - 13 = 0 4x - 6y - 6 = 0 ^ 1 h # 2: 15x - 6y - 39 = 0 ^ 2 h # 3: + 33 = 0 ^ 3 h - ^ 4 h: -11x 33 = 11x 3=x

^1h ^2h ^3h ^4h

Substitute x = 3 into ^ 1 h: You could use a computer spreadsheet to solve these simultaneous equations.

2 ^ 3 h - 3y - 3 = 0 - 3y + 3 = 0 3 = 3y 1=y So the point of intersection is ^ 3, 1 h . 2. Show that the lines 3x - y + 1 = 0, x + 2y + 12 = 0 and 4x - 3y - 7 = 0 are concurrent.

Solution Solve any two simultaneous equations: 3x - y + 1 = 0 x + 2y + 12 = 0 4x - 3y - 7 = 0 6x - 2y + 2 = 0 ^ 1 h # 2: 2 + 4 : 7 x + 14 = 0 ^ h ^ h

^1h ^2h ^3h ^4h


7x = -14 x = -2 Substitute x = -2 into ^ 1 h: 3 ^ -2 h - y + 1 = 0 -y - 5 = 0 -5 = y So the point of intersection of (1) and (2) is ^ -2, -5 h . Substitute ^ -2, -5 h into (3): 4x - 3y - 7 = 0 LHS = 4 ^ -2 h - 3 ^ - 5 h - 7 = -8 + 15 - 7 =0 = RHS So the point lies on line (3) ` all three lines are concurrent.

Equation of a line through the intersection of 2 other lines To find the equation of a line through the intersection of 2 other lines, find the point of intersection, then use it with the other information to find the equation. Another method uses a formula to find the equation.

If a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are 2 given lines then the equation of a line through their intersection is given by the formula (a 1 x + b 1 y + c 1) + k (a 2 x + b 2 y + c 2) = 0 where k is a constant

Proof Let l 1 have equation a 1 x + b 1 y + c 1 = 0. Let l 2 have equation a 2 x + b 2 y + c 2 = 0. Let the point of intersection of l 1 and l 2 be P ^ x 1, y 1 h . Then P satisfies l 1 i.e. a 1 x 1 + b 1 y 1 + c 1 = 0 P also satisfies l2 i.e. a 2 x 1 + b 2 y 1 + c 2 = 0 Substitute P into (a 1 x + b 1 y + c 1) + k (a 2 x + b 2 y + c 2) = 0 (a 1 x 1 + b 1 y 1 + c 1) + k (a 2 x 1 + b 2 y 1 + c 2) = 0 0 + k ^0h = 0 0=0 ` if point P satisfies both equations l 1 and l 2 then it satisfies l 1 + kl 2 = 0.

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EXAMPLE Find the equation of the line through ^ -1, 2 h that passes through the intersection of lines 2x + y - 5 = 0 and x - 3y + 1 = 0.

Solution Using the formula: a 1 = 2, b 1 = 1, c 1 = -5

a 2 = 1, b 2 = -3, c 2 = 1

^ a1 x + b1 y + c1 h + k ^ a2 x + b2 y + c2 h = 0 ^ 2x + y - 5 h + k ^ x - 3y + 1 h = 0 Since this line passes through ^ -1, 2 h, substitute the point into the equation: ^ -2 + 2 - 5 h + k ^ -1 - 6 + 1 h = 0 -5 - 6k = 0 -5 = 6k 5 - =k 6 So the equation becomes: 5 ^ 2x + y - 5 h - ^ x - 3y + 1 h = 0 6 6 ^ 2x + y - 5 h - 5 ^ x - 3 y + 1 h = 0 12x + 6y - 30 - 5x + 15y - 5 = 0 7x + 21y - 35 = 0 x + 3y - 5 = 0 Another way to do this example is to find the point of intersection, then use both points to find the equation.

Substitute the value of k back into the equation.

7.7 Exercises 1.

Find the point of intersection of straight lines (a) 3x + 4y + 10 = 0 and 2x - 3y - 16 = 0 (b) 5x + 2y + 11 = 0 and 3x + y + 6 = 0 (c) 7x - 3y = 16 and 5x - 2y = 12 (d) 2x - 3y = 6 and 4x - 5y = 10 (e) x - 3y - 8 = 0 and 4x + 7y - 13 = 0 (f) y = 5x + 6 and y = - 4x - 3 (g) y = 2x + 1 and 5x - 3y + 6 = 0

(h) 3x + 7y = 12 and 4x - y - 1 6 = 0 (i) 3x - 5y = - 7 and 2x - 3y = 4 (j) 8x - 7y - 3 = 0 and 5x - 2y - 1 = 0 2.

Show that the lines x - 2y - 11 = 0 and 2x - y - 10 = 0 intersect at the point ^ 3, -4 h .

3.

A triangle is formed by 3 straight lines with equations 2 x - y + 1 = 0, 2 x + y - 9 = 0


and 2x - 5y - 3 = 0. Find the coordinates of its vertices. 4.

Show that the lines x - 5y - 17 = 0, 3x - 2y - 12 = 0 and 5x + y - 7 = 0 are concurrent.

5.

Show that the lines x + 4y + 5 = 0, 3x - 7y + 15 = 0, 2x - y + 10 = 0 and 6x + 5y + 30 = 0 are concurrent.

6.

Find the equation of the straight line through the origin that passes through the intersection of the lines 5x - 2y + 14 = 0 and 3x + 4y - 7 = 0 .

7.

Find the equation of the straight line through ^ 3, 2 h that passes through the intersection of the lines 5x + 2y + 1 = 0 and 3x - y + 16 = 0.

8.

Find the equation of the straight line through ^ -4, -1 h that passes through the intersection of the lines 2x + y - 1 = 0 and 3x + 5y + 16 = 0.

9.

Find the equation of the straight line through ^ -3, 4 h that passes through the intersection of the lines 2x + y - 3 = 0 and 3x - 2y - 8 = 0 .

10. Find the equation of the straight line through ^ 2, -2 h that passes through the intersection of the lines 2x + 3y - 6 = 0 and 3x + 5y - 10 = 0. 11. Find the equation of the straight line through ^ 3, 0 h that passes through the intersection of the lines x - y + 1 = 0 and 4x - y - 2 = 0 .

12. Find the equation of the straight line through ^ -1, -2 h that passes through the intersection of the lines 2x + y - 6 = 0 and 3 x + 7 y - 9 = 0. 13. Find the equation of the straight line through ^ 1, 2 h that passes through the intersection of the lines x + 2y + 10 = 0 and 2x - y + 5 = 0. 14. Find the equation of the straight line through ^ -2, 0 h that passes through the intersection of the lines 3x + 4y - 7 = 0 and 3 x - 2 y - 1 = 0. 15. Find the equation of the straight line through ^ 3, -2 h that passes through the intersection of the lines 5x + 2y - 13 = 0 and x - 3y + 11 = 0. 16. Find the equation of the straight line through ^ -3, -2 h that passes through the intersection of the lines x + y + 1 = 0 and 3x + 2y = 0 . 17. Find the equation of the straight line through ^ 3, 1 h that passes through the intersection of the lines 3x - y + 4 = 0 and 2x - y + 12 = 0. 18. Find the equation of the straight line with gradient 3 that passes through the intersection of the lines 2x + y - 1 = 0 and 3x + 5y + 16 = 0. 19. Find the equation of the straight line with gradient 2 that passes through the intersection of the lines 5x - 2y - 3 = 0 and 7x - 3y - 4 = 0 .

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20. Find the equation of the straight line parallel to the line 3x - y - 7 = 0 that passes through the intersection of the lines 3x - 2y - 10 = 0 and 4x + y - 17 = 0.

21. Find the equation of the straight line perpendicular to the line x + 5y - 1 = 0 that passes through the intersection of lines 3x - 5y - 3 = 0 and 2x + 3y + 17 = 0.

Perpendicular Distance The distance formula d = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 is used to find the distance between two points. Perpendicular distance is used to find the distance between a point and a line. If we look at the distance between a point and a line, there could be many distances.

So we choose the shortest distance, which is the perpendicular distance.

The perpendicular distance from _ x 1, y 1 i to the line ax + by + c = 0 is A distance is always positive, so take the absolute value.

given by d =

Proof

| ax 1 + by 1 + c | a2 + b2


Let d be the perpendicular distance of _ x 1, y 1 i from the line ax + by + c = 0. - ax 1 - c c c o C = c 0, - m R = e x 1, A = b- a , 0 l b b c2 c2 + a2 b2

In D ACO, AC =

c2 b2 + c2 a2 a2 b2

= = PR = y 1 - e =

c a2 + b2 ab

- ax 1 - c

b ax 1 + by 1 + c

o

b Why?

D ACO is similar to D PRQ `

`

To find A and C, substitute y = 0 and x = 0 into ax + by + c = 0.

PQ PR = AO AC AO . PR PQ = AC ax 1 + by 1 + c c a2 + b2 c d=a# ' b ab c _ ax 1 + by 1 + c i ab = # ab c a2 + b2 ax 1 + by 1 + c = a2 + b2

All points on one side of the line ax + by + c = 0 make the numerator of this formula positive. Points on the other side make the numerator negative. Usually we take the absolute value of d. However, if we want to know if points are on the same side of a line or not, we look at the sign of d.

EXAMPLES 1. Find the perpendicular distance of ^ 4, - 3 h from the line 3x - 4y - 1 = 0.

Solution x 1 = 4, y 1 = - 3, a = 3, b = - 4, c = - 1 | ax 1 + by 1 + c | d= a2 + b2 | 3 ] 4 g + ] - 4 g ] -3 g + ] -1 g | = 3 2 + ] -4 g2 CONTINUED

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=

| 12 + 12 - 1 |

25 23 = 5 = 4 .6 So the perpendicular distance is 4.6 units. 2. Prove that the line 6x + 8y + 20 = 0 is a tangent to the circle x 2 + y 2 = 4.

Solution There are three possibilities for the intersection of a circle and a straight line.

The centre of the circle x 2 + y 2 = 4 is ^ 0, 0 h and its radius is 2 units. A tangent is perpendicular to the centre of the circle. So we prove that the perpendicular distance from the line to the point ^ 0, 0 h is 2 units (the radius). | ax 1 + by 1 + c | d= a2 + b2 | 6 (0) + 8 (0) + 20 | = 62 + 82 | 20 | = 100 20 = 10 =2 ` the line is a tangent to the circle. 3. Show that the points ^ -1, 3 h and ^ 2, 7 h lie on the same side of the line 2 x - 3 y + 4 = 0.


Solution To show that points lie on the same side of a line, their perpendicular distance must have the same sign. We use the formula without the absolute value sign. d=

ax 1 + by 1 + c a2 + b2

^ - 1, 3 h : 2 ]-1 g - 3 ]3 g + 4 d= 22 + ] - 3 g 2 -2 - 9 + 4 = 4+9 -7 = 13 ^ 2, 7 h : 2 ]2 g - 3 ]7 g + 4 d= 2 2 + ] -3 g 2 4 - 21 + 4 = 4+9 - 13 = 13

Since the perpendicular distance for both points has the same sign, the points lie on the same side of the line.

7.8 Exercises 1.

Find the perpendicular distance between (a) ^ 1, 2 h and 3x + 4y + 2 = 0 (b) ^ - 3, 2 h and 5x + 12y + 7 = 0 (c) ^ 0, 4 h and 8x - 6y - 1 = 0 (d) ^ - 3, - 2 h and 4x - 3y - 6 = 0 (e) the origin and 12x - 5y + 8 = 0.

2.

Find, correct to 3 significant figures, the perpendicular distance between (a) ^ 1, 3 h and x + 3y + 1 = 0 (b) ^ -1, 1 h and 2x + 5y + 4 = 0 (c) ^ 3, 0 h and 5x - 6y - 12 = 0 (d) ^ 5, - 3 h and 4x - y - 2 = 0 (e) ^ - 6, - 3 h and 2x - 3y + 9 = 0.

3.

Find as a surd with rational denominator the perpendicular distance between (a) the origin and the line 3x - 2y + 7 = 0 (b) ^ -1, 4 h and 2x + y + 3 = 0 (c) ^ 3, -1 h and 3x + 14y + 1 = 0 (d) ^ 2, - 6 h and 5x - y - 6 = 0 (e) ^ - 4, - 1 h and 3 x - 2 y - 4 = 0.

4.

Show that the origin is equidistant from the lines 7x + 24y + 25 = 0, 4x + 3y - 5 = 0 and 12x + 5y - 13 = 0.

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Equidistant means that two or more objects are the same distance away from another object.

5.

Show that points A ^ 3, - 5 h and B ^ -1, 4 h lie on opposite sides of 2x - y + 3 = 0.

6.

Show that the points ^ 2, - 3 h and ^ 9, 2 h lie on the same side of the line x - 3y + 2 = 0.

14. Find the perpendicular distance between ^ 0, 5 h and the line through ^ - 3, 8 h parallel to 4x - 3y - 1 = 0. 15. The perpendicular distance between the point ^ x, -1 h and the line 3x - 4y + 7 = 0 is 8 units. Find two possible values of x.

7.

Show that ^ - 3, 2 h and ^ 4, 1 h lie on opposite sides of the line 4 x - 3 y - 2 = 0.

8.

Show that ^ 0, - 2 h is equidistant from the lines 3x + 4y - 2 = 0 and 12x - 5y + 16 = 0.

16. The perpendicular distance between the point ^ 3, b h and the line 5x - 12y - 2 = 0 is 2 units. Find the values of b.

9.

Show that the points ^ 8, - 3 h and ^ 1, 1 h lie on the same side of the line 6x - y + 4 = 0.

17. Find m if the perpendicular distance between ^ m, 7 h and the line 9x + 12y + 6 = 0 is 5 units.

10. Show that ^ - 3, 2 h and ^ 4, 1 h lie on opposite sides of the line 2x + y - 2 = 0. 11. Show that the point ^ 3, - 2 h is the same distance from the line 6x - 8y + 6 = 0 as the point ^ - 4, -1 h is from the line 5x + 12y - 20 = 0. 12. Find the exact perpendicular distance with rational denominator from the point ^ 4, 5 h to the line with x-intercept 2 and y-intercept -1. 13. Find the perpendicular distance from ^ - 2, 2 h to the line passing through ^ 3, 7 h and ^ -1, 4 h .

18. Prove that the line 3x - 4y + 25 = 0 is a tangent to the circle with centre the origin and radius 5 units. 19. Show that the line 3x - 4y + 12 = 0 does not cut the circle x 2 + y 2 = 1. 20. The sides of a triangle are formed by the lines with equations 2x - y - 7 = 0, 3x + 5y - 4 = 0 and x + 3y - 4 = 0. (a) Find the vertices of the triangle. (b) Find the exact length of all the altitudes of the triangle.


Test Yourself 7 1.

Find the distance between points ^ - 1, 2 h and ^ 3, 7 h .

2.

What is the midpoint of the origin and the point ^ 5, - 4 h ?

3.

Find the gradient of the straight line (a) passing through ^ 3, -1 h and ^ - 2, 5 h (b) with equation 2x - y + 1 = 0 (c) making an angle of 30c with the x-axis in the positive direction (d) perpendicular to the line 5 x + 3 y - 8 = 0.

4.

Find the equation of the linear function (a) passing through ^ 2, 3 h and with gradient 7 (b) parallel to the line 5x + y - 3 = 0 and passing through ^ 1, 1 h (c) through the origin, and perpendicular to the line 2x - 3y + 6 = 0 (d) through ^ 3, 1 h and ^ - 2, 4 h (e) with x-intercept 3 and y-intercept –1.

5.

Find the perpendicular distance between ^ 2, 5 h and the line 2x - y + 7 = 0 in surd form with rational denominator.

6.

Prove that the line between ^ -1, 4 h and ^ 3, 3 h is perpendicular to the line 4x - y - 6 = 0.

7.

Find the x- and y-intercepts of 2x - 5y - 10 = 0.

8.

(a) Find the equation of the straight line l that is perpendicular to the line 1 y = x - 3 and passes through ^ 1, -1 h . 2 (b) Find the x-intercept of l. (c) Find the exact distance from ^ 1, -1 h to the x-intercept of l.

9.

Prove that lines y = 5x - 7 and 10x - 2y + 1 = 0 are parallel.

10. Find the equation of the straight line passing through the origin and parallel to the line with equation 3x - 4y + 5 = 0. 11. Find the point of intersection between lines y = 2x + 3 and x - 5y + 6 = 0. 12. The midpoint of ^ a, 3 h and ^ - 4, b h is ^ 1, 2 h . Find the values of a and b. 13. Show that the lines x - y - 4 = 0, 2x + y + 1 = 0, 5x - 3y - 14 = 0 and 3x - 2y - 9 = 0 are concurrent. 14. A straight line makes an angle of 153c 29l with the x-axis in the positive direction. What is its gradient, to 3 significant figures? 15. The perpendicular distance from ^ 3, - 2 h to the line 5x - 12y + c = 0 is 2. Find 2 possible values of c. 16. Find the equation of the straight line through ^ 1, 3 h that passes through the intersection of the lines 2x - y + 5 = 0 and x + 2y - 5 = 0. 17. The gradient of the line through ^ 3, - 4 h and ^ x, 2 h is −5. Evaluate x. 18. Show that the points ^ - 2, 1 h and ^ 6, 3 h are on opposite sides of the line 2 x - 3 y - 1 = 0. 19. Find the equation of the line with x-intercept 4 that makes an angle of 45c with the x-axis. 20. Find the equation of the line with y-intercept - 2 and perpendicular to the line passing through ^ 3, -2 h and ^ 0, 5 h .

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Challenge Exercise 7 1.

If points ^ - 3k, 1 h, ^ k - 1, k - 3 h and ^ k - 4, k - 5 h are collinear, find the value of k.

2.

Find the equation, in exact form, of the line passing through _ 3 , -2 i that makes an angle of 30c with the positive x-axis.

3.

Find the equation of the circle whose centre is at the origin and with tangent x - 3y + 9 = 0.

4.

ABCD is a rhombus where A = ^ - 3, 0 h, B = ^ 0, 4 h, C = ^ 5, 4 h and D = ^ 2, 0 h . Prove that the diagonals are perpendicular bisectors of one another.

joining ^ -1, 3 h and ^ 2, - 4 h makes with the x-axis in the positive direction. 10. Find the equation of the line that passes through the point of intersection of lines 2x + 5y + 19 = 0 and 4x - 3y - 1 = 0 that is perpendicular to the line 3x - 2y + 1 = 0. 11. Prove A ^ 2, 5 h, B ^ - 4, 5 h and C ^ -1, 2 h are the vertices of a right-angled isosceles triangle. 12. Find the coordinates of the centre of a circle that passes through points ^ 7, 2 h, ^ 2, 3 h and ^ -4, -1 h .

5.

Prove that the points _ -1, 2 2 i, _ 3 , - 6 i and _ - 5 , 2 i all lie on a circle with centre the origin. What are the radius and equation of the circle?

6.

Find the exact distance between the parallel lines 3x + 2y - 5 = 0 and 3x + 2y = 1.

14. Find the equation of the straight line through ^ 3, -4 h that is perpendicular to the line with x-intercept and y-intercept −2 and 5 respectively.

7.

A straight line has x-intercept A ^ a, 0 h and y-intercept B ^ 0, b h, where a and b are positive integers. The gradient of line AB is -1. Find +OBA where O is the origin and hence prove that a = b.

15. Find the exact equation of the straight line through the midpoint of ^ 0, - 5 h, and ^ 4, -1 h that is perpendicular to the line that makes an angle of 30c with the x-axis.

8.

Find the exact perpendicular distance between the line 2x + 3y + 1 = 0 and the point of intersection of lines 3x - 7y = 15 and 4x - y = - 5.

16. Point P ^ x, y h moves so that it is equidistant from points A ^ 1, 4 h and B ^ - 2, 7 h . By finding the distances AP and BP, find the equation of the locus of P.

9.

Find the magnitude of the angle, in degrees and minutes, that the line

13. If ax - y - 2 = 0 and bx - 5y + 11 = 0 intersect at the point ^ 3, 4 h, find the values of a and b.

Maths in Focus - Margaret Grove - ch7

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