] [ A = 3 square units
Rectangular Rectangular Coordinate or Cartesian System
A=
y P(x,y)
Division of Line Segment Let P (x, y) be a point on the line joining P 1(x1,y1) and P2(x2,y2) and located in such a way that segment P 1P is a given fraction k of P 1P2 that is P 1P =
x y x
o
y
Point is denoted by the ordered pair x and y as (x,y) X-coordinate called abscissa The distance from the origin to the projection of the point unto the xaxis Y-coordinate called ordinate The distance from the origin to the projection of the point unto the yaxis Origin The intersection of x and y-axis
P x, x,
–
–
Distance between Two Points P 1(x1,y1) and P2(x2,y2)
P 1 ( x 1 1(x 1 ,y 1 1 )
o
x
If k = then the formula above becomes midpoint formula
X0 = (x1 + X2)
y0 = (y1 + y0)
Sample Problem:
Find the coordinate of the point which is of the way from a (1, -1) and B(7, 5)
P 2 ( x 2 2(x 2 ,y 2 2 )
y
Solution: x = x1 + k (x2 x1) x = 1 + 1/3(7 1) x=3 Thus, point required is P(3, 1)
– –
d
( x 1 1(x 1 ,y 1 1 ) o P 1
x
–
y = y1 + k (y2 y1) y = -1 + 1/3(5+1) y=1 Answer
x x y y x
d=
Area of Polygon (Non-overlapping) of n-sides Given Vertices Given vertices (x1,y1), (x2,y2), . . . ,(x n,yn) oriented counter clockwise
-
-
[ + + + A = [x xy xy x y x y yx yx yx y x] Sample Problem: Find the area of polygon joining the points (3, 0), (2, 3), (-1, 2), (-2, -1) and (0, 2) Solution: Note: The points points are given already in counter counter clockwise clockwise orientation Thus:
(
Projections of a Line Segment on the Coordinate Axes
-
x x x x+ A = *y y y y
A=
P 2 ( x 2 2(x 2 ,y 2 2 )
–
)
y B 2 ( 0,y 2 2(0,y 2 )
B 1 ( 0,y 1 1(0,y 1 ) o
P 2 2 x 2 2 ,
P 1 ( x 1 1(x 1 ,y 1 1 ) A1 x 1 1 ,0
2
x A2 (x (x 2 2 ,0)
The direction segment, A 1A2 is the projection of the segment P 1P2 on the x-axis The direction segment, B 1B2 is the projection of the segment P 1P2 on the y-axis Obviously, A1A2 = x2 x1 B1B2 = y2 y1
– –
Sample Problem: Given two points P1(2, -1) and P2((6, 5) find the projections of P 1P2 on the x and y axis Solution: a) Projection on x-axis = x2 x1 = 6 2 = 4 b) Projection on y-axis = y2 y1 = 5 (-1) = 6
If the line L is inclined to the right,
– – – –
m>0
y
Slope, Parallel and Perpendicular Lines
x
L
y
If the line is parallel to the x-axis,
P 2( x 2 ,y 2 )
d
y
P 1( x 1 ,y 1 )
m=0 x
L x If the line is parallel to the y-axis,
Definition: The Inclination ( of a line L is defined as the least positive angle made by the line with the positive x-axis The slope (m) of a line is defined as the tangent of its inclination
and m is undefined
y
m = tan
m=
=
– –
Condition for Parallelism Two lines L1 and L2 are perpendicular if their slopes are equal (m 1 = m2) Condition for Perpendicularity Two lines L1 and L2 are perpendicular with each other if the slope to the reciprocal of the slope of the other m= or m1m2 =
Notes:
Angle Between two Concurrent Lines L Let be the inclinations of L1 and L2 respectively and let be the angle between the two lines
d
y
L 1
L 2
If the line L is inclined to the left,
m<0
x
y
x L
x
Sample Problem: Given three points A(-1, 1), B(5, -1), C(4, 3), find the angle made by segments AB and AC.
Solution: First, find the slope of AB and AC mAB = mAC =
tan = ()
1.
Two Point Form
y P 2( x 2 ,y 2 )
⁄ Ѳ = t ⁄ Ѳ ⇒
Answer
Supplementary Problems 1. A point P(x, 3) is equidistant from points A(1, 5) and B(-1, 2). Find x. Ans: 2. Find the locus of the point P(x, y) such that the distance from P to (3, 0) is twice its distance from (1, 0). Ans: 3. Find the length of the segment joining the two midpoints of the sides of the triangle if the length of the third side opposite to it is 30cm. Ans: 15cm 4. A line from P(1, 4) to Q(4, -1) is extended to a point R so that PR = 4PQ. Find the coordinate of R. Ans: R(13, -16) 5. Two vertices of a triangle are (0, - 8) and (6, 0). If the medians intersect at (9, 3), find the third vertex of the triangle. Ans: (-3, -1) 6. The area of a triangle with vertices (6, 2), (x, 4), and (0, -4) is 26. Find x. Ans: 7. Find the length of the median from A to the triangle ABC given vertices A(1, 6), B(-1, 3) and C(3, -3). Ans: 6 8. If the midpoint of a segment is (5, 2) and one endpoint is (7, -3), what are the coordinates of the other end? Ans: (3, 7) 9. Given vertices of a triangle ABC. A(1, 5), B(- 1, 1), and C(6, 3). Find the intersection of the median. 10. Find the inclination of the line 2x + 5y = 10
2.
–
A and B, not zero at the same time
–
y P(x,y)
Ѳ
x
t Ѳ 3.
Slope-Intercept Form y = mx +b where m = slope b = y intercept
y (0, b) P(x,y) x
4.
Two-Intercept Form
y
where a = x intercept b = y intercept
P 2( 0, b) P(x,y) P 1( a, 0)
Given
– – by m,
y y1 = m (x x1)
⁄ d ⁄
is a locus of points which has constant slope Theorems: Every straight line can be represented by a first degree equation. The locus of an equation of the first degree is always a straight line
– – (x – x1)
x
Point-Slope Form In (1) replacing
x y x
the curved traced by an arbitrary point as it moves in a plane is called locus of a point the locus of an equation is a curve containing only those points whose coordinates satisfy the equation
y y1 =
P 1( x 1 ,y 1 )
⁄
A, B, C are constants
–
P(x,y)
tan =
By similarity of triangles
ρ orl itercept
= segment from the origin perpendicular to the required lines = normal angle = inclination of the normal Intercept
Ѳ
xcosѲ ysiѲ
x
y
NρcosѲ, ρsiѲ ρ Ѳ ρsiѲ ρcosѲ ⁄ x t
where a = x intercept = 3 b = y intercept = 5
xy Simplifying, 5x + 3y – 15 = 0
Given the line Ax + By + C = 0 The normal form is
x y √ √ √ The sign of the radical must be chosen such that the last term will become egtive sice ρ A.
Equation of x-axis Equation of the horizontal line Equation of the y-axis Equation of a vertical line
B.
1.
y=0 y=b x=0 x=a
–
where x1 = -1 y1 = -3 y+3= (x + 1)
y y1 =
2.
–
– – Fid the equtio of lie hvig orl gle d orl legth of 1 Solution: Normal form: where
7.
point (5, 1) Solution: Using the point-slope form: where x1 = 5 y1 = 1
–
6.
x2= 2 y2 = 4
Find the equation of the line having slope of and passing through the
xcosѲ ysiѲ ρ Ѳ ρ xcos ysi = 1 Simplifying, ⁄ ⁄ x + √ y – 2 = 0 Reduce the equation 3x – 4y + 10 = 0 to normal form Solution: A = 3,
–
–
– –
Simplifying, x 3y 2 = 0
4.
Find the equation of the line with slope of -2 and y intercept of 3 Solution: Using slope-intercept form: y = mx +b, where m = -2, b = 3
Find the equation of a line with x and y intercept equal to 3 and respectively Solution: Using the intercept form:
0
– ⁄ ⁄
Choose the sign of radical to make C negative Thus,
y 1 = (x 5)
3.
B = -4, C = 10
Normal form is y y1 = m (x x1) m=
⁄ ⁄ y = 3x ⁄
–
– – (x – x1)
– –
–
Here slope m = 3 Using the same slope m = 3 y y1 = m (x x1) where x1=2, y1 = -1, m = 3 y + 1 = 3 (x 2) Simplifying,
where a is constant
Simplifying, 7x 3y 2 = 0
Find the equation of a line passing through (2, -1) and parallel to the line 12x 4y + 5 = 0 Solution: Since the required line is parallel to the given line their slopes is equal For the given line : 12x 4y + 5 = 0 or y=
–
where b is constant
Find the equation of a line passing through (-1, 3) and (2, 4) Solution:
Using two points formula
5.
Find the equations of the line/s satisfying the given conditions 1. Passing through (1, -2) and perpendicular to the line through (2, -1) and (-3, 2) 2. 3.
With x intercept of 5 and passing through (3, 4). Passing through (-3, 4) and with equal intercepts.
4.
ig gle of with the x xis d pssig through , With slope ⁄ crosses the first quadrant and forms with the axes a
5.
triangle with perimeter of 15.
Passing through (7, -4) and at a distance of 1 unit from point (2, 1) 6. 7.
Using the formula:
d=
Passing through the midpoint of the segment joining the points (1, 3) and (5, 1) and parallel to the line 2x y +5 = 0 Find the value of the parameter k so that the line 3x 5ky + 5 = 0 a) will pass through (0, 1) b) will be parallel to x +2y = 5 c) will be perpendicular to 4x +3 y = 2 d) has the y intercept equal to 3
–
⁄
–
⁄
⁄
–
8. Find the equations of the lines parallel to the line x + 2y 5 = 0 and passing at a distance 2 from the origin.
√
√
9. Find the equation of the perpendicular bisector of the line segment joining (2, 5) and (4, 3) 10. Given the vertices of a triangle ABC, A(2, 0), B(3, -2), C(7, 5) a) Find the equation of the median from A b) Find the equation of the altitude from B c) Find the intersection of the medians from B to C. 11. Find the normal intercept and the normal angle of line 5x + 12y 39 = 0
–
= √ √
(the point (3, -1) and the origin are on the opposite side of the line)
2. Find the distance between parallel lines 8x + 15y + 18 = 0and 8x + 15y + 1 = 0 Solution: A = 8, B = 15, C 2 = 18, C1 = 1 d = = =
√
√
The distance from a point P(x 0, y0) to a line Ax + By + C = 0 is give by the formula: d= √ where the sign of the radical is chosen to be the opposite that of C
1. If d > 0, the origin and P lie o n opposite sides of the given line. 2. If d < 0, the origin and P lie o n the same side of the line.
Let the parallel lines be given by the equations L1 : Ax + By + C 1 = 0 L 1 y L2 : Ax + By + C 2 = 0 L 2 The distance d between the two lines is given by the formula:
√
d=
y
y d d
d x 0
– –
1. Find the distance from point (3, -1) to the line 3x 4y 3 = 0 Solution: Here, A = 3, B = -4, C = -2 P0(x0, y0 -1)
⇔,
x 0
d>0
0
P(x 0 , y 0 ) d < 0
Regardless of the location of the point P 0(x0, y0), the distance being always positive the formula can be expressed using the absolute value as: d= √
|
|
x
Let
Ax + By + C = 0 and Dx + Ey + F = 0 be the two intersecting lines, where A, B, C,
D, E d F re costts d ≠ E F ≠
The equation of the family of lines p assing through the intersection of the two given lines is given by, (Ax + By + C) + k(Dx + Ey + F) = 0 where k is an arbitrary constant
1. Find the equation of the line passing through the intersection of the lines x y + 5 = 0 and x + 5y 4 = 0 and passing through the point (1, 0) Solution: The equation of the family is (x y + 5) + k(x + 5y 4) = 0 Since the required line is a member of the family, substitute the point (1, 0) in the locus of the family and then solve for k [(1) 0 + 5] + k[1 + 5(0) 4] = 0 6 3k = 0, k = 2 Substituting k in the equation of the family of lines above, (x y + 5) + 2(x + 5y 4) = 0 Simplifying, 3x + 9y 3 = 0
–
–
–
–
–
–
–
–
–
f(x, -y) = f(x, y) = 0 If an equation of a curve does not change upon replacement of x by x then the locus is symmetric with respect to the yaxis f(-x, y) = f(x, y) = 0 If an =equation of a curve does not change upon replacement of x by x and y by y, then the locus is symmetric with respect to the origin f(-x, -y) = f(x, y) = 0
–
–
a straight line which the curve f(x, y) = 0 approaches indefinitely near as its tracing point approaches to infinity. To find the vertical asymptote, solve the equation for y in terms of x and set the linear factors of the denominator equal to zero. To find the horizontal asymptote, solve the equation for x in terms of y and set the linear factors of the denominator equal to zero.
–
–
Circle is the locus of a point which moves so that it is always equidistant from a point. Fixed point is called the center and fixed distance is called radius.
x-intercept directed distance from the origin to the point where the curve crosses the x-axis To find the x-intercept of a curve, set y = 0, then solve for x y-intercept the directed distance from the origin to the point where curve crosses y-axis To find the y-intercept of a curve, set x = 0, then solve for y
A. In formal form Consider a circle of radius r with center at C (u, k) Let P(x, y) be a point in the circle By Pythagorean Theorem
–
y r C(h , k)
If the equation of a curve does not change upon replacement of y by y then the locus is symmetric with respect to the x-axis.
–
–
0
P(x, y) y-k
x-h
x
Center at the origin C(0, 0)
B. General Form Expanding the form (x h)2 + (y k)2 = r2 becomes x2 + y2 2xh +h2 +k 2 r2 = 0 This is of the form:
–
–
–
–
B. If two circles are tangent, their radical axis is the common tangent to the circles at their point of tangency. y
Radical Axis
where D, E, F are constants not all zero at a time. By the equation coefficients: 2h = D; h=
D⇒ bsciss of ceter 2k = E k = E ⇒ ordite of ceter h +k – r = F; r = 2
2
2
C. The radical axis of two circles is perpendicular to their ,ine of centers. Radical Axis
Consider the two non-concentric circles x2 + y2 + D1x + E1y + F1 = 0 x2 + y2 + D2x + E2y + F2 = 0 The equation represents a circle for any value of k except for k = -1 if k = -1, the equation of the family of circles above becomes (D1 D2)x + (E 1 E2)y + (F 1 F2) = 0 This represent a straight line called RADICAL AAXIS of two circles
–
–
–
y A
–
AB line of centers
B
x
0
D. All tangents drawn to the circles from a point on their radical axis have equal lengths y
A. If two circles intersect at two distinct points, their radical axis is the common chord of the circles.
Radical Axis P
T1 and T2 are points of tangency
T1 T2
y
Common Chord
0
x
0
x
Condition for Orthogonality The two non-concentric circles
x2+y2+D1x+E1y+F1=0 x2+y2+D2x+E2y+F2=0 meet at the right angles (orthogonal) if
=
x
0
–
1. Find the center and radius of circle x 2 + y2 + 8x 10y + 32 =0 Solution Reduce the given equation to standard form by completing the square in x and y: (x2 + 8x + 16) + ( y 2 10y + 25) = -32 + 16 + 25
–
–
⇔ –
–
(x + 4)2 + (y 5)2 (x h)2 + (y k)2 = r2 Here, C(h, k) = (-4, 5); r=3 2. Find the equation of the circle passing through the points of intersection of the circles x2 + y2 = 16 and x 2 + y2 10y = 0 and passing through (1, 0) Solution The equation of the family is x2 + y2 25 + k(x2 + y2 10y) = 0 To determine the particular member of the family, Solve for k by substituting the coordinates of the point (1,0) 1 + 0 25 + k(1 + 0 0) = 0 k = 24 Substituting k in the equation of the family of circles, x2 + y2 25 + 24(x2 + y2 10y) = 0 25x2 + 25y2 240y 25 = 0 Simplifying, 5x2 + 5y2 48y 5 = 0 3. Find the equation of the radical axis of two circles. x2 + y2 + 10x 8y + 25 = 0 and x 2 + y2 2x +4y + 1 = 0 Solution By getting the difference
–
–
–
–
–
–
–
–
–
–
–
–
–
– –
–
–
The locus of a point that moves in a plane such that its distance from a fixed point equals its distance from a fixed line
Fixed point is called focus Fixed line is called directrix Axis – the line passing through the focus and perpendicular to the directrix Vertex – The midpoint of the segment of the axis from the focus to the directrix Latus rectum – a segment passing throught the focus and perpendicular to the axis of parabola Focal distance – distance from vertex to focus = a
–
1. Find the center of the radius of the circle whose equation x 2 + y2 4x 6y- 12 = 0 (ECE Board Problem Oct 1981) 2. Find the area of the circle whose equation x 2 + y2 = 6x 8y (ECE Board Problem Mar 1981) ∏ Find the equation of thee circle whose center is at (3, -5) and whose radius is 4 units. For problems 3 9, determine the equation of the circle given the following conditions 3. Passes through the point (2, 3), (6, 1), and (4, -3)
– –
6. Find the equation of the line tangent to the circle x 2 + y2 -8x 8y + 7 = 0 at the point ( 1, 0)
A. Vertex at V9h, k), Vertical axis (x h)2 = 4a(y k)
Simplifying,
– –
4. Center on the y-axis, and passes through the o rigin and point (4, 2) 5. Given the endpoints of the diameter (5, 2), (-1, 2)
–
– If is positive ⇒cocve upwrd If a is negative (- ⇒cocve dowwrd
y 1. Equation of axis x = h 2. Focus: F (h, k+a) 3. End of Latus Rectum L(h 2a, k+a) R(h+2a, k+a) 4. Equation of Directrix 0 Y = k a B. Vertex at V(h, k), Horizontal Axis
–
–
axis L
F
R
V(h ,k) directrix
x
If is positive ⇒cocve to the right If a is negative (- ⇒cocve to the left
1. Equation of the axis: y = k 2. Focus: F(h+a, k) 3. Ends of Latus Rectum L(h+a, k+2a) R(h+a, k 2a) 4. Equation of Directrix
y directrix L V F (h ,k)
–
R x=h-a 0
x
7. Passes through the points of intersection of the circles x 2 + y2 = 5, x 2 + y2 x + y = 4, and t hrough the point (2, -3)
–
– –
8. Center on the line x 2y 9 = 0and passes through the points (7, -2) and (5, 0)
