ANALYTIC GEOMETRY PRELIMS REVIEWER (2
nd
Sem; 2011)
x = -6 +
I. Distance of points y=2+ 1.
2
d=
2
b.
2 (8 - 2) = 6 3
(x2 - x1) + (y2 - y1)
Q(0,6)
SAMPLES: a.
b.
d=
(5 - (-3)) + (-2-1)
d=
73 units
thrice its own length.
2
AP 4 = AB 1 x = -4 + 4(0 + 4) = 12
If the distance between (3,y) and (8,7) is 13, Find y. 2
13 =
(8-3) + (7-y)
y = 3 + 4(11 - 3) = 35
2 2
[ 13 =
The segment joining A(-4,3) and B(0,11) is extended through B a distance equal to
Find the distance of (5,-2) and (-3,1) 2
P(12,35)
2
25 + 49 - 14y + y ]
169 = 74 – 14y + y
2
c.
2
0 = y – 14y – 95
Find the point which divides the segment from ( -5,4) to (6,-2) in the ratio 2:3.
(y – 19)(y + 5) k=
y = 19 or y = -5
2 3 = 5 5
The points may be (3,19) or (3,-5). c.
x = -5 +
2 3 (6 + 5) = – 5 5
y = -4 +
2 16 (-2 + 4) = – 5 5
x = -5 +
3 8 (6 + 5) = 5 5
y = -4 +
3 24 (-2 + 4) = – 5 5
If the point (x,4) is equidistant from (3,4) and (5,-2), find x. [
2 (3 + 6) = 0 3
2
2
(3 - x) + (4 - 4) = 2
2
2
2
(5 - x) + (-2-4) ]
9 – 6x + x = 61 –10x + x
2
4x = 52
x = 13
3 16 P( – ) – , – – 5 5
P(13,4)
II. Division of a Line Segment
d.
8 24 P( , – ) 5 5
The segment joining (1,-3) and (4,-6) is extended a distance equal to 1/6 of its own length. Find the
2.
AP k= AB
3.
x = x1 + k(x 2 – x1)
4.
y = y1 + k(y2 – y1)
terminal point/s. The points may pass through either A or B kasi wala naming spinecify.
SAMPLES: a.
Find the trisection points of the line joining A(-6,2) and B(3, 8). Let’s represent the 2 other points with P & Q.
For point P: k=
1 3
x = -6 + y=2+
1 (3 + 6) = -3 3
1 (8 - 2) = 4 3
P(-3,4) For point Q: k=
2 3
k=
7 6
x=1+
7 9 (4 – 1) = 6 2
y = -3 +
x=4+
7 13 (-6 + 3) = – 6 2
P(
9 13 , – ) – 2 2
Q(
1 5 , – ) 2 2
7 1 (1 – 4) = 6 2
y = -6 +
7 5 (-3 + 6) = – 6 2
e.
The segment joining (2,0) and (-2,-3) is extended a distance equal to k=
f.
4 of its own length. 5
90° < θ < 180°
m = – (negative)
9 (again, there are 2 possible a nswers here) 5
x = -2 +
9 26 (2 + 2) = 5 5
y = -3 +
9 12 (0 + 3) = 5 5
x=2+
9 26 (-2 – 2) = – 5 5
y=0+
9 27 (-3 + 0) = – 5 5
P(
26 12 , ) 5 5
6.
m1 = m 2
26 27 Q( – , – ) 5 5 7.
The segment joining (2,-1) and (9,3) is divided into two segments, one of which is
–1 = m2 m1
3 as long as the 4
other . Find the point/s of division. k=
3 7 8. tanΦ =
3 x = 2 + (9-2) = 5 7 y = -4 +
x=9+
3 (3 + 4) = -1 7
P(5,-1)
3 (2 – 9) = –6 7
3 y = 3 + (-4 – 3) = 0 7
SAMPLES: Q( – 6, 0)
a.
The vertices of a triangle are at (2,5), (-5,3) and (-2,4). Find the interior angle at (2,5). *note: Always move
III. Slope of a Line 5.
m2 - m1 1 + (m2m1)
counterclockwise when assigning which is slope 1
y2 - y1 m= x 2 - x1
and slop 2.
m1 =
5-3 2 = 2 +5 7
tan90° = ∞
m2 =
-4 - 5 9 = -2 - 2 4
m=∞
9 2 – 4 7 55 = tanΦ = 46 9 2 1 + 4 7
* Tangent of the angle of inclination:
55 Φ = Arctan 46
tan0° = 0 m=0
Φ = 50.09°
0° < θ < 90°
m = + (positive)
b.
A line with an angle of inclination of 45° passes
a.
Find the area of the given coordinates (5, 7), (2,
5 9 through the points – – . If the ordinate of a 2 2
3), (-5, 2)
point on the line is 6, which is its abscissa?
*If we plot the points, we can now substitute
9 2 m= 5 x+ 2
the given coordinates starting from any point
m = tan45° = 1
(x3, y3) = (5, 7)
6+
[for instance we start from (-5, 2)] (x1, y1) = (-5, 2)
9 2 1= 5 x+ 2 6+
2 [x +
5 9 =6+ ] 2 2
(x2, y2) = (2, 3)
A=
1 [ (x1y2 + x2y3 + x3y1) – (y1x2 + y2x3 + y3x1) ] 2
A=
1 [ (-5*3 + 2*7 + 5*2) – (2*2 + 3*5 + 7*-5) ] 2
A = 12.5 sq units
2x + 5 = 12 9 2x = 16
V. Linear Equation
x=8 - expressed as Ax + By + C = 0 where A and B cannot be both c.
A line through (-6, -7) & (x,7) is perpendicular to a line through (1,-4) and (-5,2). Find x. 2+4 m2 = = -1 -5 - 1 m1 = –
1 m2
m1 = –
1 -1
equal to zero.
General Form Ax + By + C = 0
Standard Forms 1.
y – y1 = m(x – x1)
m1 = 1 1=
7+7 x+6
Two-point form
x=8
IV. Area of a triangle using the coordinates
Point-slope form
y – y1 =(
2.
y2 - y1 )(x – x1) x2 - x1
Slope-intercept form y = mx + b Where m is the slope and b is the y-intercept.
3.
Intercept form x y + =1 a b Where a is the x-intercept and b is the y-intercept.
1 A = [ (x1y2 + x2y3 + x3y1) – (y1x2 + y2x3 + y3x1) ] 2 4.
Normal form
Note: The arrangement of the given points is starting from any point going counter clockwise.
SAMPLES:
xcosα + ysinα = p Where p is the distance of a point from the origin.
tanα =
Shortcuts:
m2 - m1 1 + m2m1
tanα = 1
For any given line
α = arctan(1) = 45 degrees
Ax + By + C = 0 4. By = -Ax – C
A line has an equation of 3x – ky – 8 = 0. Find the value of k if this line makes an angle of 45 degrees with the line 2x + 5y – 17 = 0 (graph first)
A C y=- xB B
Let m1 = -2/5
m2 = 3/k
α = 45 degrees
A m=B
Solving for k in this formula m(perpendicular) =
B A tanα =
SAMPLES: 1.
m2 - m1 1 + m2m1
And obtain k = 7 or
Find the x-intercept of a line which is perpendicular to a line 3x + 4y + 8 = 0 and passes through a point
Let m1 = 3/k
(2, 1).
m2 = -2/5
α = 45 degrees
m(perpendicular) =
B 4 = A 3
Solving for k with the same f ormula, we obtain
k = -9/7. y – 1 =
4 (x – 2) 3
0 – 1 =
4 8 x3 3
x=
2.
(x, 0) coordinate of x-intercept
5 4
A line passes through A(-2, 5) and B(k, 1) and has x-
5.
Find the equation of the altitude from vertex B to side AC of the triangle whose vertices are at A(3, 7), B(-2, 1) and C(5, -8). (Graph first)
mAC =
7+8 = -15/2 3-5
m(perpendicular) = 2/15
at P(-2, 1)
intercept of 3. Find k. C(3, 0) coordinate of the given x-intercept
y – 1 =
2 (x + 2) 15
15y – 15 = 2x + 4 5 5-1 mAC = = -1 = = mAB -5 -2-k
2x – 15y + 19 = 0 Normal Form
3.
k=2
xcosα + ysinα = p
Find the smallest angle between the lines 2x + y – 8 =
Where α is the inclination of the normal form from
0 and x + 3y + 4 = 0 (graph first by x and y intercept
the positive x-axis and p is the distance of the line
form)
from the origin.
m1 = -2 m2 = -1/3
Conversion of the General Form to Normal Form Ax + By + C = 0
Ax + By = -C
The answer for this problem will give two equations in this case.
Divide both side by ±√ ( 2 + 2 ) m = -2 since it is parallel to the given equation at Use + when C is + and use – when C is -.
point (1, 1)
SAMPLE:
The equation will be 2x + y = 3
Transform x – y + 3 = 0 into normal form.
Divide both sides of the equation by
√ ( 2 + 2 )
or √ 5 then add an the right side of the equation x – y = -3
by ±3.
Divide both side by
−√ ( 2 + 2 ) = −√ (12 + [−1]2 )
Final Answer: 2x + y = 3 ± 3 √
or −√ 2 3.
− −3 = −√ 2 −√ 2 1 1 3 − + = √ 2 √ 2 √ 2 Where cosα is
Find the equation of the line perpendicular to 2x + 4y + 9 = 0 and passing at a distance of ±1 from the point (2, 0). The answer for this problem will also give two
1
equations in this case.
1
− √ 2 ,sinα is √ 2 and it is located in the m = 4/3 since it is parallel to the given equation
second quadrant in which the reference angle is 135 degrees. The value of p is
3
√ 2
at point (2, 0)
.
The equation will be 4x – 3y = 8
The equation in normal form is 0 0 xcos135 + ysin135 = √
Divide both sides of the equation by
√ ( 2 + 2 )
or 5 then add an the right side of the equation
APPLICATION SAMPLES:
by ±1.
1.
Two sides of the square lie on x + 3y – 9 = 0 and
Final Answer:
x + 3y + 7 = 0. Find the perimeter and area.
4x – 3y – 3 = 0 and 4x – 3y – 13 = 0
(Graph first)
V. Perpendicular Distance from a point to a line The side of the square is the sum of the values of p in the given equations. The value of p in the equation x + 3y – 9 = 0 is
9
√ 10
while the value of p
in the equation x + 3y + 7 = 0 is the square is
16
√ 10
7
√ 10
. The side of
Find the equation of the line parallel to 2x + y – 9 = 0 and passing at a distance of ±3 from the point (1, 1).
by this formula: d=
\ + + \
( 2 + ^2)
.
Final Answer: √ Perimeter: units Area: 25.6 sq units 2.
The distance of a point (m, n) to a line Ax + By + C = 0 is given
SAMPLE: Find the perpendicular distance from the point (5, 6) to the line -2x + 3y + 4 = 0 By substituting the values of m, n, A, B & C, we get
√
Try solving these problems: 1.
Find the equation of the line through (5, -2) with intercepts numerically equal but opposite signs.
Answer: x – y – 7 = 0 2.
Find the equation of the line through (6, -8) with the y-intercept twice the x-intercept.
Answer: 2x + y – 4 = 0 3.
Find k if the distance between (k, 0) and (0, 2k) is 10 units.
Answer: k = ±√ 4.
Find the angle of inclination of the line whose slope is 2. 0
Answer: 63.43 5.
Find the slope of the line with the angle of 0
inclination of 137 .
Answer: m = -0.933 6.
Find the distance from the point (-3, 7) to th e line y = (6x/5) + 2. − Answer: √
7.
What is the equation of the line perpendicular to the line joining (4, 2) and (3, -5) and passing through (4, 2)?
Answer: x + 7y -18 = 0 8.
If 4x – ky – 6 = 0 and 6x + 3y + 2 = 0 are perpendicular, what is the value of k?
Answer: k = 8 9.
A line passes through (-3, 9) and (4, 4). Another line passes through (9, -1) and (4, -8).Are the lines parallel or perpendicular?
Answer: Perpendicular
