Coordinate Geometry and Quadratic Functions - Desklib

This study material covers topics like gradient, equation of line, solving inequalities and equations, centre and radius of circle, quadratic equations with graph, etc. for the subject of Coordinate Geometry and Quadratic Functions.

Coordinate geometry and quadratic
functions
1

TABLE OF CONTENTS
QUESTION 1..................................................................................................................................3
i) Gradient....................................................................................................................................3
ii) Equation of line.......................................................................................................................3
QUESTION 2..................................................................................................................................3
i) Gradient....................................................................................................................................3
ii) Equation of line.......................................................................................................................3
iii) Line AB..................................................................................................................................4
iv) Point of intersection on x axis................................................................................................4
QUESTION 3..................................................................................................................................4
(a) Solving inequalities................................................................................................................4
(b) Solving equations...................................................................................................................6
QUESTION 4..................................................................................................................................7
Centre and radius of circle...........................................................................................................7
QUESTION 5..................................................................................................................................8
QUESTION 6..................................................................................................................................9
Finding point of intersection........................................................................................................9
QUESTION 7................................................................................................................................10
Quadratic equations...................................................................................................................10
i).................................................................................................................................................10
ii)................................................................................................................................................10
iii)...............................................................................................................................................11
QUESTION 8................................................................................................................................11
Quadratic equations with graph.................................................................................................11
a)................................................................................................................................................11
b)................................................................................................................................................12
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QUESTION 1
i) Gradient
P (0, -5)
Q (-2, 8)
Gradient for given two points (x1,y1) and (x2, y2) can be obtained by formula: (y2- y1) / (x2 –
x1)
Gradient = [8 – (-5) ] / [-2-0] = 13/-2
Gradient = -6.5
ii) Equation of line
P (0, -5)
Q (-2, 8)
Slope of given points = -6.5
Let slope of perpendicular line is m
As both lines are perpendicular their product is -1
-6.5 m = -1
m = 1/6.5 = 0.15
Perpendicular line passes through origin (0,0) so equation of perpendicular line will be:
y= 0.15x
QUESTION 2
A = (1, -4) B = (3, -7)
i) Gradient
Gradient m = [-7 – (-4)] / [3-1]
= -3/2 = -1.5
Gradient = -1.5
ii) Equation of line
Equation of line = (y – y1) = m (x- x1)
= (y +4) = -1.5 (x- 1)
Line AB : y = -1.5 x – 2.5
3

iii) Line AB
iv) Point of intersection on x axis
The point at which line crosses the x axis can be determined by putting y= 0 in the equation of
line AB.
y = -1.5 x – 2.5
When we put y = 0
1.5x = -2.5
x = - 1.66
Thus line crosses x axis at -1.66
QUESTION 3
(a) Solving inequalities
i)
15x -4 ≥ 11 (x-2)
Solution
15x -4 ≥ 11x – 22
4x ≥ -18
x ≥ -4.5
4

ii)
x2 -5x - 6 < 0
Solution
x2 -5x - 6 < 0
x [x- 6] +1 [x + 1]
(x- 6) (x+1) < 0
x > -1 and x < 6
-1 < x < 6
5

(b) Solving equations
x + y = -5
y= x
Solution
On substituting y= x in other equation
x + x= -5
2x = -5
x = -2.5
as y = x
so y = -2.5
Solution: x = -2.5 and y = -2.5
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QUESTION 4
Centre and radius of circle
x2 – 8x + y2 – 2y +9 = 0
Solution
The equation of circle is:
(x - a)2 + (y - b)2 = r2
Where r is the radius of circle and coordinates (a,b) represents the centre of circle
Rearranging the given equation in this form:
x2 – 8x + y2 – 2y = -9
At first we need to rearrange x2 – 8x in complete square form:
For ax2 + bx +c we have: a = 1 ; b = -8 and c = 0
Vertex of parabola:
a (x+ d)2 + e
d = -8 / 2*1 = -4
e = c – b2 / 4a
e = 0 – 64 / 4 = -16
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(x- 4)2 – 16
Thus x2 – 8x can be substituted as (x- 4)2 – 16
Hence we can write:
(x- 4)2 – 16 + y2 – 2y = -9
(x- 4)2 + y2 – 2y = 7
Similarly, for y2 – 2y:
a = 1; b = -2; c = 0
d = -2 / 2*1 = -1
e = -1
y2 – 2y = (y- 1)2
Now equation becomes
(x- 4)2 + (y- 1)2= 8
This represent the form: (x- h)2 + (y- k)2= r2
On comparing we have: centre (h, k) = (4,1) and radius = 2√2
QUESTION 5
Solution
Equation of circle: x2 + y2 = 4
Origin O (0,0)
Point P (√2 , √2 )
Gradient OP: (0,0) (√2 , √2 )
= √2 / √2 = 1
Gradient OP = 1
The equation of line passing through point P:
y – √2 = m (x – √2)
y – √2 = mx – m√2
mx = y – √2 + m√2)
mx - y + √2 - m√2) = 0
centre of circle = (0,0) and radius = 2
For tangent equations radius = perpendicular from centre so product of gradient of tangent and
gradient of OP must be equals to -1.
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Let gradient of tangent = m
m * Gradient OP = -1
m * 1 = -1
m = -1
Equation of tangent to the circle at P (√2 , √2 )
y – √2 = -1 (x – √2)
y = -x + √2 + √2
y = -x + 2√2
QUESTION 6
Finding point of intersection
x – y – 5 = 0
(x- 4)2 + (y- 1)2= 4
Solution
x = y +5
Substitute value of x into equation of circle we get:
(y +5 - 4)2 + (y- 1)2= 4
(y +1)2 + (y- 1)2= 4
y2 +1+ 2y + y2 + 1 – 2y = 4
2 y2 + 2 = 4
2 y2 = 2
y = 1 and y = -1
Using these values of y in equation x = y +5 we have:
For y = 1; x = 6
For y = -1 x = 4
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QUESTION 7
Quadratic equations
i)
x2 +5x – 6 = 0
Solution
x2 +5x – 6 = 0
x2 +6x - x – 6 = 0
x [x +6] -1[x + 6]
[x +6] [x -1]
x = -6 or x = +1
ii)
2x2 +5x = 0
Solution
x (2x + 5) = 0
x = 0 and (2x + 5) = 0
(2x + 5) = 0 which gives; x = -5/2
10

x = 0 and x = -5/2
iii)
x2 - 4x + 11 = 0
Solution
Formula of solving quadratic equation roots:
For given quadratic equation: ax2 + bx + c= 0
x = [ -b +- √(b2 – 4ac)] / 2a
In the given equation x2 - 4x + 11 = 0
a = 1 b = -4 c = 11
On substituting values in formula we have:
x = [ 4 +- ((-4)2 – 4(1)(11))] / 2
x = [4 +- 2√7 i ] / 2
= 2 +- √7 i
x = [2 + √7 i] and x = [2 - √7 i]
QUESTION 8
Quadratic equations with graph
a)
x2 + 4x - 12 ≥ 0
Solution
x2 + 4x - 12 ≥ 0
x2 +6x - 2x - 12 ≥ 0
(x - 2) (x + 6) ≥ 0
x ≥ 2 or x ≤ -6
11

b)
x2 - 3x – 10 < 0
Solution
x2 - 5x +2x - 10 < 0
(x - 5) (x + 2) < 0
x < 5 or x > -2
-2 < x < 5
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