Comprehensive Solution: Coordinate Geometry and Quadratic Functions

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Added on  2023/06/17

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Coordinate geometry and quadratic
functions
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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
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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
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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
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(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
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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
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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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