# Introduction to Coordinate Geometry

```Introduction to Coordinate Geometry
A(-3,-4) B(4,3)
Calculate slope gradient/slope of a line given points, equation and graphical
method
Gradient of a line given 2 points on the line
Ex 6A
Q1
b) C (2,-3) and D (1, 7)
y2 = 7, y1= -3
x2= 1, x1= 2
m = y2-y1 / x2-x1
= 7-(-3) / 1-2
= 7+3 / -1
= 10/-1
= -10
A (2, t) B (7, 2t2+7)
AB
X1 = 2
Y1 = t
M=
2=
X2 = 7
Y2 = 2t2+7
= t2+7) – t / 7-2
10 = 2t2 + 7 – t
2t2 +7 -10 –t = 0
2t2 - t - 3 = 0
Use two point distance formula to calculate distance between any two ordered
pairs
Q1. D) G (-10, 2) and H (-4, -7)
X1=-10
X2 = -4
Y1=2
Y2 = -7
GH =√(X1-X2)2 + (Y1-Y2)2
GH = √(-10-(-4))2 + (2-(-7))2
GH = √(-10+4)2 + (2+7)2
GH = √(-6)2 + 81
GH = √36 + 81
GH = √117 = 10.82
Q2) A (p, 0) B (0, p)
D = 10
X1= p
X2= 0
Y1= 0
Y2=p
AB = √ (x1-x2)2 + (y1-y2)2
10 = √2p2
10 = √ (p-0)2 + (0-p)2
102 = (√2p2)2
10= √p2 + (-p)2
100 = 2p2
10 = √p2 + p2
50 = p2
= p2
√50 = √p2
p = √50
p = 7.071
Q3 a) R (0, y) point lies
on y-axis. PR = QR
b) S (x, 0) point lies
on x-axis. PS = QS
P (-2, 6) and Q (9, 3)
a) P (-2, 6) R (0, y)
PR =√ (0-(-2)) 2 + (Y-6)2
QR = √(0-9)2 + (Y-3)2
PR =QR
b) S (x, 0)
PS = √(X-(-2))2 + (0-6)2
QS = √(X-9)2 + (0-3)2
PS =QS
Q4) MW = NW
W (0, y) M (3, 7) N (11, -6)
M
W
N
Q5) i) area = &frac12; * 12 *8
= 48unit2
Perimeter = sum of all
the lengths of the
triangle
D
√ (8-2)2 + (-2-6)2
i) area = &frac12; * 3 * 3
= 4.5 units2
K
ii) A(-2,1) C (3,4)
AC
iii) D (0, 4)
iv) area = &frac12; * b* h
12 = &frac12; * b * 3
12 * 2 = b* 3
24 = 3*b
24/3 = b
B = 8unit
CK = √(t-3)2 + (4-4)2
8 = √(t-3)2 + 0
(8)2 = (√(t2 + 9 – 6k))2
64 = t2 + 9 -6k
T2 - 6k -55 = 0
Understand meaning and effect of negative, positive and zero gradient
Find equation of a straight line using slope and y-intercept
Q1) y = -x + c
(1, 2)
2 = -1 + c
C = 2+1 = 3
Y = -x + 3
Q3) f) K (-7,-5) and L (-1,-1)
Step1: y = mx +c | m= -1-(-5) / -1-(-7)
m= -1+5 / -1+7 m= 4/6 m=2/3
Step2: y = 2/3x + c
Step3: -5 = 2/3 (-7) + c | -5 = -14/3 + c
Step4:
y = 2/3x – 1/3
c = -5+14/3 c = -1/3
Q2) Y = 4X + K
(-3, 3)
3 = 4(-3) + K
3 = -12+ K
K= 3+12 = 15
Y = 4X + 15
Q3) d) G (-6, -5) and H (4, 4)
Step1: y = mx + c | m = 4 – (-5) / 4 – (-6)
M = 9 / 10
Step2: y = 9/10 x + c
Step3: 4 = 9/10 (4) + c | 4 = 18/5 + c
C = 4 - 18/5 = 2/5
Step4:
y = 9/10 x + 2/5
Q5) (0, 0)
m=2
Step1: m=2
Step2: y = 2x + c
Step3: y = 2x + c | 0 = 2(0) + c c = 0
Step4: y = 2x + 0 | y = 2x
Q6) a)
y = mx + c
Y = 0x + c | y = 0 + c
Y=c
b)
y=1
Y = mx + c
(1.5, 0)
c)
|
x = 1.5
(0, -1) and (1, 0)
C = -1
M = 0 – (-1) / 1 – 0 = 1/1 =
1
Y = 1x + (-1)
Y = 1x – 1
d) Y= -1/2 x + 1
(0, 1) (2, 0)
Relationships between gradients of parallel and perpendicular lines to find
Calculate midpoint
B
A
T
Q)
A(5, 2) B (7, 5)
T is a point equidistant from A and B.
Find the coordinates of T?
T(
,
T (6, 3.5)
)
T(
,
)
Q1) a) m = 8
b) m = -1/2
Q2) a) m1
* m2 = -1
3 * m2 = -1
M2 = -1/3
b) -3/4 * m2 =1
M2 = -1 / (*-3/4)
M2 = -1* -4/3
M2 = 4/3
2y = 8x + 5
M=4
2Y = 3/2x + 4
M = &frac34; = 0.75
Area of triangle given 3 points/vertices
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