Correction exercises COORDINATE GEOMETRY
1) The point A has coordinates(1 ; 7) and the point B has coordinates (5 ; 1)
a) (i) Find the gradient of the line (AB)
𝑦 −𝑦
7−1
6
−𝟑
The gradient (or slope) of the line (AB) is given by the formula : m =𝑥 𝐴 −𝑥 𝐵 = 1−5=−4= 𝟐
𝐴
𝐵
(ii) Show that the line (AB) has equation 3x + 2y = 17
The equation of a straight line is given by the formula y = m x + p.
We know m and the coordinates of B so we can write 𝑦𝐵 =
1=
−3
2
× 5+ p
-> 1 =
−15
2
The equation of (AB) is y =
2
2
So
y=
−3
2
x+
17
2
->
15
+p
-> 1 + 2 = p ->
−3
2
17
2
x+
2
p=2+
−3
2
15
2
=
𝑥𝐵 + p
17
2
. We are asked to write it in another form (without any fraction)
2y = - 3 x + 17
-> 3x + 2y = 17
b) The line (AB) intersects the line with equation x – 4y = 8 at the point C. Find the coordinates of C.
C belongs to the line (AB) so we can write 3𝑥𝐶 + 2𝑦𝐶 = 17
C also belongs to another line, so it gives 𝑥𝐶 – 4𝑦𝐶 =8
We consider both equations to solve the following system with the elimination method
3𝑥𝐶 + 2𝑦𝐶 = 17
𝑥𝐶 – 4𝑦𝐶
=8
If we want to find the value of 𝑥𝐶 , we are going to multiply the first equation by 2 so that the
6𝑥𝐶 + 4𝑦𝐶 = 34
coefficients of y are opposites
Now we can add both lines, which gives
𝑥𝐶 – 4𝑦𝐶 = 8
7 𝑥𝐶
= 42 -> 𝑥𝐶 =
42
7
-> 𝒙𝑪 = 6
Now we substitute 𝑥𝐶 in either of the original equation to get the value of 𝑦𝐶 .
2
For example in the 2nd one it gives 6 – 4𝑦𝐶 = 8 -> – 4𝑦𝐶 = 8 – 6 -> – 4𝑦𝐶 = 2 -> 𝑦𝐶 = − 4 -> 𝒚𝑪 =
The coordinates of C are (6 ;
−𝟏
𝟐
−𝟏
)
𝟐
c) Find an equation of the line through A which is perpendicular to (AB)
We know that if 2 lines are perpendicular, the product if their slope is equal to -1. Here it means that
m × 𝑚(𝐴𝐵) = -1. From the very first question, we have he slope of the (AB):
m×
−3
2
=-1
Our line is y =
2
-> m = - 1×
2
3
2
−3
=+
𝟐
𝟑
2
−3
.
2
So it gives
𝑥 + p. We know that the point A is on the line, so 𝑦𝐴 = 3 𝑥𝐴 + 𝑝. With A coordinates:
2
2
7= 3 × 1 + 𝑝 -> 7= 3 + 𝑝 -> 7- 3 = 𝑝 ->
The equation of the line is y =
2
3
𝑥+
19
3
21
3
2
- 3 = 𝑝 ->
𝟏𝟗
𝟑
= 𝒑
y
8
7
A
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
-1
0
-1
-2
d2
-3
B
1
2
3
4
5
6
C
7
8
x
1
2) The line 𝑙1 passes through the point (9 ; -4) and has gradient (slope) 3
a) Find an equation for 𝑙1 in the form ax + by + c = 0 where a, b and c are integers
The equation of a straight line is given by the formula y = m x + p.
1
We know m and the coordinates of one point so we can write - 4 = 3 × 9+ p
1
-> - 4 = 3 × 3 × 3 + p
-> - 4 = 3 + p -> - 4 – 3 = p -> p = - 7
1
The equation of (𝑙1 ) is y = 3 x - 7 . We are asked to write it in another form (ax + by + c = 0)
1
3
So we put every term on the left hand side of the equal sign: - x + y+ 7= 0
We are also asked to give the equation with integers so we multiply every term by 3
- 1 x + 3 y + 21 = 0
b) The line 𝑙2 passes through the origin O and has gradient – 2. The lines 𝑙1 and 𝑙2 intersect at the point P.
Calculate the coordinates of P
* First of all, we have to find the equation of 𝑙2 . We know its slope: - 2 and the point (0;0)
So, the formula y = m x + p becomes 0 = - 2 × 0 + p -> 0 = p
The equation of 𝑙2 is y = - 2𝑥
* P belongs to the line (𝑙1 ) so we can write - 1𝑥𝑃 + 3𝑦𝑃 + 21= 0
P also belongs to the line (𝑙1 ), so it gives 𝑦𝑃 = - 2 𝑥𝑃
We consider both equations to solve the following system with the substitution method
− 1𝑥𝑃 + 3𝑦𝑃 + 21 = 0
𝑦𝑃 = −2𝑥𝑃
We substitute 𝑦𝑃 by − 2𝑥𝑃 in the first equation: − 1𝑥𝑃 + 3 × (−2𝑥𝑃 ) + 21 = 0
-> − 1𝑥𝑃 − 6𝑥𝑃 + 21 = 0
-> − 7𝑥𝑃 + 21 = 0
-> − 7𝑥𝑃 = - 21 -> 𝒙𝑷 =
We substitute 𝑥 by 3 in the second equation So, the coordinates of P are ( 3; - 6).
We can check this result on the following drawing
−21
−7
= +3
y
2
1
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
-4
-5
-6
P
-7
-8
L1
-9
L2
6
7
8
9
10
11x
3) The points A (1 ; 7), B (20;7) and C (𝑥; 𝑦) form the vertices of a triangle ABC, as shown on the diagram below.
The point D (8;2) is the mid-point of [AC]
a) (i) Find the coordinates of C
The formula of a mid-point is
𝑥𝐷 =
8=
𝑥 𝐴 +𝑥 𝐶
2
1+𝑥 𝐶
2
8 ×2 = 1 +𝑥𝐶
𝑦𝐷 =
2=
𝑦 𝐴 +𝑦 𝐶
2
7+𝑦 𝐶
2
2 ×2 = 7 + 𝑦𝐶
16 = 1 +𝑥𝐶
4 = 7 + 𝑦𝐶
16 - 1 = 𝑥𝐶
4 - 7 = 𝑦𝐶
𝑥𝐶 = 15
𝑦𝐶 = −3
The coordinates of C are ( 15; - 3)
(ii) Calculate the distance between the points A and C
The formula of distance AC is (𝑥𝐶 − 𝑥𝐴 )2 + (𝑦𝐶 − 𝑦𝐴 )2 with A (1;7) and C(15; -3)
So AC = (15 − 1)2 + (−3 − 7)2 = (14)2 + (−10)2 = 196 + 100 = 296 = 4 × 74 = 2 𝟕𝟒
b) The line 𝑙, which passes through D and is perpendicular to (AC), intersects (AB) at E. Find an equation for 𝑙, in
the form ax + by + c = 0, where a, b and c are integers.
if 2 lines are perpendicular, the product of their slope is equal to -1. Here, 𝑚𝑙 × 𝑚(𝐴𝐷) = -1
𝑦 𝐴 −𝑦 𝐷
𝑥 𝐴 −𝑥 𝐷
Let us calculate the slope of (AD) m =
Since 𝑚𝑙 × 𝑚(𝐴𝐷) = -1, we have
7−2
5
= 1−8=−7
𝑚𝑙 × ( -
5
7
)= -1 -> 𝑚𝑙 = - 1×
−7
5
=
7
5
7
We know now that for the line 𝑙, y = 5 𝑥 + p. Let’s use the coordinates of D to work out the value of p
7
5
𝑦𝐷 = 𝑥𝐷 + 𝑝
7
2=5×8+ 𝑝
->
2=
56
5
7
+ 𝑝 -> 2−
So an equation of 𝑙 is y = 5 𝑥 -
46
5
56
5
= 𝑝
->
10
5
−
56
5
= 𝑝
-> 𝑝 = -
46
5
.
We are asked to give the equation with integers so we multiply every term by 5: 5y = 7 𝑥 – 46.
We want the form ax + by + c = 0, so we put every term on the left hand side:
The equation of 𝒍 is - 𝟕𝒙 + 5y + 46 = 0
c) Find the exact 𝑥- coordinates of E
E belongs to the line (𝐴𝐵). We notice that 𝑦𝐴 = 𝑦𝐵 = 7. So the equation of (AB) is 𝑦 = 7 and we can
say that 𝒚𝑬 = 7
E belongs to the line 𝑙 too. So we can write - 7𝑥𝐸 + 5𝑦𝐸 + 46 = 0
We substitute 𝑦𝐸 by 7 in the equation of 𝑙: - 7𝑥𝐸 + 5 × 7 + 46= 0 -> - 7𝑥𝐸 + 35 + 46= 0
-> - 7𝑥𝐸 + 81= 0
-> - 7𝑥𝐸 = - 81
-> 𝑥𝐸 =
−81
−7
-> 𝒙𝑬 =
𝟖𝟏
𝟕
so E coordinates are ( 7;
𝟖𝟏
)
𝟕
We can check our results on the following drawing
y
14
13
12
11
10
9
8
7
6
5
4
3
2
1
-10-9 -8 -7 -6 -5 -4 -3 -2 -1 0
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 x
y
8
7
6
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0
-1
-2
-3
-4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 x