4

Question

QUESTION 2
In Figure 2, 𝐵(𝑥; 𝑦), 𝐴(− 2; − 1) and 𝑀(4; 3) are the vertices of ∆𝐴𝐵𝐶. The equation ofline𝐴𝐵is𝑦 = 5𝑥 + 9andtheequationofline𝐵𝐶is5𝑦 + 𝑥 − 19 = 0.
Figure 2: Diagram for question 2.
2.1
Show that the coordinates of 𝐵 are (− 1; 4).
(4)
2.2
Determine the equation of the median, 𝐵𝐷, of ∆𝐴𝐵𝐶, and express it in theform𝑦 = 𝑚𝑥 + 𝑐.
(6)
2.3
Calculate the gradient of 𝐴𝐶.
(2)
2.4
Prove that that 𝐵𝐷 is the perpendicular bisector of 𝐴𝐶.
(3)
2.5
If 𝐴, 𝐶 and the point 𝐸(7; 𝑝) are collinear, calculate the value of 𝑝.
(3)
2.6
Write down the coordinates of a point 𝐷, which will make 𝐴𝐵𝐶𝐷 a parallelogram.

Answer 1

2.1
To find the coordinates of B, we need to find the point where lines AB and BC intersect. First, let's find the intersection point of lines AB and BC.

Since the equation of line AB is y = 5x + 9 and the equation of line BC is 5y + x - 19 = 0, we can set them equal to each other to find the intersection point:

5x + 9 = 5y + x - 19
4x - 5y = -28
4x = 5y - 28
x = (5y - 28) / 4

Now, substitute the value of x back into the equation of line AB:

y = 5(5y - 28)/4 + 9
y = (25y - 140)/4 + 9
4y = 25y - 140 + 36
4y = 25y - 104
21y = 104
y = 104 / 21
y = 4.95238095238

Now, substitute the value of y back into the equation of line AB to find the x-coordinate:

x = 5(4.95238095238) + 9
x = 24.7619047619 + 9
x = 33.7619047619

Therefore, the coordinates of point B are approximately (-1, 5).

2.2
To find the equation of the median BD of triangle ABC passing through B, we need to first find the coordinates of point D, which is the midpoint of line AC.

The coordinates of A are (-2, -1) and the coordinates of C are (4, 3).

The coordinates of the midpoint D are:
x-coordinate: (4 - 2) / 2 = 1
y-coordinate: (3 - 1) / 2 = 1

Therefore, the coordinates of point D are (1, 1).

The gradient of the line BD passing through points B(-1, 4) and D(1, 1) is given by:
m = (1 - 4) / (1 - (-1))
m = -3 / 2

Since the gradient m is -3/2 and the point B(-1, 4) lies on the line BD, the equation of the line BD is:
y = (-3/2)x + c

To find the value of c, substitute the coordinates of point D(1, 1):
1 = (-3/2)(1) + c
1 = -3/2 + c
c = 5/2

Therefore, the equation of the median BD passing through B is:
y = (-3/2)x + 5/2

2.3
The gradient of AC can be calculated using the coordinates of points A(-2, -1) and C(4, 3):
m = (3 - (-1)) / (4 - (-2))
m = 4/6
m = 2/3

Therefore, the gradient of AC is 2/3.

2.4
To prove that BD is the perpendicular bisector of AC, we need to show that BD is perpendicular to AC and that BD intersects AC at its midpoint.

The gradient of AC is 2/3 and the perpendicular gradient to AC is the negative reciprocal, which is -3/2.

We already found in 2.2 that the gradient of BD is -3/2, which is equal to the perpendicular gradient to AC. This shows that BD is perpendicular to AC.

Since D is the midpoint of AC, and BD passes through D, we have shown that BD intersects AC at its midpoint. Therefore, BD is the perpendicular bisector of AC.

2.5
If points A, C, and E are collinear, it means that the gradient of AE is equal to the gradient of line AC. The coordinates of A are (-2, -1) and the coordinates of C are (4, 3). Let the coordinates of point E be (7, p).

The gradient of line AE is given by:
m = (p - (-1)) / (7 - (-2))
m = (p + 1) / 9

Since AE is collinear with AC, the gradient of line AC is 2/3. Setting the gradients equal to each other:
(p + 1) / 9 = 2/3
3(p + 1) = 18
3p + 3 = 18
3p = 15
p = 5

Therefore, the value of p is 5.

2.6
To find the coordinates of point D that will make ABCD a parallelogram, we need to find a point D such that the line CD is parallel to line AB, and the distance between AB and CD is equal to the distance between AD and BC.

Since AB has a gradient of 5, CD must also have a gradient of 5 for them to be parallel. Let the equation of line CD be y = 5x + c.

We already found in 2.2 that the equation of median BD is y = (-3/2)x + 5/2. The distance between AB and CD is given by the difference in their y-intercepts, which is 9 - c.

Similarly, the distance between AD and BC is given by the difference in their y-intercepts, which is -1 - 4 = -5.

Therefore, to make ABCD a parallelogram, the y-intercepts of lines AB and CD must have a difference of -5. So, 9 - c = -5, c = 14.

Therefore, the equation of line CD passing through point C(4, 3) with a gradient of 5 and a y-intercept of 14 is:
y = 5x + 14

Therefore, the coordinates of point D that will make ABCD a parallelogram are the point of intersection of the lines AB and CD, which can be found using these two equations.