Question

The coordinates of the midpoint of PQ is (-2,1). If the coordinates of P are (-4,2). (i) Find the coordinates of Q. (ii) find the distance PQ. (iii) plot the coordinates of PQ on the XOY-Plane and the angle it makes with the positive X-axis. (iv) find the gradient (v) find the equation of the line PQ.

Answer 1

(i) Let the coordinates of Q be (x, y).
Using the midpoint formula, we have:
(-4 + x)/2 = -2
-4 + x = -4
x = 0

(2 + y)/2 = 1
2 + y = 2
y = 0

Therefore, the coordinates of Q are (0, 0).

(ii) The distance between two points (x1, y1) and (x2, y2) is given by:
√((x2 - x1)^2 + (y2 - y1)^2)

In this case:
√((-4 - 0)^2 + (2 - 0)^2)
= √(16 + 4)
= √20
= 2√5

So, the distance PQ is 2√5.

(iii) The coordinates of PQ are (-2,1). This point lies in the second quadrant and makes an angle of approximately 153.43° with the positive X-axis.

(iv) The gradient of a line passing through two points (x1, y1) and (x2, y2) is given by:
m = (y2 - y1)/(x2 - x1)

In this case:
m = (0 - 2)/(0 - (-4))
m = -2/4
m = -1/2

So, the gradient of line PQ is -1/2.

(v) To find the equation of a line given a point (x1, y1) and gradient m, the equation of the line is given by:
y - y1 = m(x - x1)

Substitute the values of (-4, 2) and -1/2 into the equation:
y - 2 = -1/2(x + 4)
y - 2 = -1/2x - 2
y = -1/2x

So, the equation of line PQ is y = -1/2x.