Question

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

Answer 1

(i) Let the coordinates of Q be (x, y).
Using the midpoint formula:
(x - 4)/2 = -2
(x - 4) = -4
x = 0
(y + 2)/2 = 1
y + 2 = 2
y = 0
Therefore, the coordinates of Q are (0, 0).

(ii) To find the distance of PQ, we can use the distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
d = √((-2 - (-4))^2 + (1 - 2)^2)
d = √(2^2 + 1)
d = √(4 + 1)
d = √5

(iii) Plotting the coordinates of PQ on the XY plane, we have P(-4, 2) and Q(0, 0). The line segment PQ forms an angle with the positive X-axis, which can be calculated using trigonometry.

(iv) The gradient of a line is given by the formula:
m = (y2 - y1)/(x2 - x1)
m = (0 - 2)/(0 - (-4))
m = -2/4
m = -1/2
Therefore, the gradient of PQ is -1/2.

(v) To find the equation of the line PQ, we can use the point-slope form:
y - y1 = m(x - x1)
y - 2 = -1/2(x - (-4))
y - 2 = -1/2x + 2
y = -1/2x + 4

Therefore, the equation of the line PQ is y = -1/2x + 4.