Question

P(-1, 4) and Q( 3, 2) are the end points of a line segment PQ. Determine
a. the gradient of PQ
b. The coordinates of the midpoint of PQ *
c. Another line segment, JK, is perpendicular to PQ and passes through the midpoint of PQ. Determine the equation of the line JK.

Answer 1

a. The gradient of PQ = -1/2

b. The coordinates of the midpoint of PQ = M(1, 3).

c. The equation of line JK is y - 3 = 2(x - 1)

Recall:

Gradient/slope (m) = change in y / change in x

Midpoint formula = $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} )$

Given:

P(-1, 4) and Q( 3, 2)

a. Gradient of PQ = (2 - 4) / (3 - (-1)) = -2/4

Gradient of PQ = -1/2

b. Coordinates of midpoint of PQ = $M(\frac{-1 + 3}{2}, \frac{4 + 2}{2} )$

M(1, 3).

c. Since the line of JK is perpendicular to PQ, the slope of JK will be the negative reciprocal of the slope of PQ, which is 2.

The midpoint, M(1, 3) is also on a point on line JK.

To write an equation of line JK, substitute m = 2, and (x1, y1) = (1, 3) into y - y1 = m(x - x1).

• Thus:

y - 3 = 2(x - 1)

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