Question

Find (a) PQ to the nearest tenth and (b) the coordinates of the midpoint of PQ.

Answer 1

Answer/Step-by-step explanation:

Given:

P(2, 6)

Q(-6, 1)

Required:

a. PQ

b. Coordinate of the midpoint of PQ

SOLUTION:

a. $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$P(2, 6) = (x_1, y_1)$

$Q(-6, 1) = (x_2, y_2)$

$PQ = \sqrt{(-6 - 2)^2 + (1 - 6)^2}$

$PQ = \sqrt{(-8)^2 + (-5)^2} = \sqrt{64 + 25}$

$PQ = \sqrt{89} = 3.1$ (to nearest tenth)

b. Coordinate of the midpoint of PQ

$M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

Let $P(2, 6) = (x_1, y_1)$

$Q(-6, 1) = (x_2, y_2)$

Thus:

$M(\frac{2 +(-6)}{2}, \frac{6 + 1}{2})$

$M(\frac{-4}{2}, \frac{7}{2})$

$M(-2, \frac{7}{2})$