Question

Given that point A is the midpoint of line segment ED, where E( -1,9) and A(4,5), what is the location of point D?

Answer 1

Answer:

D(9, 1)

Explanation:

Since A is the midpoint of ED, then;

A = ($\frac{X_{1}+X_{2} }{2}$, $\frac{Y_{1}+Y_{2} }{2}$)

A(4, 5) ⇒ 4 =  $\frac{X_{1}+X_{2} }{2}$ and 5 = $\frac{Y_{1}+Y_{2} }{2}$

$\frac{X_{1}+X_{2} }{2}$ = 4

$X_{1}$ + $X_{2}$ = 8

But $X_{1}$ = -1, then;

-1 + $X_{2}$ = 8

$X_{2}$ = 9

$\frac{Y_{1}+Y_{2} }{2}$ = 5

$Y_{1}$ + $Y_{2}$ = 10

But $Y_{1}$ = 9, then;

9 + $Y_{2}$  = 10

$Y_{2}$  = 1

Therefore, point D has coordinates (9, 1).