Question

If the distance between p(-7,-3) and q(y,4) is7units, find the value of y

Answer 1

Step-by-step explanation:

The distance between two points

(a,b) and (c,d)

is given by

d =√[(a-c)² + (b-d)²]

Plug in the given values

7 = √[(-7-y)² + (-3-4)²]

Squaring both sides

49 = (-7-y)² + (-7)²

49 = (-7-y)² + 49

(-7-y)² = 0

-7-y = 0

y = -7