Question

A point P(2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

• A. 5
• B. 1
• C. 3
• D. 7

Answer 1

The correct option is D 7

Given, the point P(2, -1) is equidistant from the points A(a, 7) and B(-3, a).
So, we have
$PA=PB$
$\Rightarrow P{A}^2=P{B}^2$
By using the distance formula, we have
$(a-2{)}^2+(7+1{)}^2=(-3-2{)}^2+(a+1{)}^2$
$a^2+4–4a+64=25+a^2+1+2a$
$68–4a=26+2a$
6a = 42
a = 7
Hence, the value of a is 7.