Question

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

Answer 1

P(x1 , y1) = (a, 7)

Q(x2 , y2) = (- 3 , a)

Equidistant means it must be midpoint

R(x , y) = (2 , -1)

$(2. - 1) = ( \frac{x1 + x2}{2} . \frac{y1 + y2}{2} ) \\ \\ (2. - 1) = ( \frac{a + ( - 3))}{2} . \frac{7 + a}{2} ) \\ \\ (2. - 1) = ( \frac{a - 3}{2} . \frac{7 + a}{2} )$

$\frac{a - 3}{2} = 2 \\ \\ a - 3 = 4 \\ a = 4 + 3 \\ a = 7$

$\frac{7 + a}{2} = - 1 \\ \\ 7 + a = - 2 \\a = - 2 - 7 \\ a = - 9$