Question

Find the values of x for which the distance between the points P(4,-5) and Q(12,x) is 10 units.

Answer 1

$PQ=10$

$P{Q}^2=100$

$(12-4{)}^2+[x-(-5){]}^2=100$

$8^2+(x+5{)}^2=100$

$(x+5{)}^2=36$

$x+5=\pm 6$

$x=1$ or $-11$