Question

Let P and Q be the points on the line joining $A(-2,5)$ and $B(3,1)$ such that $AP=PQ=QB$ , then the mid point of PQ is

• A. $(\frac{1}{2},3)$
• B. $(\frac{-1}{4},4)$
• C. $(2,3)$
• D. $(-1,4)$

Answer 1

Answer: (A)

$(\frac{1}{2},3)$

Given points $A(-2,5),B(3,1)$

Here $AP=PQ=QB$

Hence point P divides AB in ratio $2:1$

By section formula $x=\frac{mb+a}{m+1}$

Point $P(\frac{2\times 3-2}{3},\frac{2\times 1+5}{3})$

$P(\frac{4}{3},\frac{7}{3})$

Since all three distances are equal then Q divides AB in opposite ratio of P i.e. $1:2$

$Q(\frac{-1}{3},\frac{11}{3})$

Mid point of PQ

$(\frac{\frac{4}{3}-\frac{1}{3}}{2},\frac{\frac{7}{3}+\frac{11}{3}}{2})$

$(\frac{\frac{3}{3}}{2},\frac{\frac{18}{3}}{2})$

$(\frac{1}{2},\frac{6}{2})$

$(\frac{1}{2},3)$