Question

The coordinate of a point $P$ on the line segment joining $A(1,2)$ and $B(6,7)$ such that $AP=\frac{2AB}{5}$, is

• A. $(3,4)$
• B. $(1,2)$
• C. $(-1,2)$
• D. None of these

Answer 1

Answer: (A)

$(3,4)$

Here, point $P$ is on $AB$ such that $AP=\frac{2AB}{5}\Rightarrow \frac{AP}{AB}=\frac{2}{5}$.

Using componendo and dividendo, we have,

$\frac{AP}{(AB-AP)}=\frac{2}{(5-2)}$

$\frac{AP}{PB}=\frac{2}{3}$

This means that $P$ divides $AB$ in the ratio $2:3$.

Therefore, coordinates of P will be,

$\Rightarrow P(\frac{2\times 6+3\times 1}{2+3},\frac{2\times 7+3\times 2}{2+3})$

$\Rightarrow P(3,4)$

Hence, the required point is $P(3,4)$.