Question

Find the ratio in which the point $P(\frac{3}{4},\frac{5}{12})$ divides the line segment, joining the points $A(\frac{1}{2},\frac{3}{2})$ and $B(2,-5)$.

• A. $1:3$
• B. $1:5$
• C. $2:3$
• D. $2:5$

Answer 1

Answer: (B)

$1:5$

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ is

$P(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let point $P$ divide $AB$ in the ratio $1:k$

Then, by section formula,

$(\frac{3}{4},\frac{5}{12})\equiv (\frac{k\left(\frac{1}{2}\right)+1\left(2\right)}{k+1},\frac{k\left(\frac{3}{2}\right)+1(-5)}{k+1})$

Equating $x$ and $y$ coordinates, we get

$\frac{k/2+2}{k+1}=\frac{3}{4},\frac{3k/2-5}{k+1}=\frac{5}{12}$

$\Rightarrow 2k+8=3k+3,18k-60=5k+5$

$\Rightarrow k=5$

Hence $P$ divides $AB$ in the ratio $1:5$