Question

Find the ratio in which the point $(2,y)$ divides the line segment joining the points $(-4,3)$ and $(6,3)$ and hence find the value of $y$ :

• A. $2:3,y=3$
• B. $3:2,y=4$
• C. $3:2,y=3$
• D. $3:2,y=2$

Answer 1

Answer: (C)

$3:2,y=3$

Let the given point $C(2,y)$ divides $A(-4,3)$ and $B(6,3)$ in the required ratio be $k:1$
Using the section formula $-$
$x$ = $\frac{x_{2}k+x_1\left(1\right)}{k+1}$ and $y$ = $\frac{y_{2}k+y_1\left(1\right)}{k+1}$

Then, $\frac{6k-4\left(1\right)}{k+1}=2$

$\Rightarrow 6k-4=2k+2$
$\Rightarrow 6k-2k=4+2$
$\Rightarrow 4k=6$
$\Rightarrow k=\frac{6}{4}$

$\Rightarrow k=\frac{3}{2}$

Hence the ratio is $3:2$

Thus, substituting we get the value of $y$ as,

$y=\frac{3\left(3\right)+2\left(3\right)}{3+2}$
$y=\frac{9+6}{5}$

$y=\frac{15}{5}=3$

The value of $y$ is $3$