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The midpoint P of the line segment joining points $\mathrm{A}(-10,4)$ and $\mathrm{B}(-2,0)$ lies on the line segment joining the points $C(-9,-4)$ and $D(-4, y)$. Find the ratio in which P divides $C D$. Also, find the value of $\mathrm{y}$.
The midpoint P of the line segment joining points $\mathrm{A}(-10,4)$ and $\mathrm{B}(-2,0)$ lies on the line segment joining the points $C(-9,-4)$ and $D(-4, y)$. Find the ratio in which P divides $C D$. Also, find the value of $\mathrm{y}$.
CBSE | Class 10 | Coordinate Geometry | Exercise 16.2 | Maths | RS Aggarwal
The midpoint P of the line segment joining points $\mathrm{A}(-10,4)$ and $\mathrm{B}(-2,0)$ lies on the line segment joining the points $C(-9,-4)$ and $D(-4, y)$. Find the ratio in which P divides $C D$. Also, find the value of $\mathrm{y}$.

The midpoint of $\mathrm{AB}$ is $\left(\frac{-10-2}{2}, \frac{4+10}{2}\right)=\mathrm{P}(-6,2)$

Let $k$ be the ratio in which P divides $C D$. So

$(-6,2)=\left(\frac{\mathrm{k}(-4)-9}{\mathrm{k}+1}, \frac{\mathrm{k}(\mathrm{y})-4}{\mathrm{k}+1}\right)$

$\Rightarrow \frac{\mathrm{k}(-4)-9}{\mathrm{k}+1}=-6$ and $\frac{\mathrm{k}(\mathrm{y})-4}{\mathrm{k}+1}=2$

$\Rightarrow \mathrm{k}=\frac{3}{2}$ Now, substituting $k=\frac{3}{2}$ in $\frac{\mathrm{k}(\mathrm{y})-4}{\mathrm{k}+1}=2$, we get

$\frac{\mathrm{y} \times \frac{3}{2}-4}{\frac{3}{2}+1}=2$

$\Rightarrow \frac{3 y-8}{5}=2$

$\Rightarrow \mathrm{y}=\frac{10+8}{3}=6$

Therefore, the required ratio is $3: 2$ and $\mathrm{y}=6$.

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