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The mid-point $P$ of the line segment joining the points $A (-10, 4)$ and $B (-2, 0)$ lies on the line segment joining the points $C (-9, -4)$ and $D (-4, y)$. Find the ratio in which $P$ divides $CD$. Also, find the value of $y$.
Given:
The mid-point $P$ of the line segment joining the points $A (-10, 4)$ and $B (-2, 0)$ lies on the line segment joining the points $C (-9, -4)$ and $D (-4, y)$.
To do:
We have to find the ratio in which $P$ divides $CD$ and the value of $y$.
Solution:
Let \( P \) divides \( C(9,-4) \) and \( D(-4, y) \) in the ratio \( m: n \)
P is the mid-point of the line segment joining the points $A (-10, 4)$ and $B (-2, 0)$
This implies,
\( P=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \)
\( =\left(\frac{-10-2}{2}, \frac{4+0}{2}\right) \)
\( =\left(\frac{-12}{2}, \frac{4}{2}\right) \)
\( =(-6,2) \)
P lies on CD.
Using the section formula, if a point $( x,\ y)$ divides the line joining the points
$( x_1,\ y_1)$ and $( x_2,\ y_2)$ in the ratio $m:n$, then
$(x,\ y)=( \frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n})$
This implies,
\( (-6,2)=(\frac{m(-4)+n(-9)}{m+n},\frac{3 \times y+2 \times(-4)}{3+2}) \)
On comparing, we get,
\( -6=\frac{m(-4)+n(-9)}{m+n} \)
\( \Rightarrow -6 m-6n=-4 m-9n \)
\( \Rightarrow -6 m+4 m=-9n+6n \)
\( \Rightarrow -2 m=-3 n \)
\( \Rightarrow \frac{m}{n}=\frac{-3}{-2} \)
\( =\frac{3}{2} \)
\( 2=\frac{3 \times y+2 \times(-4)}{3+2} \)
\( \Rightarrow 2=\frac{3 y-8}{5} \)
\( \Rightarrow 10=3 y-8 \)
\( \Rightarrow 3 y=10+8 \)
\( \Rightarrow 3 y=18 \)
\( \Rightarrow y=\frac{18}{3} \)
\( \Rightarrow y=6 \)
The ratio in which $P$ divides $CD$ is $3:2$ and the value of $y$ is $6$.
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