Question

Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, -3). Also, find the value of x.

Answer 1

Hint: In this question, we are given coordinates of end points of a line and y coordinates of a point dividing the line segment. We have to find x coordinate of the point and ratio in which that point divides the line segment. For this, we will use a section formula. Section formula used to find the coordinates of a point dividing the line segment having end points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ in the ratio m:n is given by $\left( x,y \right)=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$.

Complete step-by-step solution:
As per the question, we have to find the value of x and m:n. Using $\left( x \right)=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right)$ we will find value of x and using $\left( y \right)=\left( \dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$ we will find value of m:n.
Here, we are given end points of line segment AB as A(12,5) and B(4,-3). Point P(x,2) divides the line segment AB in some ratio. Let us suppose that this ratio is m:n. As we know, section formula used to find coordinates of a point dividing the line segment having end points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ in ratio $m:n$ given by $\left( x,y \right)=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$.
From given question, y = 2, ${{x}_{1}}=12,{{y}_{1}}=5,{{x}_{2}}=4\text{ and }{{y}_{2}}=-3$

So section formula for the line AB where point P divides AB in ratio m:n becomes
\[\Rightarrow \left( x,2 \right)=\left( \dfrac{4m+12n}{m+n},\dfrac{-3m+5n}{m+n} \right)\]
Comparing coordinates we get:
\[\begin{align}
  & \Rightarrow x=\dfrac{4m+12n}{m+n}\cdots \cdots \cdots \cdots \cdots \left( 1 \right) \\
 & \Rightarrow 2=\dfrac{-3m+5n}{m+n}\cdots \cdots \cdots \cdots \cdots \left( 2 \right) \\
\end{align}\]
Let us use (2) to find the ratio m:n we get:
\[\Rightarrow 2=\dfrac{-3m+5n}{m+n}\]
Cross multiplying we get:
\[\begin{align}
  & \Rightarrow 2\left( m+n \right)=-3m+5n \\
 & \Rightarrow 2m+2n=-3m+5n \\
\end{align}\]
Taking m values one side and n values on other we get:
\[\begin{align}
  & \Rightarrow 2m+3m=5n-2n \\
 & \Rightarrow 5m=3n \\
\end{align}\]
Dividing both side by 5n we get:
\[\begin{align}
  & \Rightarrow \dfrac{5m}{5n}=\dfrac{3n}{5n} \\
 & \Rightarrow \dfrac{m}{n}=\dfrac{3}{5} \\
\end{align}\]
Therefore, m:n = 3:5 . . . . . . . . . . . . . . . . . (3).
Using (1) and (3), let us find value of x, equation (1) is given as $\Rightarrow x=\dfrac{4m+12n}{m+n}$
From (3) let us put values of m and n, we get:
\[\begin{align}
  & \Rightarrow x=\dfrac{4\left( 3 \right)+12\left( 5 \right)}{3+5} \\
 & \Rightarrow x=\dfrac{12+60}{8} \\
 & \Rightarrow x=\dfrac{72}{8} \\
 & \therefore x=9 \\
\end{align}\]
Therefore, the value of x is equal to 9. Hence, ratio $m:n$ is equal to 3:5, and the value of x is equal to 9.

Note: Students should carefully apply the section formula for finding coordinates of point dividing the line in the ratio m:n. We have used ratio values of m and n in equation (1), because the common ratio of m, n will be cancelled out automatically in the formula $x=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right)$.