Question

A line perpendicular to the line segment joining the points (1,0) and (2,3) divides it in the ratio 1: n. Find the equation of the line.

Answer 1

Hint: In order to solve this problem, we need to understand that the line is divided in a certain ratio and the coordinates of that point are given by section formula. Section formula says that if the segment is divided in ratio m:n between the two coordinates namely $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ then the coordinates that divides are $C\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$ . We also need to know how to find the slope of the line. The formula for the slope with the same two coordinates $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is $slope\left( {{m}_{1}} \right)=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .

Complete step-by-step answer:
We are given a with endpoints (1,0) and (2,3) divided in the ratio of 1: n.
We need to find the coordinates of the point that divides the segment.
We can use the section formula for this.

Section formula says that if the segment is divided in ratio m:n between the two coordinates namely
$A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ then the coordinates that divides are $C\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$ .
Now for this question let $A\left( {{x}_{1}},{{y}_{1}} \right)=A\left( 1,0 \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)=B\left( 2,3 \right)$ .
Applying the formula, we get the coordinates of C as follows,
$C\left( \dfrac{\left( 1\times 2 \right)+\left( n\times 1 \right)}{1+n},\dfrac{\left( n\times 0 \right)+\left( 1\times 3 \right)}{1+n} \right)$
Solving this the coordinates of points C will be,
$C\left( \dfrac{n+2}{1+n},\dfrac{3}{n+1} \right)$
Now we need to find the slope of this line.
The formula for the slope with the same two coordinates $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is $slope\left( {{m}_{1}} \right)=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
Substituting the values, we get,
${{m}_{1}}=\dfrac{3-0}{2-1}=3$ .
Now we need to find the slope of the line perpendicular to this line.
But, the product of the slopes of two lines perpendicular to each other is -1.
Hence the slope of other line $\left( {{m}_{2}} \right)$ is as follows,
$\begin{align}
  & {{m}_{1}}.{{m}_{2}}=-1 \\
 & 3\times {{m}_{2}}=-1 \\
 & {{m}_{2}}=\dfrac{-1}{3}. \\
\end{align}$
Now we have to slope and the one point of the line that we have to calculate.
The formula to the equation of the line is as follows,
$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$
Substituting $m=\dfrac{-1}{3},{{y}_{1}}=\dfrac{3}{n+1},{{x}_{1}}=\dfrac{n+2}{n+1}$ we get,
$\left( y-\dfrac{3}{n+1} \right)=\dfrac{-1}{3}\left( x-\dfrac{n+2}{n+1} \right)$
Solving this equation, we get,
$\begin{align}
  & 3\left( \dfrac{y\left( n+1 \right)-3}{n+1} \right)=-1\left( \dfrac{x\left( n+1 \right)-\left( n+2 \right)}{n+1} \right) \\
 & 3yn+3y-9=-1\left( xn+x-n-2 \right) \\
 & \left( 1+n \right)x+3\left( 1+n \right)y=n+11 \\
\end{align}$
Therefore, the equation of the line is $\left( 1+n \right)x+3\left( 1+n \right)y=n+11$ .

Note: In this problem, while using the section formula the order is very important as the ratio is not the same, therefore, it is very important of which number multiplies to which number. While calculating the slope can start from any point because as it is the ratio the sign gets neutralized.