Question

Point $p$ divides the line segment joining $A\left(-1,3\right)$ and $B\left(9,8\right)$ in ratio $k:1$. If $p$ lies on the line $x-y+2=0$, then the value of $k$ is

Answer 1

Step 1. Determine the value of $x\mathrm{and}y$.

It is given that $p$ divides the line segment in ratio $k:1$ and the points $A\left(-1,3\right)$ and $B\left(9,8\right)$.Step 1.

Consider $\begin{array}{rcl}p& =& \left(x,y\right),A\left(-1,3\right)=\left(x_1,y_1\right),\mathrm{and}B\left(9,8\right)=\left(x_2,y_2\right)\end{array}$.

Apply the section formula: $\mathbf{x}\mathbf{=}\frac{{\mathbf{mx}}_{\mathbf{2}}\mathbf{+}{\mathbf{nx}}_{\mathbf{1}}}{\mathbf{m}\mathbf{+}\mathbf{n}}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\frac{{\mathbf{my}}_{\mathbf{2}}\mathbf{+}{\mathbf{ny}}_{\mathbf{1}}}{\mathbf{m}\mathbf{+}\mathbf{n}}$

Since $p$ divides the line segment in ratio $k:1$, $m+n=k+1$.

Substitute the value of $\left(x_1,y_1\right),\left(x_2,y_2\right)$, and $m+n$ into the formula.

$\begin{array}{rcl}x& =& \frac{k\left(9\right)+1\left(-1\right)}{k+1}\\ & \Rightarrow & x=\frac{9k-1}{k+1}\\ y& =& \frac{k\left(8\right)+1\left(3\right)}{k+1}\\ & \Rightarrow & y=\frac{8k+3}{k+1}\end{array}$

Hence, the value is $\begin{array}{rcl}x& =& \frac{9k-1}{k+1}\mathrm{and}y=\frac{8k+3}{k+1}\end{array}$.

Step 2. Determine the value of $k$.

Substitute the value of $x$ and $y$ into $x-y+2=0$.

$\begin{array}{rcl}\frac{9k-1}{k+1}-\frac{8k+3}{k+1}+2& =& 0\\ & \Rightarrow & 9k-1-\left(8k+3\right)+2\left(k+1\right)=0\\ & \Rightarrow & k-4+2k+2=0\\ & \Rightarrow & 3k=2\\ & \Rightarrow & k=\frac{2}{3}\end{array}$

Hence, the value of $k$ is $\frac{2}{3}$.