Question

Find the equation of the straight line which divides the join of the points (2, 3) and (−5, 8) in the ratio 3 : 4 and is also perpendicular to it.

Answer 1

Let the required line divide the line joining the points $A\left(2,3\right)\mathrm{and}B\left(-5,8\right)$ at P (x₁, y₁).
Here, AP : PB = 3 : 4

$\therefore P\left(x_1,y_1\right)=\left(\frac{4\times 2-5\times 3}{3+4},\frac{4\times 3+3\times 8}{3+4}\right)=\left(-1,\frac{36}{7}\right)$

Now, slope of AB = $\frac{8-3}{-5-2}=-\frac{5}{7}$
Let m be the slope of the required line.
Since, the required line is perpendicular to the line joining the points $A\left(2,3\right)\mathrm{and}B\left(-5,8\right)$

$$\begin{gathered} \therefore m\times \mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{joining}\mathrm{the}\mathrm{points}A\left(2,3\right)\mathrm{and}B\left(-5,8\right)=-1 \\ \Rightarrow m\times \left(\frac{-5}{7}\right)=-1 \\ \Rightarrow m=\frac{7}{5} \end{gathered}$$

Substituting $m=\frac{7}{5},x_1=-1\mathrm{and}{y}_1=\frac{36}{7}$ in $y-y_1=m\left(x-x_1\right)$ we get,

$$\begin{gathered} y-\frac{36}{7}=\frac{7}{5}\left(x+1\right) \\ \Rightarrow 35y-180=49x+49 \\ \Rightarrow 49x-35y+229=0 \end{gathered}$$

Hence, the equation of the required line is $49x-35y+229=0$