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 $$A(2,3)\text{ and }B(-5,8)$$ at P (x₁, y₁).
Here, AP : PB = 3 : 4

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

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 $$A(2,3)\text{ and }B(-5,8)$$

$$\therefore m\times \text{Slope of the line joining the points }A(2,3)\text{ and }B(-5,8)=-1$$

$$\Rightarrow m\times \left(\frac{-5}{7}\right)=-1$$

$$\Rightarrow m=\frac{7}{5}$$

Substituting

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

$$y-\frac{36}{7}=\frac{7}{5}(x+1)$$

$$\Rightarrow 35y-180=49x+49$$

$$\Rightarrow 49x-35y+229=0$$

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