Question

Find the equation to the straight line which passes through the point (–4, 3) and is such that the portion of it between the axes is divided by the point in the ratio 5 : 3.

Answer 1

Let the equation of the required line be $\frac{x}{a}+\frac{y}{b}=1$. .....(1)
The line intersects the x - axis at (a, 0) and the y - axis at (0, b).
The point (–4, 3) divides the line between the two axes in the ratio 5 : 3.
Using section formula,
$$\begin{gathered} \left(\frac{5\times 0+3a}{5+3},\frac{5\times b+3\times 0}{5+3}\right)=\left(-4,3\right) \\ \Rightarrow \left(\frac{3a}{8},\frac{5b}{8}\right)=\left(-4,3\right) \\ \Rightarrow \frac{3a}{8}=-4 \\ \Rightarrow a=\frac{-32}{3} \end{gathered}$$
Also,
$$\begin{gathered} \frac{5b}{8}=3 \\ \Rightarrow b=\frac{24}{5} \end{gathered}$$
Thus, the two points obtained are $\left(\frac{-32}{3},0\right)\mathrm{and}\left(0,\frac{24}{5}\right)$
Putting these values in (1) we get the required equation of the line.
$$\begin{gathered} \frac{x}{\left({\displaystyle \frac{-32}{3}}\right)}+\frac{y}{\left({\displaystyle \frac{24}{5}}\right)}=1 \\ \Rightarrow \frac{3x}{-32}+\frac{5y}{24}=1 \\ \Rightarrow \frac{-9x+20y}{96}=1 \\ \Rightarrow 9x-20y+96=0 \end{gathered}$$