Find the equation of the line which passes through

Let $A B$ be a line passing through a point $(-4,3)$ and meets $x$-axis at $A(a, 0)$ and $y$-axis at $B(0, b)$

Using the section formula for internal division, we have

$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right)$ ...........(i)

Here, $m_{1}=5, m_{2}=3$

$\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(\mathrm{a}, 0)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0, \mathrm{~b})$

Substituting the above values in the above formula, we get

$\Rightarrow \mathrm{x}=\frac{5(0)+3(\mathrm{a})}{5+3}, \mathrm{y}=\frac{5(\mathrm{~b})+3(0)}{5+3}$

$\Rightarrow-4=\frac{3 a}{8}, 3=\frac{5 b}{8}$

$\Rightarrow-32=3 a$ or $24=5 b$

$\Rightarrow \mathrm{a}=-\frac{32}{3} \mathrm{Or} \mathrm{b}=\frac{24}{5}$

We know that intercept form of the line is

$\frac{x}{a}+\frac{y}{b}=1$

Substituting the value of $a$ and $b$ in above equation, we get

$\frac{x}{-\frac{32}{3}}+\frac{y}{\frac{24}{5}}=1$

On simplification we get

$\Rightarrow-\frac{3 x}{32}+\frac{5 y}{24}=1$

Taking LCM

$\Rightarrow \frac{-72 x+160 y}{(32)(24)}=1$

On cross multiplication we get

⇒ -72x + 160y = 768

⇒ -36x + 80y = 384

⇒ 18x – 40y + 192 = 0

⇒ 9x – 20y + 96 = 0

Hence, the required equation is 9x – 20y + 96 = 0