Find the equation of the line which passes through


Find the equation of the line which passes through the point (– 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.


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


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


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


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