# Find the equation of the line passing through the point

Question:

Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes, whose sum is 9.

Solution:

To Find: The equation of the line passing through the point (2, 2) and cutting off intercepts on the axes, whose sum is 9.

Given : Let $\mathrm{a}$ and $\mathrm{b}$ be two intercepts of $\mathrm{x}$-axis and $\mathrm{y}$-axis respectively.

sum of the intercepts is $9, i . e, a+b=9$

$\Rightarrow a=9-b$ or $b=9-a$

Formula used:

The equation of a line is given by:

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

The given point (2, 2) passing through the line and satisfies the equation of the line.

$\frac{2}{a}+\frac{2}{9-a}=1$

$2(9-a)+2 a=9 a-a^{2}$

$18-2 a+2 a=9 a-a^{2}$

$a^{2}-9 a+18=0$

$a^{2}-6 a-3 a+18=0$

$a(a-6)-3(a-6)=0$

$(a-3)(a-6)=0$

$a=3, a=6$

when $a=3, b=6$ and $a=6, b=3$

case $1:$ when $a=3$ and $b=6$

Equation of the line : $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$

$\frac{x}{3}+\frac{y}{6}$

Hence, 2x + y = 6 is the required equation of the line.

case 2 : when a=6 and b=3

Equation of the line : $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$

$\frac{x}{6}+\frac{y}{3}=1$

Hence , x + 2y = 6 is the required equation of the line.

Therefore, $2 x+y=6$ is the required equation of the line when $a=3$ and $b=6$.And, $x+2 y$ $=6$ is the required equation of the line when $a=6$ and $b=3$.