Find the equation of the line which passes through the point

To Find:The equation of the line that passes through the point (22, -6) and intercepts on the x-axis exceeds the intercept on the y-axis by 5.

Given : let x-intercept be a and $y$-intercept be $b$.

According to the question : $a=b+5$

Formula used: And the given point satisfies the equation of the line, so

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

$\frac{22}{b+5}+\frac{-6}{b}=1$

$22 b-6 b-30=b^{2}+5 b$

$11 b-30=b^{2}$

$b^{2}-11 b+30=0$

$b^{2}-6 b-5 b+30=0$

$b(b-6)-5(b-6)=0$

$(b-5)(b-6)=0$

The values are $\mathrm{b}=5, \mathrm{~b}=6$

When $\mathrm{b}=5$ then $\mathrm{a}=10$

and $b=6$ then $a=11$

case 1 : when $b=5$ and $a=10$

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

$\frac{x}{10}+\frac{y}{5}=1$

$\frac{x+2 y}{10}=1$

Hence, x + 2y = 10 is the required equation of the line.

case 2 : when b=6 and a=11

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

$\frac{x}{11}+\frac{y}{6}=1$

$\frac{6 x+11 y}{66}=1$

Hence, 6x + 11y = 66 is the required equation of the line.

Therefore, $x+2 y=10$ is the required equation of the line when $b=5$ and $a=10 .$ And $6 x+$ $11 \mathrm{y}=66$ is the required equation of the line when $\mathrm{b}=6$ and $\mathrm{a}=11$.