Question

A triangle of area $24sq$. units is formed by a straight line and the coordinate axes in the first quadrant. Find the equation of the straight line, if it passes through $(3,4)$.

Answer 1

Given that,

Area of triangle $=24squnits$

We know that,

Equation of line in intercept form

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

If it passes through $(3,4)$

Then,

$\frac{3}{a}+\frac{4}{b}=1$

$4a+5b=ab........\left(1\right)$

Given that

Area of triangle

$=\frac{1}{2}ab$

$\frac{1}{2}ab=24$

$ab=48$

$b=\frac{48}{a}.......\left(2\right)$

Put the equation (1) and we get,

$4a+3\left(\frac{48}{a}\right)=48$

$4a^2+3\times 48=48a$

$4a^2-48a+3\times 48=0$

$4(a^2-12a+3\times 12)=0$

$a^2-12a+36=0$

${(a-6)}^2=0$

$a=6$

Then, put the value of $a=6$ in equation (2)

$ab=48$

$6b=48$

$b=8$

Hence, the equation of line is

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

$\frac{x}{6}+\frac{y}{8}=1$

Hence, this is the answer.