Question

The area of the triangle formed by the coordinate axes and a line is $6$ square units and the length of its hypotenuse is $5$ units. Find the equation of the line.

Answer 1

Let X'OX and YOY' be the coordinate axes, and let AB be the part of the given line intercepted by the axes.

Let $OA=a$ and $OB=b.$

Then, Area of triangle $AOB=\frac{1}{2}\times base\times height=\frac{1}{2}ab=6$ (Given)

$\Rightarrow ab=12$ .....(i)

and, $a^2+b^2=25$ .....(ii) [AB $=$ hypotenuse $=25$, given]

Putting $b=\frac{12}{a}$ in (ii), we get

$a^2+\frac{144}{a^2}=25$

$\Rightarrow a^4-25a^2+144=0$

$\Rightarrow p^2-25p+144=0$ [Assuming $a^2=p$]

$\Rightarrow p^2-16p-9p+144=0$

$\Rightarrow p(p-16)-9(p-16)=0$

$\Rightarrow (p-16)(p-9)=0$

Putting $p=a^2$, we get

$\Rightarrow (a^2-16)(a^2-9)=0$

$\Rightarrow a^2=16\text{or}{a}^2=9$

$\Rightarrow a=4\text{or}a=3[\because \text{length is always positive}]$

Now, $(a=4\Rightarrow b=3)$ and $(a=3\Rightarrow b=4)$.

Hence, the required equation of the line is

$\frac{x}{4}+\frac{y}{3}=1\text{or,}\frac{x}{3}+\frac{y}{4}=1$

i.e.,$3x+4y-12=0\text{or}4x+3y-12=0$