Question

A triangle of area 24 Sq. Units is formed by a straight line and the co-ordinate axes in the first quadrant. Find the equation of that straight line if it passes through (3,4)

Answer 1

Let the equation of required line in intercept form be $\frac{x}{a}+\frac{y}{b}=1$

Since the line apsses through point $(3,4)$ :
 $\frac{3}{a}+\frac{4}{b}=1$

Also area of triangle formed with axes must be $24$ :
$\frac{1}{2}ab=24\implies b=\frac{48}{a}$

Plug this value in earlier equation  and get
 $\frac{3}{a}+\frac{4}{48/a}=1$
 $\frac{3}{a}+\frac{a}{12}=1$
 $a^2+36=12a$
 $a^2-12a+36=0$
 $(a-6)^2=0$
 $a=6$
 $\therefore\,b=\frac{48}{a}=\frac{48}{6}=8$

Thus the equation of straight line is $\frac{x}{6}+\frac{y}{8}=1$

Answer 2

Answer:

Answer is x/6+y/8=1 is the answer