Question

A line has slope $-3/4$, positive y-intercept and forms a triangle of area $24sq.units$ with coordinate axes. Then, the equation of the line is

• A. $3x+4y+24=0$
• B. $3x+4y-24=0$
• C. $3x+4y-25=0$
• D. $3x+4y+25=0$

Answer 1

The correct option is A $3x+4y+24=0$

The equation of line is $y=mx+c$ and the slope is given as $-\frac{3}{4}$, so,

$y=-\frac{3}{4}x+c$

The area of the triangle is given as,

$A=24$

$\frac{1}{2}\left(base\right)\left(height\right)=24$

$\frac{1}{2}\left(x-intercept\right)\left(c\right)=24$

The $x-intercept$ will be,

$0=-\frac{3}{4}x+c$

$x=\frac{4}{3}c$

Then,

$\frac{1}{2}\left(\frac{4}{3}c\right)\left(c\right)=24$

$\frac{2}{3}{c}^2=24$

$c^2=36$

$c=\pm 6$

Therefore, the equation of line is,

$y=-\frac{3}{4}x+6$

$4y=-3x+24$

$3x+4y-24=0$

And,

$y=-\frac{3}{4}x-6$

$4y=-3x-24$

$3x+4y+24=0$