Question

The area of the triangle bounded by the straight line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$,$\left(a,b,c\ne 0\right)$ and the coordinate axes is equal to:

• A. $\frac{1}{2}\left|\frac{{\mathrm{c}}^2}{\mathrm{ab}}\right|$
• B. $\frac{1}{2}\left|\frac{{\mathrm{a}}^2}{\mathrm{ab}}\right|$
• C. $\frac{1}{2}\left|\frac{{\mathrm{b}}^2}{\mathrm{ab}}\right|$
• D. $1$

Answer 1

The correct option is A

$\frac{1}{2}\left|\frac{{\mathrm{c}}^2}{\mathrm{ab}}\right|$

Explanation for correct option :

Calculating the area of the triangle :

Given,

Equation of straight line is$\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$

And coordinate axes are, $\mathrm{x}=0;\mathrm{y}=0$

when, $x=0,y=-\frac{c}{b}$

when, $y=0,x=-\frac{c}{a}$

Therefore, $x$- intercept $=-\frac{c}{a}$, $y$ - intercept $=-\frac{c}{b}$

We know that,

$\mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

Therefore, the area of the triangle bounded by the straight line and coordinate axes will be:

$=\frac{1}{2}\times$x intercept $\times$y intercept

$$\begin{gathered} =\frac{1}{2}\times \frac{\mathrm{c}}{\mathrm{a}}\times \frac{\mathrm{c}}{\mathrm{b}} \\ =\frac{{\mathrm{c}}^2}{2\mathrm{ab}} \end{gathered}$$

Hence, option (A) is the correct answer.