Question

Find the area of the triangle formed by the line $3x+4y=12$ and the co-ordinate axes.

Answer 1

Given the equation of the straight line is $3x+4y=12$.

This straight line can be written as in the intercept form as,

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

From the intercept form of the given line, it can be concluded that the line intersects the co-ordinate axes at the points $(4,0)$ and $(0,3)$.

Now we get a triangle whose base is of length $4\text{units}$ and length of perpendicular of length $3\text{units}$.

So, the area of the triangle will be,

$\Rightarrow \frac{1}{2}\times 3\times 4$

$\Rightarrow 6\text{sq. units}$