Question

Draw the graph of the equation $\frac{x}{4}+\frac{y}{3}=1$. Also, find the area of the triangle formed by the line and the coordinate axes.

Answer 1

We are given, the equation

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

For graph, let $x=0$ we get $y=34$

at $y=0$ we get $x=4$

So, the prints an equation(s) are

$(0,3)$ & $(4,0)$

$X-intercept$ will be $OA=\sqrt{(4-0{)}^2+(0-0{)}^2}$

$OA=4$

$Y-intercept$ will be $OB=\sqrt{(0-0{)}^2+(0-3{)}^2}$

$OA=3$

So, the Area of $OAB$ is

$△OAB=\frac{1}{2}\times OA\times OB$

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

$△OAB=6(unit{)}^2$