Draw the graph of the equation x/3 + y/4 = 1. Also, find the area of the triangle formed by 3/4 the line and the coordinates axes.
We are given. x/3 + y/4 = 1
4x + 3y =12
$y=\frac{12-4 x}{3}$
Now, substituting x = 0 in
$\mathrm{y}=\frac{12-4 \mathrm{x}}{3}$
we get y = 4
Substituting x = 3 in
$y=\frac{12-4 x}{3}$
we get y = 0
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
The region bounded by the graph is ABC which forms a triangle.
AC at y axis is the base of triangle having AC = 4 units on y axis.
BC at x axis is the height of triangle having BC = 3 units on x axis.
Therefore, Area of triangle ABC, say A is given by A = (Base × Height)/2
A= (AC × BC)/2
A = (4 × 3)/2
A = 6 sq. units