Question

Determine the equation of the line in the given graph.

Answer 1

Recall that the general equation of a straight line is $y=ax+b$, where $a,b$ are real constants. The given graph shows that $a\ne 0$. (If $a=0$, then $y=b$ represents a straight line parallel to $y$-axis).

If $x=0$, then $y=b$. Thus $(0,b)$ is a point on the straight line.
Similarly, taking $y=0$, you get $ax+b=0$ or $x=-\frac{b}{a}$. Thus $(-\frac{b}{a},0)$ is also a point on the line.

Looking at the graph, we see that it cuts $y$-axis at $(0,4)$ and the x-axis at $(3,0)$.

So, we conclude that $(0,b)=(0,4)$ and $(-\frac{b}{a},0)=(3,0)$
Thus, $b=4$ and $-\frac{b}{a}=3$, which gives $a=-\frac{4}{3}$.

Hence, the equation of the given graph is $y=-\frac{4}{3}x+4$.

This may also be written as $3y+4x=12$