# How do you write an equation of a line that has an x-intercept of 3 and a y-intercept of 4?

Apr 11, 2015

One quick way to do this is to use the fact that $3 \setminus \cdot 4 = 12$ to say an equation for this line is $4 x + 3 y = 12$ (since, when $y = 0$, it follows that $x = 3$ and when $x = 0$ it follows that $y = 4$).

You could also find the slope of the line containing the points $\left(x , y\right) = \left(3 , 0\right)$ and $\left(x , y\right) = \left(0 , 4\right)$ as $\setminus \frac{\setminus \Delta y}{\setminus \Delta x} = \setminus \frac{4 - 0}{0 - 3} = - \frac{4}{3}$ so that the equation has the form $y = - \frac{4}{3} x + b$, where $b$ is the $y$-intercept so the equation becomes $y = - \frac{4}{3} x + 4$. This is equivalent to the last answer because you can multiply everything by 3 and rearrange to get $4 x + 3 y = 12$.

Apr 11, 2015

The x-intercept is the value of x when y=0. The y-intercept is the value when x=0.

The x-intercept of 3 corresponds to y=0, and the point on the line is (3,0).
The y-intercept of 4 corresponds to x=0, and the point on the line is (0,4).

Since we have two points on the line, we can find the slope using the formula $m = \frac{\Delta y}{\Delta x} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

For the line running through the points $\left(3 , 0\right)$ and $\left(0 , 4\right)$, ${y}_{2} = 4 \mathmr{and} {y}_{1} = 0$, and ${x}_{2} = 0 \mathmr{and} {x}_{1} = 3$.

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{0 - 4}{0 - 3} = - \frac{4}{3}$

Now that we have the slope, we can write the equation for the line in point-slope form.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$ =

$y - 0 = - \frac{4}{3} \left(x - 3\right)$