# How do you find the general form of the line with x intercept of -4, y intercept of 5?

Jul 2, 2016

Slope-intercept form $y = - \frac{5}{4} x - 5$

Point-slope form $y - 0 = - \frac{5}{4} \left(x + 4\right)$

Standard form $4 y + 5 x + 20 = 0$

#### Explanation:

To find the equation of the line for a line with an x intercept of $- 4$ and a y intercept of $5$, we can begin with the points of interception.

x intercept $= \left(- 4 , 0\right)$
y intercept $= \left(0 , 5\right)$

The formula for the slope of a line based upon two coordinate points is

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

For the coordinate points $\left(5 , 7\right) \mathmr{and} \left(9 , 14\right)$
${x}_{1} = - 4$
${x}_{2} = 0$
${y}_{1} = 0$
${y}_{2} = 5$

$m = \frac{5 - 0}{0 - \left(- 4\right)}$

$m = \frac{5}{-} 4$

The slope is $m = - \frac{5}{4}$

The point slope formula would be written as
$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$m = - \frac{5}{4}$
${x}_{1} = - 4$
${y}_{1} = 0$

$y - 0 = = \frac{5}{4} \left(x - \left(- 4\right)\right)$

$y = - \frac{5}{4} x - 5$

$\left(4\right) y = \left(4\right) \left(- \frac{5}{4} x\right) - 5 \left(4\right)$

$4 y = - 5 x - 20$

$4 y + 5 x + 20 = 0$