How do you write an equation of the line that passes through (-3,4) and (1,0)?

2 Answers

$y = - x + 1$

Explanation:

The line equation is of the form
$y = a x + b$

As the line passes through this two points
$\left({x}_{0} , {y}_{0}\right) = \left(- 3 , 4\right)$
$\left({x}_{1} , {y}_{1}\right) = \left(1 , 0\right)$

they both obey the equation
${y}_{1} = a {x}_{1} + b \implies 0 = a + b \implies a = - b$
${y}_{0} = a {x}_{0} + b \implies 4 = - 3 a + b$

Therefore:
$4 = - 3 a - a \implies 4 = - 4 a \implies a = - 1 \mathmr{and} b = 1$

So, the equation of the line that passes through those points is
$y = - x + 1$

Sep 2, 2016

We will write the equation in slope-intercept form, $y = m x + c$. I have calculated below that the slope of the line is -1 and its y-intercept is 1.

The equation of the line can be written: $y = - x + 1$

(we don't write the number '1' in front of the $x$, since $1 x = x$)

Explanation:

First step: find the slope (gradient) of the line:

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

It doesn't matter which of the two given points we decided is 'Point 1' $\left({x}_{1} , {y}_{1}\right)$, but let's choose the point $\left(1 , 0\right)$, so ${x}_{1} = 1$ and ${y}_{1} = 0$.

Similarly for the other point, $\left(- 3 , 4\right)$, so ${x}_{2} = - 3$ and ${y}_{2} = 4$.

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

Now that we know the slope, we can use it and the value of one of the points we were given to find the y-intercept of the line. This is the point where the line crosses the y-axis. The y-axis is the line $x = 0$, so if we substitute $m = - 1$ and, for example, ${x}_{1} = 1$ and ${y}_{1} = 0$, into the equation:

$y = m x + c$

$0 = - 1 \left(1\right) + c$

Rearranging, $c = 1$.