A line passes through (6 ,2 ) and (5 ,7 ). A second line passes through (3 ,8 ). What is one other point that the second line may pass through if it is parallel to the first line?

Mar 13, 2018

The point $\left(1 , 18\right)$ is one possible point on the second line.

(but there is literally an infinite number of possible points)

Explanation:

The lines are parallel so they will have the same gradient (slope). First we find the gradient of the first line:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{7 - 2}{5 - 6} = \frac{5}{- 1} = - 5$

To find the y-intercept we can write the equation of the line in gradient-intercept form:

$y = m x + c$

We rearrange this equation and plug in the $x$ and $y$ values of the point we have:

$c = y - m x = 8 - \left(- 5\right) \left(3\right) = 8 + 15 = 23$

The equation of the second line, then, is $y = - 5 x + 23$. Any set of $x$ and $y$ that makes that equation true is a point on the second line.

Let's use $x = 1$, then $y = - 5 \left(x\right) + 23 = 18$, so the point $\left(1 , 18\right)$ is on the second line.