# 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

#### Answer:

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.