# A line passes through (8 ,2 ) and (6 ,7 ). A second line passes through (3 ,4 ). What is one other point that the second line may pass through if it is parallel to the first line?

Apr 15, 2018

Any point on the line $y = \frac{23 - 5 x}{2}$ is acceptable. For example $\left(1 , 9\right)$ is a correct answer.

#### Explanation:

To determine the slope, $m$ of a line we can use the formula

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

where $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ are any two distinct points on the line. Here, the two points for the first line are $\left(8 , 2\right)$ and $\left(6 , 7\right)$ so the slope of the line is

$m = \frac{7 - 2}{6 - 8} = - \frac{5}{2}$

The second line must also abide by this slope definition, yet we only know one point on the line. Let's call the coordinates of the second point $\left(x , y\right)$. The slope formula for the second line then gives

$m = - \frac{5}{2} = \frac{y - 4}{x - 3}$

Let's solve this equation for y by multiplying both sides of this equation by $\left(x - 3\right)$ and then adding 4 to both sides of this equation.

$y = - \frac{5}{2} \left(x - 3\right) + 4 = - \frac{5}{2} x + \frac{15}{2} + 4 = - \frac{5}{2} x + \frac{23}{2}$

So our line has a slope of -5/2 and a $y$-intercept of $\frac{23}{2}$.

The problem statement asks for ANY other point on this line. This means we can pick any $x$-value we want (except for 3) and then calculate the value of y. I'm going to choose $x = 1$. If $x = 1$, then

$y = - \frac{5}{2} \left(1\right) + \frac{23}{2} = \frac{18}{2} = 9$

So $\left(1 , 9\right)$ is a point on the second line.