# A line passes through (6 ,2 ) and (2 ,1 ). A second line passes through (3 ,2 ). What is one other point that the second line may pass through if it is parallel to the first line?

May 23, 2017

$\left(1 , \frac{3}{2}\right)$ or graph{y = x/4 + 1/2 [-10, 10, -5, 5]} any point on the line $y = \frac{x}{4} + \frac{5}{4}$

#### Explanation:

First, find the gradient of the first line from the two points given.

gradient $= \frac{r i s e}{r u n}$

gradient $= \frac{y 1 - y 2}{x 1 - x 2}$

$\left(6 , 2\right) = \left(x 1 , y 1\right)$ and $\left(2 , 1\right) = \left(x 2 , y 2\right)$

gradient $= \frac{2 - 1}{6 - 2}$

gradient $= \frac{1}{4}$
This is the first line: ($y = \frac{x}{4} + \frac{1}{2}$) - the equation was found by substituting a coordinate into $y = \frac{x}{4} + c$ to find c. (but you don't need to find the equation, this graph is just for explanation purposes)
graph{y = x/4 + 1/2 [-10, 10, -5, 5]}

The second line is parallel, meaning it has the same gradient, which is $\frac{1}{4}$. We also have one point $\left(3 , 2\right)$.

Substitute the point $\left(3 , 2\right)$ and gradient of $\frac{1}{4}$in to $y = m x + c$ where $m =$gradient to find the value of c.

$2 = \frac{1}{4} \left(3\right) + c$

$c = \frac{5}{4}$

Therefore the equation for the second line is $y = \frac{x}{4} + \frac{5}{4}$, as shown:
graph{y = x/4 + 5/4 [-10, 10, -5, 5]}

Now, just choose any random x-value and find the corresponding y-value to get a point on that line.