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?

1 Answer
May 23, 2017

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

Explanation:

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

gradient =(rise)/(run)

gradient =(y1-y2)/(x1-x2)

(6,2) = (x1,y1) and (2,1) = (x2,y2)

gradient =(2-1)/(6-2)

gradient =(1)/(4)
This is the first line: ( y = x/4 + 1/2 ) - the equation was found by substituting a coordinate into y = 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 1/4. We also have one point (3,2).

Substitute the point (3,2) and gradient of 1/4in to y = mx + c where m=gradient to find the value of c.

2 = 1/4(3) +c

c =5/4

Therefore the equation for the second line is y = x/4 + 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.