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

1 Answer
David G.
Mar 13, 2018

The point (1,18)(1,18) 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=(y_2-y_1)/(x_2-x_1)=(7-2)/(5-6)=5/(-1)=-5m=y2−y1x2−x1=7−25−6=5−1=−5

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

y=mx+cy=mx+c

We rearrange this equation and plug in the xx and yy values of the point we have:

c=y-mx=8-(-5)(3)=8+15=23c=y−mx=8−(−5)(3)=8+15=23

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

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

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