A line passes through (4 ,7 )(4,7) and (2 ,8 )(2,8). A second line passes through (1 ,5 )(1,5). What is one other point that the second line may pass through if it is parallel to the first line?

1 Answer
Jun 20, 2016

The point (0, 11/2)(0,112).

Explanation:

The equation of a line is y=mx+qy=mx+q
The line that passes from (4,7)(4,7) and (2,8)(2,8) can be obtained substituting the coordinates as xx and yy and doing the system

7=4m+q7=4m+q
8=2m+q8=2m+q

We can subtract the second from the first to remove qq obtaining

7-8=4m-2m+\cancel(q)-\cancel(q)

-1=2m and m=-1/2.

We can also calculate q, even if it is not relevant for our problem, using one of the two equations

7=-1/2*4+q

7=-2+q

q=9

The equation of the first line is then

y=-1/2x+9

The second line is parallel, it means that the slope is the same -1/2.
So the equation of the parallel is

y=-1/2x+r where r is the intercept of the second line. To find it we know that the line passes from (1,5), then

5=-1/2*1+r

5+1/2=r

11/2=r

The equation of the parallel is

y=-1/2x+11/2.

To find another point in this line it is enough to give a value to x and calculate the y.
For example x=0, y=-1/2*0+11/2=11/2, then the point (0, 11/2) is another point of the parallel.