A line passes through (8 ,5 )(8,5) and (6 ,4 )(6,4). A second line passes through (3 ,5 )(3,5). What is one other point that the second line may pass through if it is parallel to the first line?

1 Answer
Sep 22, 2016

Any point other than (7,6)(7,6) and satisfying the condition x-2y+7=0x2y+7=0 could be such a point. Examples are (-5,1)(5,1), (7,7)(7,7) and (-1,3)(1,3).

Explanation:

A line passing through (x_1, y_1)(x1,y1) and (x_2, y_2)(x2,y2) has a slope of (y_2-y_1)/(x_2-x_1)y2y1x2x1. Hence slope of line joining (8,5)(8,5) and (6,4)(6,4) is

(4-5)/(6-8)=(-1)/(-2)=1/24568=12=12

As the second line passing through (3,5)(3,5) is parallel to the above, its slope too is 1/212.

Now, equation of a line passing through (x_1,y_1)(x1,y1) and having a slope of mm is (y-y_1)=m(x-x_1)(yy1)=m(xx1). Hence, the equation of line passing through (3,5)(3,5) and having a slope of 1/212 is

(y-5)=1/2(x-3)(y5)=12(x3) or 2y-10=x-32y10=x3 i.e. x-2y+7=0x2y+7=0

Hence, any point satisfying the condition x-2y+7=0x2y+7=0 (other than (3,5)(3,5) will satisfy this.

Let x=-5x=5 then y=1y=1, hence point could be (-5,1)(5,1) or

let x=7x=7 then y=7y=7, hence other point could be (7,7)(7,7) or

let x=-1x=1 then y=3y=3, hence point could be (-1,3)(1,3)