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)y2−y1x2−x1. Hence slope of line joining (8,5)(8,5) and (6,4)(6,4) is
(4-5)/(6-8)=(-1)/(-2)=1/24−56−8=−1−2=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)(y−y1)=m(x−x1). 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)(y−5)=12(x−3) or 2y-10=x-32y−10=x−3 i.e. x-2y+7=0x−2y+7=0
Hence, any point satisfying the condition x-2y+7=0x−2y+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)