The equation of the line passing through (1, 1) and parallel to the line 2x+3y-7=0
is [RPET 1996]
A) 2x+3y-5=0 B) 3x+2y-5=0 C) 3x-2y-7=0 D) 2x+3y+5=0
Answers
Answered by
8
2x+3y-7=0; Rearrange this in to y=mx+c form;
3y=-2x+7; y=(-2/3)x+7/3;
Slope m1=-2/3=m2; because parallel line m1=m2;
Equation of linepassing through (1,1) and has slope m2;
(y-1)=(-2/3)(x-1);
By solving this we get 2x+3y-5=0; Option A is correct
3y=-2x+7; y=(-2/3)x+7/3;
Slope m1=-2/3=m2; because parallel line m1=m2;
Equation of linepassing through (1,1) and has slope m2;
(y-1)=(-2/3)(x-1);
By solving this we get 2x+3y-5=0; Option A is correct
Similar questions