Equation of a line passing through (1, -2) and perpendicular to the line 3x-5y+7=0
is [RPET 2003]
A) 5x+3y+1=0 B) 3x+5y+1=0 C) 5x-3y-1=0 D) 3x-5y+1=0
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3x-5y=7; Slope of this line is m1=3/5;
Slope of line which is perpendicular to the line 3x-5y+7=0; m2=-5/3;
Equation of line which passing through (1,-2) and has slope is -5/3;
(y+2)=-5/3(x-1);
By solving this we get 5x+3y+1=0; option a
Slope of line which is perpendicular to the line 3x-5y+7=0; m2=-5/3;
Equation of line which passing through (1,-2) and has slope is -5/3;
(y+2)=-5/3(x-1);
By solving this we get 5x+3y+1=0; option a
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