find the coordinates of the foot of the perpendicular from the origin on the straight line 3x+2y=13
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the line has a slope m
3x+2y=13;
2y=13-3x
y=13/2-3x/2
therefore m=-3/2
therefore slope of perpendicular(m1) =1/(-m)
m1=2/3
eq of perp=3y-2x=0;
Solving this we get
x=9/2; y=-3
3x+2y=13;
2y=13-3x
y=13/2-3x/2
therefore m=-3/2
therefore slope of perpendicular(m1) =1/(-m)
m1=2/3
eq of perp=3y-2x=0;
Solving this we get
x=9/2; y=-3
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