Asked by Shirley
How do you find the distance from the point to the line given the equation and the coordinates?
4x-2y-6=0 (1,1)
4x-2y-6=0 (1,1)
Answers
Answered by
bobpursley
put the line in slope intercept form
y=mx+b
then,you know the slope of the perpendicular line is the negative reciprocal of m (new slope of perpendicular line=-1/m
So for the perpendicular line...
y=-1/m x + b
put in the point 1,1
1=-/m + b you know m, solve for b.
y=mx+b
then,you know the slope of the perpendicular line is the negative reciprocal of m (new slope of perpendicular line=-1/m
So for the perpendicular line...
y=-1/m x + b
put in the point 1,1
1=-/m + b you know m, solve for b.
Answered by
Reiny
there is a little formula for this question:
for Ax + By + C = 0
the distance from a point (p,q) to that line is
│Ap + Bq + C│/√(a^2+b^2)
so distance = │4-2-6│/√(16+4)
= 4/√20
= 4√20 /20
= 4√5/5
for Ax + By + C = 0
the distance from a point (p,q) to that line is
│Ap + Bq + C│/√(a^2+b^2)
so distance = │4-2-6│/√(16+4)
= 4/√20
= 4√20 /20
= 4√5/5
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