I want a line perpendicular to that line that passes through the point.
(There are quick formulas for this but I do not remember them)
the line in slope intercept form:
2 y = 2 x + 3
m = 2/2 = 1
so slope of out perpendicular is -1/1
= -1
so
y = - x + b
goes through (-4,-4)
-4 = 4 + b
so
y = -x -8
where does that hit our original line?
-2x +2(-x-8)) = 3
-4 x - 16 = 3
-4 x = 19
x = -19/4
y = -13/4
Find the point on the line –2x+2y–3=0 which is closest to the point (–4–4).
Please provide solution. ( , )
Thanks guys!
2 answers
quick way:
for a given point (p,q) , the shortest distance from that point to the line Ax + By + C = 0 is
|Ap + Bq + C|/)√(A^2+B^2)
so we have (p,q) = (-4,-4) and line
2x - 2y + 3 = 0
distance =|-8 +8 + 3|/(4+4)
= 3/√8
= 3/√8*√8/√8
= 3√8/8
= 6√2/8
= 3√2 /4
for a given point (p,q) , the shortest distance from that point to the line Ax + By + C = 0 is
|Ap + Bq + C|/)√(A^2+B^2)
so we have (p,q) = (-4,-4) and line
2x - 2y + 3 = 0
distance =|-8 +8 + 3|/(4+4)
= 3/√8
= 3/√8*√8/√8
= 3√8/8
= 6√2/8
= 3√2 /4
2 answers