find the distance of the line 4x+7y+5=0 from the point (1,2)measured parallel to 4x-2y+5=0
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consider the attached diagram,
to find AP we first find A
which is the intersection of the line 4x+7y+5=0 and line parallel to 4x-2y+5=0 passing through P(1,2)
line parallel to 4x-2y+5=0 passing through P(1,2) is of the form 4x-2y+k=0
so k=0 so the line is 2x-y=0
to find A (x₁, y₁) solving 2x₁=y₁ & 4x₁+7y₁+5=0
(-5/18,-5/9)
hence AP = 2.86
to find AP we first find A
which is the intersection of the line 4x+7y+5=0 and line parallel to 4x-2y+5=0 passing through P(1,2)
line parallel to 4x-2y+5=0 passing through P(1,2) is of the form 4x-2y+k=0
so k=0 so the line is 2x-y=0
to find A (x₁, y₁) solving 2x₁=y₁ & 4x₁+7y₁+5=0
(-5/18,-5/9)
hence AP = 2.86
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