Question
One side of the rectangle lies along the line 4x+7y+5=0 , two of its vertices are (-3,1)and (1,1). find
(i) the equations of the other three sides.
(ii) the angle between the two diagonals of the rectangle .
(iii) the area of the rectangle.
(i) the equations of the other three sides.
(ii) the angle between the two diagonals of the rectangle .
(iii) the area of the rectangle.
Answers
the given line has slope = -4/7
so, the perpendicular lines have slope 7/4
If we label the corners
A = (-3,1)
B
C = (1,1)
D
Then for side AB, (y-1)/(x+3) = 7/4
for side CD, (y-1)/(x-1) = 7/4
That will give you two lines; find where they intersect each other for B, and where CD intersects the given line for D.
Then find the lengths of the sides to get the diagonals and the area.
(We already know the diagonals have length 4 from AC.)
Yeah, it's some work, but this should get you started.
so, the perpendicular lines have slope 7/4
If we label the corners
A = (-3,1)
B
C = (1,1)
D
Then for side AB, (y-1)/(x+3) = 7/4
for side CD, (y-1)/(x-1) = 7/4
That will give you two lines; find where they intersect each other for B, and where CD intersects the given line for D.
Then find the lengths of the sides to get the diagonals and the area.
(We already know the diagonals have length 4 from AC.)
Yeah, it's some work, but this should get you started.
post it.
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