Asked by Weddy
The diagonals of a rectangle pqrs intersect at 5,3 given that the equation for a line pq is 4y-9x=13 and that of line ps y-4x=5 what are coordinates of p and r ,what is the equation of line rq ,what is the equation of perpendicular line drawn to meet pr at 5,3 showing solution
Answers
Answered by
oobleck
The lines given for PQ and PS intersect at (-1,1) so that must mean that point P is at (-1,1)
If X = (5,3) then X-P = (6,2)
so R = X+(6,2) = (11,5)
PR has slope 1/3, so the perpendicular to PR has slope -3, making its equation
y-3 = -3(x-5)
Now, Q and S must lie on a circle of radius √40 with center at (5,3)
So the equation of the circle is (x-5)^2 + (y-3)^2 = 40
That circle intersects PQ and PS at Q and S, respectively.
The problem here is that the alleged lines PQ and PS do not intersect the circle anywhere near the points needed to make a rectangle.
If X = (5,3) then X-P = (6,2)
so R = X+(6,2) = (11,5)
PR has slope 1/3, so the perpendicular to PR has slope -3, making its equation
y-3 = -3(x-5)
Now, Q and S must lie on a circle of radius √40 with center at (5,3)
So the equation of the circle is (x-5)^2 + (y-3)^2 = 40
That circle intersects PQ and PS at Q and S, respectively.
The problem here is that the alleged lines PQ and PS do not intersect the circle anywhere near the points needed to make a rectangle.
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