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.
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
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