A geometry problem by Siddharth Goel
Geometry
Level
2
A line cuts the x-axis at A(7,0) and the y-axis at B(0,--5). A variable line PQ is drawn perpendicular to AB cutting the x- axis at P and the y- axis at Q. If AQ and BP intersect at R, find the locus of R.
x^{2}+y^{2}-7x+5y=0
x^{3}+y^{3}-5x+7y=0
None of the above
x^{2}+y^{2}-5x+7y=0
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
A simple problem by geometry.. AP perpendicular to BQ. PQ perpendicular to AB Thus P is orthocenter of AQB. And BR passes through P. Thus R must be perpendicular to AQ. So .. Slope (BR) × Slope (AQ) = -1. Using this we get the answer.