A(1,6)and B(3,5) find the equation of the locus of point P such that segment AB subtends right angle at P (APB)=90°
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The locus of point P , x² + y² -9x -6y +23 = 0
Let the point P have coordinates (x,y) .
Given the points A(1,6)and B(3,5) .
The angle APB is a right angle and so the line segments PA and PB are perpendicular to each other .
Slope of PB = -1/ slope of PA [ Both lines are perpendicular to each other]
Slope of the line PB = (y-5)/(x-3)
Slope of the line PA = (y-1)/(x-6)
=> (y-5)/(x-3) = -1/((y-1)/(x-6))
=> (y-5)/(x-3) = -(x-6)/(y-1)
=> y² -6y +5 = -(x² -9x +18)
=> y² -6y +5 = -x² +9x -18
=> x² + y² -9x -6y +23 = 0
The locus of the point P, x² + y² -9x -6y +23 = 0
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