If A (5, 3), B (11, −5) and P (12, y) are the vertices of a right triangle right angled at P, then y=
(a) −2, 4
(b) −2, −4
(c) 2, −4
(d) 2, 4
Disclaimer: option (b) and (c) are given to be same in the book. So, we are considering option (b) as −2, −4
instead of −2, 4.
We have a right angled triangle $\triangle \mathrm{APC}$ whose co-ordinates are $\mathrm{A}(5,3) ; \mathrm{B}(11,-5)$;
$\mathrm{P}(12, y)$. So clearly the triangle is, right angled at A. So,
$\mathrm{AP}^{2}=(12-5)^{2}+(y-3)^{2}$
$\mathrm{BP}^{2}=(12-11)^{2}+(y+5)^{2}$
$\mathrm{AB}^{2}=(11-5)^{2}+(-5-3)^{2}$
Now apply Pythagoras theorem to get,
$\mathrm{AB}^{2}=\mathrm{AP}^{2}+\mathrm{BP}^{2}$
So,
$100=50+2 y^{2}+34+4 y$
On further simplification we get the quadratic equation as,
$2 y^{2}+4 y-16=0$
$y^{2}+2 y-8=0$
Now solve this equation using factorization method to get,
$(y+4)(y-2)=0$
Therefore, $y=2,-4$
So the answer is (c)
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