If A (5, 3), B (11, −5) and P (12, y)

Solution:

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)