Question

If A (5,3), B (11, -5) and P (12, y) are the vertices of a right angled triangle at P, then y is….?
$A$ -2,4
$B$ -2, -4
$C$ 2, -4
$D$ 2,4

Answer 1

Hint: In order to solve this question, we must know the concept of right angle triangle as well as its properties. Along with this, we should also know the Pythagoras theorem to get the angles.

Complete Step-by-Step solution:

Here, According to the question
Given that A, P, B are the vertices of a right angled triangle at p where
A (5,3)
B (11, -5)
P (12, y)
Now in order to apply Pythagoras theorem, we must convert the vertices.
So we will use the Distance formula –
$ \Rightarrow $ D = $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Now we get
$ \Rightarrow $AP$^2$ = ${\left( {12 - 5} \right)^2}$ + ${\left( {y - 3} \right)^2}$
$ \Rightarrow $ BP$^2$ =${\left( {12 - 11} \right)^2}$+${\left( {y + 5} \right)^2}$
$ \Rightarrow $ AB$^2$ = ${\left( {11 - 5} \right)^2}$+${\left( { - 5 - 3} \right)^2}$
So we will now use Pythagoras theorem,
$ \Rightarrow $ AB$^2$= AP$^2$+ BP$^2$
$ \Rightarrow $36+64 = 49 + y$^2$ - 6y + 9 + 1 + y$^2$+ 10y + 25
$ \Rightarrow $100 = 50 + 2 y$^2$+ 34 + 4y
$ \Rightarrow $Simplifying further we get,
$ \Rightarrow $ y$^2$+ 2y – 8 = 0
Now we have to factorise the above calculated quadratic equation to find its root,
$ \Rightarrow $(y + 4) (y – 2) = 0
Thus y = -4, y = 2
$\therefore $ Option C is correct.

Note: To solve this type of question, we should take a simplified quadratic equation so that roots can be easily solved. Along with that one must know the applications of the Pythagoras theorem. This application is appropriate for this type of question as it gives the distance between two points. Hence we will get the desired result.