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Question
If A (5, 3), B (11, −5) and P (12, y) are the vertices of a right triangle right angled at P, then y=
Options
−2, 4
−2, −4
2, −4
2, 4
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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 ΔAPC whose co-ordinates are A (5, 3); B (11,−5);
P(12 , y) . So clearly the triangle is, right angled at A. So,
`AP^2 = (12 - 5)^2 + (y -3)^2`
`BP^2 = (12-11)^2 + (y + 5)^2`
`AB^2 = ( 11 - 5)^2 + (-5 -3)^2`
Now apply Pythagoras theorem to get,
`AB^2 = AP^2 +BP^2`
So,
`100 = 50 + 2y^2 + 34 + 4y`
On further simplification we get the quadratic equation as,
`2y^2 + 4y - 16 = 0`
`y^2 + 2y - 8 = 0`
Now solve this equation using factorization method to get,
(y + 4 )(y-2 ) = 0
Therefore, y = 2 , - 4
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