Question

If $A(5,3),B(11,-5)\&P(12,y)$ are the vertices of a right triangle, right angled at $P$, then the value of $y$ is equal to

• A. -2, 4
• B. -2, -4
• C. 2, -4
• D. 2, 4

Answer 1

The correct option is C 2, -4

$\text{Given, APB is right angled triangle.}\angle P={90}^{\circ }A{P}^2=(12-5{)}^2+(y-3{)}^2=49+y^2-6y+9=y^2-6y+58\text{units}P{B}^2=(12-11{)}^2+(y+5{)}^2=1+y^2+10y+25=y^2+10y+26\text{units}A{B}^2=(11-5{)}^2+(-5-3{)}^2=36+64=100\text{units}\therefore \text{By pythagoras theorem,}A{P}^2+P{B}^2=A{B}^2{y}^2-6y+58+y^2+10y+26=100\Rightarrow 2y^2+4y-16=0\Rightarrow y^2+2y-8=0\Rightarrow (y+4)(y-2)=0\Rightarrow y=-4\left(\text{or}\right)y=2$