Question

Let $A(h,k),B(1,1)$ and $C(2,1)$ be the vertices of a right angled triangle with $AC$ as its hypotenuse. If the area of the triangle is $1$, then the set of values which $k$ can take is given by

• A. $\{1,3\}$
• B. $\{0,2\}$
• C. $\{-1,3\}$
• D. $\{3,2\}$

Answer 1

Answer: (C)

$\{-1,3\}$

$\because A(h,k),B(1,1)$ and $C(2,1)$ are the vertices of a right angled triangle ABC
Now, $AB=\sqrt{{(1-h)}^2+{(1-k)}^2}$ or $BC=\sqrt{{(2-1)}^2+{(1-1)}^2}=1$
or $CA=\sqrt{{(h-2)}^2+{(k-1)}^2}$
Now, pythagorus theorem
$A{C}^2=A{B}^2+B{C}^{2}4+h^2-4h+k^2+1-2k=h^2+1-2h+k^2+1-2k+15-4h=3-2h\Rightarrow h=1$
Now, given that area of the triangle is 1,
Then, area $\left(△ABC\right)=\frac{1}{2}\times AB\times BC$
$1=\frac{1}{2}\times \sqrt{{(1-h)}^2+{(1-k)}^2}\times 1\Rightarrow 2=\sqrt{{(1-h)}^2+{(1-k)}^2}$ ...(2)
Putting $h=1$ from equation (1), we get $2=\sqrt{{(k-1)}^2}$
Squaring both the sides , we get
$4=k^2+1-2k$ or $k^2-2k-3=0$
or $(k-3)(k+1)=0$
So, $k=-1,3$
Thus the set of values of k is $\{-1,3\}$