Question

Let $A\left(1,k\right)$, $B\left(1,1\right)$, $C\left(2,1\right)$ 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. $\left\{1,3\right\}$
• B. $\left\{0,2\right\}$
• C. $\left\{-1,3\right\}$
• D. $\left\{-3-2\right\}$

Answer 1

The correct option is C

$\left\{-1,3\right\}$

Explanation for the correct answer:

Finding the set values for $k$:

Given $A\left(1,k\right)$, $B\left(1,1\right)$, $C\left(2,1\right)$ be the vertices of a right-angled triangle.

Using the distance formula to find $BC$ and $AB$

$\text{Distance formula}=\sqrt{{\left(x_2-x_1\right)}^2-{\left(y_2-y_1\right)}^2}$

$$\begin{gathered} BC=\sqrt{{\left(2-1\right)}^2-{\left(1-1\right)}^2}=1 \\ AB=\sqrt{{\left(1-1\right)}^2-{\left(k-1\right)}^2}=\sqrt{{\left(k-1\right)}^2} \end{gathered}$$

Given that the area of the triangle is $1$

$$\begin{gathered} \frac{1}{2}\times b\times h=1 \\ \frac{1}{2}\times BC\times AB=1 \\ \frac{1}{2}\times 1\times \sqrt{{(k-1)}^2}=1 \\ \Rightarrow \sqrt{{\left(k-1\right)}^2}=2 \end{gathered}$$

Squaring both sides

$$\begin{gathered} {\left(k-1\right)}^2=4 \\ \Rightarrow k^2+1-2k=4 \\ \Rightarrow k^2-2k-3=0 \\ \Rightarrow k^2-3k+k-3=0 \\ \Rightarrow k(k-3)+1(k-3)=0 \\ \Rightarrow (k-3)(k+1)=0 \end{gathered}$$

Therefore,$k=\left\{-1,3\right\}$

Hence, option (C) is the correct answer.