Question

Let $A\left( {1,k} \right)$, $B\left( {1,1} \right)$ and $C\left( {2,1} \right)$ be the vertices of a right angled triangle with $AC$ as it’s hypotenuse. If the area of the triangle is $1$, then the set of value 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

Hint: We are to first identify the point at which the triangle is right angled. Then from the given points we can get the distance between two points in order to find the length of the base and height of the triangle, by using the formula,
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
So, after finding the height and base we can use the area of triangle, given,
Area of triangle, ${A_r} = \dfrac{1}{2} \times base \times height$
Now, from the given conditions, we can find the values of k.

Complete step by step answer:
The given points are, $A\left( {1,k} \right)$, $B\left( {1,1} \right)$, $C\left( {2,1} \right)$.
Also, the area of the triangle is given to be $1$ sq. units.
Now, given $AC$ is the hypotenuse of the triangle, so, that means the triangle is right angled at $B$.
So, the base and height of the triangle are $AB$ and $BC$.
Now, by using the distance formula, we get,
$AB = \sqrt {{{\left( {1 - 1} \right)}^2} + {{\left( {1 - k} \right)}^2}} $
$ \Rightarrow AB = \sqrt {{{\left( 0 \right)}^2} + {{\left( {1 - k} \right)}^2}} $
$ \Rightarrow AB = \sqrt {{{\left( {1 - k} \right)}^2}} $
$ \Rightarrow AB = \pm (1 - k)$
And, $BC = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {1 - 1} \right)}^2}} $
$ \Rightarrow BC = \sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2}} $
$ \Rightarrow BC = 1$
So, we have the height and base of the triangle.
So, using the formula of area of a triangle, we get,
Area of triangle, ${A_r} = \dfrac{1}{2} \times base \times height$
$ \Rightarrow {A_r} = \dfrac{1}{2} \times \left( { \pm \left( {1 - k} \right)} \right) \times 1$
$ \Rightarrow {A_r} = \dfrac{{ \pm (1 - k)}}{2}$
Now, given, the area of the triangle is $1$.
Now, substituting the value in the above equation, we get,
$\dfrac{{ \pm \left( {1 - k} \right)}}{2} = 1$
$ \Rightarrow 2 = \pm \left( {1 - k} \right) - - - - \left( 1 \right)$
Now, for $ + \left( {1 - k} \right)$, we get,
$ \Rightarrow 2 = 1 - k$
Adding $ - 1$ on both sides, we get,
$ \Rightarrow 2 - 1 = - k$
$ \Rightarrow 1 = - k$
Multiplying both sides with $ - 1$, we get,
$ \Rightarrow k = - 1$
Now, for $ - \left( {1 - k} \right)$ in equation $\left( 1 \right)$, we get,
$ \Rightarrow 2 = - \left( {1 - k} \right)$
$ \Rightarrow 2 = - 1 + k$
Now, adding $1$ on both sides, we get,
$ \Rightarrow 2 + 1 = k$
$ \Rightarrow k = 3$
Therefore, the set of values which $k$ can take is $\left\{ { - 1,3} \right\}$, the correct option is option (C).

Note:
We can also do the given problem by another method. We can find the area of the triangle only if the points of the triangle are given. Then we can use the formula,
Area of triangle, ${A_r} = \dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$
But, in the given problem if this method is used, the equation may become a bit complex, so we chose the simpler approach.