Question

Let \[A\left( {1,{\text{ }}k} \right)\], \[B\left( {1,1} \right)\]and \[C\left( {2,1} \right)\] be the vertices of a right-angled triangle with \[AC{\text{ }}\]as its hypotenuse. If the area of the triangle is 1, then the set of values that k can take is given by
(A) \[\left\{ {{\mathbf{1}},{\mathbf{3}}} \right\}\]
(B) \[\left\{ {{\mathbf{0}},{\mathbf{2}}} \right\}\]
(C) \[\left\{ {-{\mathbf{1}},{\mathbf{3}}} \right\}\]
(D) \[\left\{ {-{\mathbf{3}},-{\mathbf{2}}} \right\}\]

Answer 1

Hint: We are given the three vertices of a right-angled triangle. Using the distance formula and Pythagoras’ theorem the value of h can be found. By substituting the value of h and other known values in the area of the triangle formula we can obtain the value of k.

Formula Used:
The distance between the two points \[\left( {{x_1},{y_1}} \right)\]and \[\left( {{x_2},{y_2}} \right)\] is given by the distance formula.
\[d=\sqrt{(\mathrm{(x_2-x_1)}^2)+(\mathrm{(y_2-y_1)}^2)}\]

The Pythagoras theorem equation is expressed as, \[A{C^2} = A{B^2} + B{C^2}\] where \[AC\]= hypotenuse of the right triangle and\[AB\ ]and \[BC\]are the other two sides.

Complete step by step Solution:
\[\because A(h,k)\], \[B(1,1)\;\]and \[C(2,1)\;\]are the vertices of a right-angled triangle ABC we have,

Here \[AB = \sqrt {{{(1 - h)}^2} + {{(1 - k)}^2}} \] , \[BC = \sqrt {{{(2 - 1)}^2} + {{(1 - 1)}^2}} = 1\] and \[CA = \sqrt {{{(h - 2)}^2} + {{(k - 1)}^2}} \]
We know that Pythagoras' theorem is
\[A{C^2} = A{B^2} + B{C^2}\]
\[ \Rightarrow 4 + {h^2} - 4h + {k^2} + 1 - 2k = {h^2} + 1 - 2h + {k^2} + 1 - 2k + 1\]
\[ \Rightarrow \,5 - 4h = 3 - 2h\]
\[ \Rightarrow \,h = 1\]
We are given that area of the triangle is 1,
Hence \[area(\vartriangle ABC) = \dfrac{1}{2} \times AB \times BC\]
\[ \Rightarrow 1 = \dfrac{1}{2} \times \sqrt {{{(1 - h)}^2} + {{(1 - k)}^2}} \times 1\]
\[ \Rightarrow 2 = \sqrt {{{(1 - h)}^2} + {{(1 - k)}^2}} \]
Putting h=1 in the above equation we get,
\[ \Rightarrow 2 = \sqrt {{{(1 - k)}^2}} \]
Squaring on both sides we get,
\[ \Rightarrow 4 = {(1 - k)^2}\]
\[ \Rightarrow 4 = 1 - 2k + {k^2}\]
\[ \Rightarrow {k^2} - 2k - 3 = 0\]
On factoring by splitting the middle term we get,
\[ \Rightarrow {k^2} - 3k + k - 3 = 0\]
\[ \Rightarrow (k - 3)(k + 1) = 0\]
\[ \Rightarrow k = - 1,3\]

Hence, the correct option is C.

Note: In order to solve the given question, one must know to find the distance between two points.
The distance between the two points \[\left( {{x_1},{y_1}} \right)\]and \[\left( {{x_2},{y_2}} \right)\] is given by the distance formula.