question_answer

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       AIEEE  Solved  Paper-2007

A)  {1, 3}             

B)         {0, 2}    

C)         {-1, 3}                  

D)         {-3,-2}

Correct Answer: C

Solution :

\[\because \]A(h, k), B(1,1) and C(2,1) are the vertices of a right angled\[\Delta ABC\]. Now,  \[AB=\sqrt{{{(1-h)}^{2}}+{{(1-k)}^{2}}}\] or    \[BC=\sqrt{{{(2-1)}^{2}}+{{(1-1)}^{2}}}\] or    \[CA=\sqrt{{{(h-2)}^{2}}+{{(k-1)}^{2}}}\] Now, by Pythagoras theorem, \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] \[4+{{h}^{2}}-4h+{{k}^{2}}+1-2k\] \[={{h}^{2}}+1-2h+{{k}^{2}}+1-2k+1\] \[\Rightarrow \]               \[5-4h=3-2h\Rightarrow h=1\]        ...(i) Now, given that area of the triangle is 1. Then, area \[(\Delta ABC)=\frac{1}{2}\times AB\times BC\] \[1=\frac{1}{2}\times \sqrt{{{(1-h)}^{2}}+{{(1-k)}^{2}}}\times 1\] \[2=\sqrt{{{(1-h)}^{2}}+{{(1-k)}^{2}}}\]     ...(ii) Putting\[h=1\]from Eq. (i), we get \[2=\sqrt{{{(k-1)}^{2}}}\] Squaring both 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}.