JEE MAIN - Mathematics (2007 - No. 18)
Let A $$\left( {h,k} \right)$$, B$$\left( {1,1} \right)$$ and C $$(2, 1)$$ be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is $$1$$ square unit, then the set of values which $$'k'$$ can take is given by :
$$\left\{ { - 1,3} \right\}$$
$$\left\{ { - 3, - 2} \right\}$$
$$\left\{ { 1,3} \right\}$$
$$\left\{ {0,2} \right\}$$
Explanation
Given : The vertices of a right angled triangle $$A\left( {1,k} \right),$$
$$B\left( {1,1} \right)$$ and $$C\left( {2,1} \right)$$ and area of $$\Delta ABC = 1$$ square unit
We know that, area of night angled triangle
$$ = {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$$
$$ \Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3$$
$$B\left( {1,1} \right)$$ and $$C\left( {2,1} \right)$$ and area of $$\Delta ABC = 1$$ square unit
We know that, area of night angled triangle
$$ = {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$$
$$ \Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3$$
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