Let A(h,k), B(1,1) and C(2,1) be the vertices of a right angled triangle with AC as its hy…

Question

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 square unit, then the set of values which $k$ can take is given by:

Accepted Answer

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

$$ = {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$$

$$ \Rightarrow  \pm \left( {k - 1} \right) = 2 \Rightarrow k =  - 1,3$$

Solutions · 1