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
A
$$\left\{ { - 1,3} \right\}$$
B
$$\left\{ { - 3, - 2} \right\}$$
C
$$\left\{ { 1,3} \right\}$$
D
$$\left\{ {0,2} \right\}$$
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
$$\therefore$$ Equation of line is $${x \over 6} + {y \over 8} = 1$$
or $$4x + 3y = 24$$
4
AIEEE 2005
MCQ (Single Correct Answer)
If a vertex of a triangle is $$(1, 1)$$ and the mid points of two sides through this vertex are $$(-1, 2)$$ and $$(3, 2)$$ then the centroid of the triangle is
A
$$\left( { - 1,{7 \over 3}} \right)$$
B
$$\left( {{{ - 1} \over 3},{7 \over 3}} \right)$$
C
$$\left( { 1,{7 \over 3}} \right)$$
D
$$\left( {{{ 1} \over 3},{7 \over 3}} \right)$$
Explanation
Vertex of triangle is $$\left( {1,\,1} \right)$$ and midpoint of sides through -
this vertex is $$\left( { - 1,\,2} \right)$$ and $$\left( {3,2} \right)$$
$$ \Rightarrow $$ vertex $$B$$ and $$C$$ come out to be $$\left( { - 3,3} \right)$$ and $$\left( {5,3} \right)$$