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
A
{−1,3}
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B
{−3,−2}
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C
{1,3}
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D
{0,2}
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Solution
The correct option is A {−1,3} Given : The vertices of a right angled triangle A(1,k),B(1,1) and C(2,1) and Area of ΔABC = 1 square unit
We know that, area of right angled triangle =12×BC×AB=112(1)|(k−1)|⇒±(k−1)=2⇒k=−1,3