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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

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)|(k1)|±(k1)=2k=1,3

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