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Question

The base of a triangle lies along x=a and is of length a. If the area of the triangle is a2, then the locus of its vertex is

A
(x+a)(x3a)=0
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B
(xa)(x+3a)=0
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C
(x+a)(x+3a)=0
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D
(x+2a)(xa)=0
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Solution

The correct option is A (x+a)(x3a)=0
Suppose, the coordinates of the vertex are (h,k).
The area of the triangle is 12×a|ha|=a2.
h=a or h=3a.
Hence, the combined equation of the pair of lines is (h+a)(h3a)=0
(x+a)(x3a)=0 is the required locus.

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