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Question

Let k be an integer such that the triangle with vertices (k,3k),(5,k) and (k,2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:

A
(2,12)
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B
(1,34)
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C
(1,34)
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D
(2,12)
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Solution

The correct option is D (2,12)
Given vertices of the triangle are (k,3k),(5,k) and (k,2)
Now, the area of triangle is
12x1x2x3x1y1y2y3y1=2812k5kk3kk23k=28|(k2+15k)+(10+k2)+(3k22k)|=56|5k2+13k+10|=565k2+13k+10=56 (5k2+13k+10>0)5k2+13k46=0(5k+23)(k2)=0k=2 (k is integer)

Now vertices areA(2,6),B(5,2),C(2,2)
Equation of the altitude dropped from vertex A is x=2
Equation of the altitude dropped from vertex C is 3x+8y10=0
On solving the equations of altitude we get, Orthocentre =(2,12)

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