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Question

The three lines 4x7y+10=0,x+y=5 and 7x+4y=15 form the sides of a triangle. Then the point (1,2) is its

A
centroid
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B
incentre
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C
orthocentre
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D
none of these
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Solution

The correct option is C orthocentre
Given lines
4x7y+10=0(1)
x+y5=0(2)
7x+4y15=0(3)
On solving eq (1) and (3)
7x+4(4x+107)15=0

49x+16x+40105=0
65x=65x=1
From eq (1)
47y=10
7y=14
y=2
Point of intersection is (1,2)
Here Equation (1) and (3) are perpendicular So the lines formed right angle triangle
Hence the orthocentre of Right angle triangle is the point at which 900 angle is formed
Hence (1,2) is orthocentre

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