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Question

The orthocenter of the triangle whose sides are given by 4x7y+10=0,x+y5=0 and 7x+4y15=0 is

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

The correct option is D (1,2)
Given lines

4x7y+10=0
7y=4x+10
y=4x+107 .....(1)
x+y5=0
y=5x .....(2)
7x+4y15=0 .....(3)

Substituting (1) in (3) we get

7x+4y15=0
7x+4(4x+107)15=0
49x+16x+40105=0
65x=65x=1
From eqn(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 90 angle is formed

Hence (1,2) is orthocentre

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