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Question

# The orthocenter of the triangle whose sides are given by 4x−7y+10=0,x+y−5=0 and 7x+4y−15=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 lines4x−7y+10=0⇒7y=4x+10⇒y=4x+107 .....(1)x+y−5=0⇒y=5−x .....(2)7x+4y−15=0 .....(3)Substituting (1) in (3) we get7x+4y−15=07x+4(4x+107)−15=0⇒49x+16x+40−105=0⇒65x=65⇒x=1From eqn(1)4−7y=−10⇒−7y=−14∴y=2Point 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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