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Question

The three lines 4x−7y+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 orthocentreGiven lines4x−7y+10=0−−−−(1)x+y−5=0−−−−(2)7x+4y−15=0−−−−(3)On solving eq (1) and (3)7x+4(4x+107)−15=049x+16x+40−105=065x=65⇒x=1From eq (1)4−7y=−10−7y=−14y=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 900 angle is formed Hence (1,2) is orthocentre

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