The co-ordinates of the orthocentre of the tr | vedclass

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Question

Given lines $4 x-7 y+10=0----(1)$

$x+y-5=0----(2)$

$7 x+4 y-15=0----(3)$

On solving eq ( 1 ) and ( 3 ) $7 x+4\left(\frac{4 x+10}{7}\right)-15

Vedclass · Accepted Answer

$(1, 2)$

Vedclass · Answer

Given lines $4 x-7 y+10=0----(1)$

$x+y-5=0----(2)$

$7 x+4 y-15=0----(3)$

On solving eq ( 1 ) and ( 3 ) $7 x+4\left(\frac{4 x+10}{7}\right)-15=0$

$49 x+16 x+40-105=0$

$65 x=65 \Rightarrow x=1$

From eq (1) $4-7 y=-10$

$-7 y=-14$

$y=2$

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 $90^{0}$ angle is formed Hence (1,2) is orthocentre

The co-ordinates of the orthocentre of the triangle bounded by the lines, $4x - 7y + 10 = 0; x + y=5$ and $7x + 4y = 15$ is :

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