Co-ordinates of orthocentre of triangle bounded by lines, 4x

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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 :

$(2, 1)$

$(- 1, 2)$

$(1, 2)$

$(1, - 2)$

Solution

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

Standard 11

Mathematics

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normal

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Solution

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