Question

find the ortho center of the triangle whose
sides are given by 4x-7y+10=0,x+y=5and7x+4y=15​

Answer 1

Answer:

Given lines

4x−7y+10=0−−−−(1)

x+y−5=0−−−−(2)

7x+4y−15=0−−−−(3)

On solving eq (1) and (3)

7x+4(

7

4x+10

)−15=0

49x+16x+40−105=0

65x=65⇒x=1

From eq (1)

4−7y=−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

0

angle is formed

Hence (1,2) is orthocentre

Step-by-step explanation:

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