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Question

The orthocenter of the triangle formed by the pair of lines 2x2−xy−y2+x+2y−1=0 and the line x+y+1=0

A
(-1,0)
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B
(0, 1)
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C
(1, 1)
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D
None of these
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Solution

The correct option is C (-1,0)
Given pair of lines are2x2xyy2+x+2y1=0
and x+y+1=0 (1)

(xy)(2x+y)+x+2y1=0

Equations are xy+c1=0and2x+y+c2=0
Also
2x2xyy2+x+2y1=0=(xy+c1)(2x+y+c2)
=2x2+yx+xc22xyy2yc2+2xc1+yc1+c1c2
=2x2xyy2+x(c2+2c1)+y(c1c2)+c1c2

On comparing we get
c2+2c1=1 and c1c2=2
On solving we get c1=1andc2=1
Eqs.(1)xy+1=0 (2)
and 2x+y1=0 (3)

Here slope of line (1) and slope of line (2) are -1 and 1
Product of slope =-1
Hence line 1line 2

So the intersection of line 1 and line 2 forming a right angle which is an orthocenter
solving eqs.(1) and (2) we get, x=-1 and y=0

So orthocenter is (-1,0)

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