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Question

Two vertices of a triangle are (1,3) and (4,7). The orthocentre lies on the x+y=3. The locus of the third vertex is

A
x22xy+2y23x4y+36=0
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B
2x24xy+3y24xy+42=0
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C
4y2xy3x2+2x24y+40=0
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D
x24xy+3y22xy+40=0
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Solution

The correct option is C 4y2xy3x2+2x24y+40=0
O=(h,3h)
mBO=3h3h1=hh1
mCO=73+h4h=4+h4h
APCP
mAP=h4h+4
AmBm
mAm=h1h
Now
Apy3=h4h+4(x1)
h(y3)+4(y3)=h(x1)4(x1)
h(yx2)=4y+124x+4
h(yx2)=(4y+4x16) ………..(1)
Now
amy7=h1h(x4)
h(y7)=h(x4)(x4)
h(yx3)=(x4) …………(2)
Equation (1)÷(2)
h(yx3)h(yx2)=x44y+4x16
(yx3)(4y+4x16)=(x4)(yx2)
4y2+4xy16y4xy4x2+16x12y12x+48=xy4yx2+4x2x+8
4y23x2xy+2x24y+40=0.

1212824_1333396_ans_3ad5c28ed83c42d0ac3fa8a75fc39b46.jpg

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