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ABC is a variable triangle such that A i...

`ABC` is a variable triangle such that `A` is (1, 2), and B and C on the line `y=x+lambda(lambda` is a variable). Then the locus of the orthocentre of `DeltaA B C` is (a)`x+y=0` (b) `x-y=0` `x^2+y^2=4` (d) `x+y=3`

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A B C is a variable triangle such that A is (1, 2), and Ba n dC on the line y=x+lambda(lambda is a variable). Then the locus of the orthocentre of "triangle"A B C is x+y=0 (b) x-y=0 x^2+y^2=4 (d) x+y=3

A B C is a variable triangle such that A is (1,2) and B and C lie on line y=x+lambda (where lambda is a variable). Then the locus of the orthocentre of triangle A B C is (a) (x-1)^2+y^2=4 (b) x+y=3 (c) 2x-y=0 (d) none of these

Tangent are drawn to the circle x^2+y^2=1 at the points where it is met by the circles x^2+y^2-(lambda+6)x+(8-2lambda)y-3=0,lambda being the variable. The locus of the point of intersection of these tangents is 2x-y+10=0 (b) 2x+y-10=0 x-2y+10=0 (d) 2x+y-10=0

The line L_1-=4x+3y-12=0 intersects the x-and y-axies at A and B , respectively. A variable line perpendicular to L_1 intersects the x- and the y-axis at P and Q , respectively. Then the locus of the circumcenter of triangle A B Q is (a) 3x-4y+2=0 (b) 4x+3y+7=0 (c) 6x-8y+7=0 (d) none of these

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If points A and B are (1, 0) and (0, 1), respectively, and point C is on the circle x^2+y^2=1 , then the locus of the orthocentre of triangle A B C is x^2+y^2=4 x^2+y^2-x-y=0 x^2+y^2-2x-2y+1=0 x^2+y^2+2x-2y+1=0

Vertices of a variable triangle are (3,4); (5costheta, 5sintheta) and (5sintheta,-5costheta) where theta is a parameter then the locus of its orthocentre is a. (x+y-1)^2+(x-y-7)^2=100 b. (x+y-7)^2+(x-y-1)^2=100 c. (x+y-7)^2+(x+y-1)^2=100 d. (x+y-7)^2+(x-y+1)^2=100

From the points (3, 4), chords are drawn to the circle x^2+y^2-4x=0 . The locus of the midpoints of the chords is (a) x^2+y^2-5x-4y+6=0 (b) x^2+y^2+5x-4y+6=0 (c) x^2+y^2-5x+4y+6=0 (d) x^2+y^2-5x-4y-6=0

If a variable line drawn through the intersection of the lines x/3+y/4=1 and x/4+y/3=1 meets the coordinate axes at A and B, (A !=B), then the locus of the midpoint of AB is (A) 6xy = 7 (x+y) (B) 4 (x+y)^2 - 28(x+y) + 49 =0 (C) 7xy = 6(x+y) (D) 14 (x+y)^2 - 97(x+y) + 168 = 0

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CENGAGE ENGLISH-STRAIGHT LINES-All Questions
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