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JEE MATHEMATICS MCQ: STRAIGHT LINES

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If a variable line drown through the point of intersection of straight lines xα+yβ=1\frac{x}{\mathrm{\alpha }} + \frac{y}{\mathrm{\beta }} = 1 and xβ+yα=1\frac{x}{\mathrm{\beta }} + \frac{y}{\mathrm{\alpha }} = 1 meets the coordinate axes in A and B,then the locus of the mid point of AB is
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Correct option: B
αβ(x+y)=2xy(α+β)\mathrm{\alpha \beta }\left( {x + y} \right) = 2xy\left( {\mathrm{\alpha } + \mathrm{\beta }} \right)
Option B (αβ(x+y)=2xy(α+β)\mathrm{\alpha \beta }\left( {x + y} \right) = 2xy\left( {\mathrm{\alpha } + \mathrm{\beta }} \right)) is correct because it represents the locus of the midpoint of AB based on the given conditions.

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If the three lines x−3y=p,ax+2y=qx-3 y=p, a x+2 y=q and ax+y=ra x+y=r form a right-angled triangle then :
The orthocentre of the triangle whose vertices are

{at1t2,  a(t1+t2)},{at2t3,a(t2+t3)},  {at3t1,  a(t3+t1)}\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\} is
 The distance between the parallel lines y=2  x+4y = 2\;x + 4 and 6x=3y+56x = 3y + 5 is The distance between the pair of lines given by x2+y2+2xy−8ax−8ay−9a2=0{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0 is Find the image of the point (3,8)(3,8) with respect of the line x+3y=7x+3 y=7 assuming the line to be plane mirror: The centroid of the triangle whose three sides are given by the combined equation (x2+7xy+2y2)(y−1)=0\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0 is
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