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JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1
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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Correct option: C
(−73,23)\left( { - \frac{7}{3},\frac{2}{3}} \right)
Option C ((-7/3, 2/3)) is correct because it is the centroid calculated from the vertices of the triangle formed by the given lines.

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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: A line LL passes through the points (1,1)(1,1) and (2,0)(2,0) and another line L′L^{\prime} passes through (12,0)\left(\frac{1}{2}, 0\right) and perpendicular to LL. Then the area of the triangle formed by the lines L,L′L, L^{\prime} and y−y- axis, is NN. Find 16N16 N
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