The equation of a straight line; perpendicular to 8x - 4y = 6 and

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

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1.

The equation of a straight line; perpendicular to 8x - 4y = 6 and forming a triangle of area 6sq. units with coordinate axes, is

  • x - 2y = 6

  • 4x + 3y = 12

  • 4x + 3y + 24 = 0

  • 3x + 4y = 12


B.

4x + 3y = 12

Given eq of line is3x - 4y = 6Perpendicular to above line is

 Area of OAB= 12 × OA × OB 6 = 12 × k4 × k3 6 = k224 k2 = 144 k = ± 12 Required equation of line is4x + 3y = ± 12


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2.

If the image of - 76, - 65 in a line is 1, 2, then the equation of the line is

  • 4x + 3y = 1

  • 3x - y = 0

  • 4x - y = 0

  • 3x + 4y = 1


3.

If a line l passes through (k, 2k), (3k, 3k) and (3, 1), k  0, then the distance from the origin to the line is

  • 15

  • 45

  • 35

  • 25


4.

The area (in sq. units) of the triangle formed by the lines x2 - 3y + y = 0 and x + y + 1 = 0, is

  • 23

  • 32

  • 52

  • 125


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5.

If x2 + αy2 +2βy = α2 represents a pair of perpendicular lines, then p equals to

  • 4a

  • a

  • 2a

  • 3a


6.

A circle with centre at (2, 4) is such that the line x + y + 2 = 0 cuts a chord of length 6. The radius of the circle is

  • 41 cm

  • 11 cm

  • 21 cm

  • 31 cm


7.

The point at which the circles x2 + y2 - 4x - 4y + 7 = 0 and x2 + y2 - 12x - 10y + 45 = 0 touch each other, is

  • 135, 145

  • 25, 56

  • 145, 135

  • 125, 2 + 215


8.

The condition for the lines lx + my + n = 0 and l1a + m1y + n1 = 0 to be conjugate with respect to the circle x2 + y2 = r2, is

  • r2(ll1 + mm1) = nn1

  • r2(ll1 - mm1) = nn1

  • r2(ll1 + mm1) + nn= 0

  • r2(lm1 + l1m) = nn1


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9.

The length of the common chord of the two circles x2 + y2 - 4y = 0 and x2 + y2 - 8x - 4y + 11 = 0, is

  • 1454 cm

  • 112 cm

  • 135 cm

  • 1354


10.

The locus of the centre of the circle, which cuts the circle x2 + y2 - 20 + 4 = 0 orthogonally and touches the line x = 2, is

  • x2 = 16y

  • y2 = 4x

  • y2 = 16x

  • x2 = 4y


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