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NEET-UG Biology

Concurrence of three lines; distance of a point from a line — practice questions

21 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  1. Question 1 · difficulty L2 · understanding

    The image of the point (3, 5) in the line x - y + 1 = 0, lies on :

    • A. (x - 4) 2 + (y - 4) 2 = 8
    • B. (x - 4) 2 + (y ++ 2) 2 = 16
    • C. (x - 2) 2 + (y - 2) 2 = 12
    • D. (x - 2) 2 + (y - 4) 2 = 4
  2. Question 2 · difficulty L3 · understanding

    If the sum of squares of all real values of α\alpha, for which the lines 2xy+3=0,6x+3y+1=02 x-y+3=0,6 x+3 y+1=0 and αx+2y2=0\alpha x+2 y-2=0 do not form a triangle is pp, then the greatest integer less than or equal to pp is _________.

  3. Question 3 · difficulty L3 · understanding

    Let the equations of two adjacent sides of a parallelogram ABCD\mathrm{ABCD} be 2x3y=232 x-3 y=-23 and 5x+4y=235 x+4 y=23. If the equation of its one diagonal AC\mathrm{AC} is 3x+7y=233 x+7 y=23 and the distance of A from the other diagonal is d\mathrm{d}, then 50 d250 \mathrm{~d}^{2} is equal to ____________.

  4. Question 4 · difficulty L3 · understanding

    Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line x+22y=4x+2 \sqrt{2} y=4. If the co-ordinates of the vertex A are (α,β)(\alpha, \beta), then the greatest integer less than or equal to α+2β|\alpha+\sqrt{2} \beta| is

    • A. 5
    • B. 4
    • C. 2
    • D. 3
  5. Question 5 · difficulty L3 · understanding

    A rectangle is formed by the lines x=0,y=0,x=3x=0, y=0, x=3 and y=4y=4. Let the line L be perpendicular to 3x+y+6=03 x+y+6=0 and divide the area of the rectangle into two equal parts. Then the distance of the point (12,5)\left(\frac{1}{2},-5\right) from the line LL is equal to :

    • A. 10\sqrt{10}
    • B. 252 \sqrt{5}
    • C. 2102 \sqrt{10}
    • D. 3103 \sqrt{10}
  6. Question 6 · difficulty L3 · understanding

    Let a point A lie between the parallel lines L1\mathrm{L}_1 and L2\mathrm{L}_2 such that its distances from L1\mathrm{L}_1 and L2\mathrm{L}_2 are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC , where the points B and C lie on the lines L1\mathrm{L}_1 and L2\mathrm{L}_2, respectively, is :

    • A. 21321 \sqrt{3}
    • B. 12212 \sqrt{2}
    • C. 15615 \sqrt{6}
    • D. 27
  7. Question 7 · difficulty L3 · understanding

    Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle α\alpha with the positive x-axis and the equations of its diagonals are (3+1)x+(31)y=0(\sqrt{3}+1)x+(\sqrt{3}-1)y=0 and (31)x(3+1)y+83=0(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0. Then aa 2 is equal to :

    • A. 48
    • B. 16
    • C. 24
    • D. 32
  8. Question 8 · difficulty L3 · understanding

    Consider the lines x(3λ+1)+y(7λ+2)=17λ+5,λx(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5, \lambda being a parameter, all passing through a point P. One of these lines (say LL ) is farthest from the origin. If the distance of LL from the point (3,6)(3,6) is dd, then the value of d2d^2 is

    • A. 10
    • B. 20
    • C. 15
    • D. 30
  9. Question 9 · difficulty L3 · understanding

    A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines L1:2x+y+6=0\mathrm{L}_1: 2 x+y+6=0 and L2:4x+2yp=0,p>0\mathrm{L}_2: 4 x+2 y-p=0, p>0, at the points A and B , respectively. If AB=92A B=\frac{9}{\sqrt{2}} and the foot of the perpendicular from the point AA on the line L2L_2 is MM, then AMBM\frac{A M}{B M} is equal to

    • A. 5
    • B. 3
    • C. 2
    • D. 4
  10. Question 10 · difficulty L3 · understanding

    The vertices of a triangle are A(1,3),B(2,2)\mathrm{A}(-1,3), \mathrm{B}(-2,2) and C(3,1)\mathrm{C}(3,-1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

    • A. x+y(22)=0-x+y-(2-\sqrt{2})=0
    • B. x+y(22)=0x+y-(2-\sqrt{2})=0
    • C. x+y+(22)=0x+y+(2-\sqrt{2})=0
    • D. xy(2+2)=0x-y-(2+\sqrt{2})=0

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