CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 13 - MCQExams.com

if the points A(z),B(-z) and C(z+1) are vertices of an equilateral triangle then ---- test question
  • area of triangle is $$\left( \sqrt { 3 } /4 \right) $$sq. units
  • Re(z)=1/2
  • perimeter of the triangle is 3 units
  • Re(z)=1/4
Area of a triangle whose vertices are $$(a\cos \theta ,b\sin \theta ),(-a\sin \theta ,b\cos \theta)$$ and $$(-a\cos \theta ,-b\sin \theta )$$ is-
  • $$ab\sin \theta \cos \theta $$
  • $$a\cos \theta \sin \theta$$
  • $$\frac 12ab$$
  • ab
If three lines $$x-3y=p,ax+2y=q$$ and $$ax+y=r$$ form a right angled triangle, then 
  • $${ a }^{ 2 }-9a+18=0$$
  • $${ a }^{ 2 }-6a-12=0$$
  • $${ a }^{ 2 }-6a+9=0$$
  • $${ a }^{ 2 }-9a+12=0$$
If the algebraic sum of the perpendicular distances from the points ( 2, 0), (0, 2) and (1, 1) to a variable straight line is zero, then the line passes through the point:
  • (1, 1)
  • (-1, 1)
  • (-1, -1)
  • (1, -1)
Area of a triangle whose vertices are $$ (a \cos \theta, b \sin \theta),(-a \sin \theta, b \cos \theta)  $$ and$$ (-a \cos \theta,-b \sin \theta)  $$ is $$ - $$
  • a b sin $$ \theta \cos \theta $$
  • $$ a \cos \theta \sin \theta $$
  • $$ \frac{1}{2} a b $$
  • $$ a b $$
If the area of triangle formed by the points (2a , b) (a + b , 2b + a) and (2b , 2a) be $$\lambda$$ , then the area of the triangle whose vertices are (a + b , a b) , (3b a , b + 3a) and (3a b , 3b a) will be
  • $$\dfrac{3}{2}\lambda$$
  • $$3\lambda$$
  • $$4\lambda$$
  • none of these
If $$P , Q$$ are two points on the line $$3 x + 4 y + 15 = 0$$ such that $$O P = O Q = 9$$ then the area of $$\Delta O P Q$$ is
  • 6$$\sqrt { 2 }$$
  • 9$$\sqrt { 2 }$$
  • 12$$\sqrt { 2 }$$
  • 18$$\sqrt { 2 }$$
The angle between the tangents drawn from a point $$\left( -a,2a \right)$$ to $${ y }^{ 2 }=4$$ ax is
  • $$\dfrac { \pi }{ 4 } $$
  • $$\dfrac { \pi }{ 2 } $$
  • $$\dfrac { \pi }{ 3 } $$
  • $$\dfrac { \pi }{ 6 } $$
$$PQR$$ is an equilateral triangle such that the vertices $$Q$$ and $$R$$ lie on the lines $$x + y = \sqrt {2}$$ and $$x + y = 7\sqrt {2}$$ respectively. If $$P$$ lies between the two lines at a distance $$4$$ from one of them then the length of side of equilateral triangle $$PQR$$ is (in units).
  • $$8$$
  • $$\dfrac {4\sqrt {7}}{\sqrt {3}}$$
  • $$\dfrac {\sqrt {85}}{3}$$
  • $$\dfrac {4\sqrt {5}}{\sqrt {3}}$$
If p,q denote the lengths of the perpendicu
lars from the origin on the lines $$  x \sec \alpha-y \cos e c \alpha=a  $$ and $$  x \cos \alpha+y \sin \alpha=a \cos 2 \alpha  $$ then (Exam 2013)
  • $$

    4 p^{2}+q^{2}=d^{2}

    $$
  • $$

    p^{2}+q^{2}=a^{2}

    $$
  • $$

    p^{2}+2 q^{2}=a^{2}

    $$
  • $$

    4 p^{2}+q^{2}=2 a^{2}

    $$
The angle at which the circle $$x^2$$ + $$y^2$$ = 16 can be seen from the point (8 , 0) is 
  • $$\dfrac{\pi}{6}$$
  • $$\dfrac{\pi}{4}$$
  • $$\dfrac{\pi}{2}$$
  • $$\dfrac{\pi}{3}$$
If A+B=C and A=B=C then what should be the angle between A and B ?
  • $$0$$
  • $$\pi /3$$
  • $$2\pi /3$$
  • $$\pi $$
The distance of the points of intersection of the lines 
$$2x - 3y + 5 = 0$$ and $$3x + 4y = 0$$ from the line 
$$5x - 2y = 0$$ is 
  • $$\frac{{180}}{{17\sqrt {29} }}$$
  • $$\frac{{13}}{{7\sqrt {29} }}$$
  • $$\frac{{130}}{7}$$
  • None of these
Area of the triangle formed by the tangents at the points $$\left( {4,6} \right),\left( {10,8} \right)$$ and $$\left( {2,4} \right)$$ on the parabola $${y^2} - 2x = 8y - 20,$$is (in sq. units)
  • 4
  • 2
  • 1
  • 8
The area of the triangle inscribed in the parabola $$y^2$$ = $$4x$$ , the ordinates of whose vertices are 1 , 2 and 4 is :
  • 7/2
  • 5/2
  • 3/2
  • 3/4
Let A(-4, 0) & B(4,0). Then the number of points C=(x,y) on the circle $${x^2} + {y^2} = 16$$ lying in first quadrant $$(x,y \geqslant 0)$$ such that the area of the triangle whose vertices are A,B,C is a integer is 
  • 14
  • 15
  • 16
  • None of these
The area of the triangle formed by the straight line $$x+y=3$$ and the bisectors of the pair of straight lines $${ x }^{ 2 }-{ y }^{ 2 }+2y=1$$ is 
  • 1
  • 2
  • 3
  • 6
If the area of triangle formed by the points $$(2a, b) (a + b, 2b + a)$$ and $$(2b, 2a)$$ be $$\lambda$$, then the area of the triangle whose vertices are $$(a + b, a - b), (3b - a, b + 3a)$$ and $$(3a - b, 3b - a)$$ will be
  • $$\frac{3}{2}\lambda$$
  • $$3\lambda$$
  • $$4\lambda$$
  • none of these
Find a point P, which is equidistant from the three points A (0, 1), B (1, 0) and C (4, 3).
  • (1, 2)
  • (2, 2)
  • (2, 3)
  • (3, 3)
The inclination of the straight line passing through the point (3 , 6) and the mid-point of the line joining the points (4 , 5) and (2 , 9) is 
  • $$\dfrac{\pi}{4}$$
  • $$\dfrac{\pi}{6}$$
  • $$\dfrac{\pi}{3}$$
  • $$\dfrac{3\pi}{4}$$
  • $$\dfrac{5\pi}{6}$$
The inclination of the line $$x - y + 3 = 0$$ with the positive direction of $$x - axis$$ is 
  • $${45^0}$$
  • $${135^0}$$
  • $$ - {45^0}$$
  • $$ - {135^0}$$
If $$\alpha ,\beta $$ are the angles between the tw tangents drawn from (0, 0) and $${ x }^{ 2 }+{ y }^{ 2 }-14x+2y+25=0$$ , then $$\alpha -\beta =$$
  • $$\cfrac { \pi }{ 2 } $$
  • $$\cfrac { \pi }{ 3 } $$
  • $${ 0 }^{ 0 }$$
  • None of these.
If $${ p }_{ 1 },{ p }_{ 2 },{ p }_{ 3 }$$ are the altitudes of a triangle from its vertices $$A,B,C$$ and $$\triangle $$, the area of the triangle $$ABC$$, then $$\frac { 1 }{ { p }_{ 1 } } +\frac { 1 }{ { p }_{ 2 } } +\frac { 1 }{ { p }_{ 3 } } $$ is equal to-
  • $$\frac { s }{ \triangle } $$
  • $$\frac { s-c }{ \triangle } $$
  • $$\frac { s-b }{ \triangle } $$
  • $$\frac { s-a }{ \triangle } $$
If the sides of a triangle are $$\dfrac{x}{y}+\dfrac{y}{z} ; \dfrac{y}{z}+\dfrac{z}{x} ; \dfrac{z}{x}+\dfrac{x}{y}$$ then the area of the triangle is 
  • $$xyz$$
  • $$1$$
  • $$\sqrt{\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}}$$
  • $$\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$$
If $$m_1$$ and $$m_2$$ are the roots of the equation $$x^2 +(\sqrt 3 +2) x + \sqrt 3 - 1 = 0 $$, then the area of the triangle formed by the lines $$y=m_1x,y=-m_2x$$ and $$y=1$$ is:
  • $$\dfrac{1}{2} \Bigg ( \dfrac{-\sqrt 3 + 2}{\sqrt 3 -1} \Bigg )$$
  • $$\dfrac{1}{2} \Bigg ( \dfrac{-\sqrt 3 + 2}{\sqrt 3 +1} \Bigg )$$
  • $$\dfrac{1}{2} \Bigg ( \dfrac{\sqrt 3 + 2}{\sqrt 3 -1} \Bigg )$$
  • $$\dfrac{1}{2} \Bigg ( \dfrac{\sqrt 3 + 2}{\sqrt 3 +1} \Bigg )$$
If $$ (\sqrt{2x}+\sqrt{3y}^{2} -36(\sqrt{3x}-\sqrt{2y})^{2}=0$$ and $$\sqrt{2x}-\sqrt{3y}+4\sqrt{5}=0$$ represents an Issosceles traiangle with base angle $$tan^{-1}6$$ then its area is
  • 4
  • 8/3
  • $$8\sqrt{6}$$
  • 9
A triangle has two of its vertices at (0, 1) and (2, 2) in the cartesioan plane. Its third vertex lies on the x-axis. If the area of the triangle is 2 square units then the sum of the possible abscissae of the third vertex, is -
  • -4
  • 0
  • 5
  • 6
Let P and Q be points $$(4, 4)$$ and $$(9, 6)$$ of parabola $${y^2} = 4a\left( {x - b} \right)$$ If R be a point on the arc of the parabola between P and Q, such that the area of  $$\Delta PRQ$$ is largest, then R is
  • $$\left( {\frac{1}{4},4} \right)$$
  • $$\left( {\frac{1}{4},1} \right)$$
  • $$\left( {4,4} \right)$$
  • $$\left( {2,2\sqrt 2 } \right)$$
The perpendicular distance between the lines represented by  $${ x }^{ 2 }-4xy+{ 4y }^{ 2 }+x-2y-6=0\quad is-$$
  • $$\frac { 1 }{ \sqrt { 5 } } $$
  • $$\sqrt { 5 } $$
  • $$2\sqrt { 5 } $$
  • $$3\sqrt { 5 } $$
Area of the triangle formed by $${ (x }_{ 1 },{ y }_{ 1 })$$,$${ (x }_{ 2 },{ y }_{ 2 })$$,
$${ (3x }_{ 2 }-{ 2x }_{ 1 },{ 3y }_{ 2 }-{ 2y }_{ 1 })$$ is
  • 0 sq.units
  • $${ { x }_{ 1 } }y_{ 1 } $$sq.units
  • 3 sq.units
  • 6 sq.units
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