MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 14

The area of a triangle whose sides are

$a , b$ and

$c$ is

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$\sqrt { ( s - a ) ( s - b ) ( s - c ) }$ sq. units
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$\sqrt { s ( s + a ) ( s + b ) ( s + c ) }$ sq. units
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$\sqrt { s ( s \times a ) ( s \times b ) ( s \times c ) }$ sq. units
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$\sqrt { s ( s - a ) ( s - b ) ( s - c ) }$ sq. units

The arrangement of the areas of the triangles formed by the following points in ascending order is
i) P(0,0), Q(4,0), R(0,3)
ii) P(0,0), Q(5,0), R(0,2)
iii) P(0,0), Q(0,5), R(6,0)
iv) p(3,0), Q(0,6), R(0,0)

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i,ii,iii,iv
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ii,i,iii,iv
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ii,i,iv,iii
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iv,iii,ii,i

two medians drawn from the acute angles of a right angled triangle interect at the angle

$\pi /6$ if thelength the hypotenuse of the triangle is 3 units then the area of the triangle is

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$\sqrt { 3 }$
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3
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$\sqrt { 2 }$
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9

The area inside the parabola

$5x^2 - y = 0$ but outside the parabola

$2x^2 - y + 9 = 0$ is

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$12\sqrt{3}$ sq.units
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$6\sqrt{3}$ sq.units
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$8\sqrt{3}$ sq.units
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$4\sqrt{3}$ sq.units

The inclination of the line y=

$\sqrt { 3 } x+3$ is

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$\dfrac { \pi }{ 4 }$
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$\dfrac { 3\pi }{ 4 }$
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$\dfrac { \pi }{ 2 }$
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$\dfrac { \pi }{ 3 }$

Area of the triangle formed by co-ordinate axes and the line x + y = 5 is

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$\dfrac { 5 }{ 2 }$
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$4\dfrac { 25 }{ 2 }$ $
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$\dfrac { 75 }{ 2 }$
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$\dfrac { 125 }{ 2 }$

The points (1, 1), (-2, 7) and (3, -3)

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form a right angled triangle
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form an isosceles triangle
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are co linear
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form an equilateral triangle

Explanation

$\textbf{Step-1: Naming the given points and calculating the distance between each two of them}$

$\text{Let the given points be }$

${A(1,1), B(-2,7),C(3,-3)}$

$\text{By using the distance formula, distance between any two points}$

${(x_1,y_1)}$

$\text{and }$

${(x_2,y_2)}$

${d=\sqrt{(y_2-y_1)^{2}+(x_2-x_1)^{2}}}$

${AB= \sqrt{(7-1)^{2}+(-2-1)^{2}}}$

${=\sqrt{(6)^2+(-3)^2}}$

${=3\sqrt{5}}$

$\text{units}$

${BC= \sqrt{(-3-7)^{2}+(3+2)^{2}}}$

${=\sqrt{(-10)^2+(5)^2}}$

${=5\sqrt{5}}$

$\text{units}$

${AC= \sqrt{(1+3)^{2}+(1-3)^{2}}}$

${=\sqrt{(4)^2+(2)^2}}$

${=2\sqrt{5}}$

$\text{units}$

$\textbf{Step-2: Drawing conclusions from the calculated distances}$

$\text{We have,}$

${AB+AC=5\sqrt{5}}$

$\text{units}$

$\because$

${AB+AC=BC,}$

$\text{we can say that the three points are collinear}$

$\textbf{Hence, the points}$

$\mathbf{ A(1,1), B(-2,7)}$

$\textbf{and}$

$\mathbf{C(3,-3)}$

$\textbf{are collinear}$

The centroid and two vertices of a triangles are (4,-8), (-9,7), (1,4) then the area of the triangle is

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333 sq.units
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166.5 sq.units
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111 sq.units
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55.5 sq.units

Circles

${ x }^{ 2 }+{ y }^{ 2 }=1$ and

${ x }^{ 2 }+{ y }^{ 2 }-8x+11=0$ cut off equal intercepts on a line through the points

$\left( -2,\dfrac { 1 }{ 2 } \right)$ The slope of the line is

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$\dfrac { -1+\sqrt { 29 } }{ 14 }$
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$\dfrac { 1+\sqrt { 7 } }{ 4 }$
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$\dfrac { -1-\sqrt { 29 } }{ 14 }$
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none of these

$\triangle ABC,A(1,2);B(5,5),\angle ACB={ 90 }^{ 0 }$ . If area of

$\triangle ABC$ is to be 6.5 sq. units, then the possible number of points for C is

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1
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2
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0
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4

A line forms a triangle of area

$54\sqrt { 3 }$ sq-units with the coordinate axes. Then the equation of the line if the perpendicular drawn from the origin to the line makes an angle of

${ 60 }^{ \circ }$ with the x-axis is

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$x-\sqrt { 3 } y=18$
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$x+\sqrt { 3 } y=18$
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$x+\sqrt { 3 } y=-18$
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None of these

If the area of the triangle formed by the line -y = 0, x + y = 0 and x - c = 0 is 16 units, c is

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1
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2
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4
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8

The obtuse angle between the line y=-2 and y=x+2 is-

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$120^{\circ}$
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$135^{\circ}$
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$150^{\circ}$
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$160^{\circ}$

Let

$A(h,k),B(1,1)$ and

$C(2,1)$ be the vertices of a right- angled triangle with

$AC$ as its hypotenuse. If the area of the triangle is 1, then the set of values which

$k$ can take is given by

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{1, 3}
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{0, 2}
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{-1, 3}
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{-3, 2}

The plot of

$\sqrt v$ vs Z is

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Straight line
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Exponential curve
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Hyperbolic
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Curve with -ve slope

If sides of a triangle are

$y=m x+a, y=n x+b$ and

$x=0,$ then its area is -

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$\frac{1(a-b)^{2}}{2(m-n)}$
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$\frac{1}{2} \frac{(a-b)^{2}}{m+n}$
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$\frac{1(a+b)^{2}}{2(m-n)}$
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none of these

A line is drawn form

$A\left ( -2, 0 \right )$ to intersect the curve

$y^{2} = 4x$ in

$P$ and

$Q$ in the first quadrant such that

$\frac{1} {AP} + \frac{1} {AQ} < \frac{1} {4}$ , then slope of the line is always

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$> \sqrt{3}$
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$< \frac{1} {\sqrt{3}}$
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$> \sqrt{2}$
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$> \frac{1} {3}$

Angles made with the

$x-$ axis by two lines drawn through the point

$(1, 2)$ cutting the line

$x+y=4$ at a distance

$\sqrt 6/3$ from the point

$(1, 2)$ are

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$\dfrac{\pi}{12}$ and $\dfrac{5\pi}{12}$
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$-\dfrac{7\pi}{12}$ and $-\dfrac{11\pi}{12}$
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$\dfrac{\pi}{8}$ and $\dfrac{3\pi}{8}$
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$None\ of\ these$

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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