MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 2

The diagram shows two points,

$M$ and

$N$ on a Cartesian plane.
Calculate the distance between

$M$ and

$N$ .

0%
$\sqrt{10}$ units
0%
$\sqrt{12}$ units
0%
$\sqrt{20}$ units
0%
$\sqrt{52}$ units

Explanation

Coordinates of

$M$ and

$N$ are

$(1,3)$ &

$(5,1)$ respectively.

$MN=\sqrt{(1-5)^{2}+(3-1)^{2}}=\sqrt{20}$ units

A student moves

$\sqrt {2x} km$ east from his residence and then moves x km north. He then goes x km north east and finally he takes a turn of

$90^{\circ}$ towards right and moves a distance x km and reaches his school. What is the shortest distance of the school from his residence?

0%
$(2\sqrt {2} + 1) x\ km$
0%
$3x\ km$
0%
$2\sqrt {2}x\ km$
0%
$3\sqrt {2}x\ km$

Explanation

In triangle

$BCD$

$BD^{2} = (x)^{2} + (x)^{2} = 2x^{2}$

$BD = x\sqrt {2}$
As

$AE = BD$ ,

$AE = x\sqrt {2}$

$\therefore OE = 2\sqrt {2}x$
Also

$DE = x$

$\therefore OD = \sqrt {x^{2} + (2\sqrt {2x})^{2}} = \sqrt {9x^{2}} = 3x$

A line is of length

$10$ and one end is

$(2,-3)$ . If the abscissa of the other end is

$10$ , then find its ordinate.

0%
$3$ or $-9$
0%
$-3$ or $-9$
0%
$3$ or $9$
0%
None of these

Explanation

Coordinates are

$(10,y)$

$\Rightarrow \sqrt{(10-2)^{2}+(y+3)^{2}}=10$

$\Rightarrow (y+3)^{2}=6^{2}\Rightarrow y+3=\pm 6$

$y=3$ or

$-9$

On the Cartesian plane.

$PQR$ is an isosceles triangle.
The perimeter of

$\triangle{PQR}$ is

0%
$16$ units
0%
$18$ units
0%
$20$ units
0%
$22$ units

Explanation

$QR=6$

$PQ=\sqrt{(5-2)^{2}+4^{2}}$

$=\sqrt{3^{2}+4^{2}}=5$

$PR=\sqrt{(5-8)^{2}+4^{2}}=5$

Perimeter =

$PQ+QR+PR$

$\Rightarrow 5+5+6=16$

The area of the triangle formed by three vertices

$O(0, 0), A(1, 0), B(0, 1)$ is _____ sq. units.

0%
$0$
0%
$\dfrac{1}{2}$
0%
$1$
0%
$2$

Explanation

Points are given as

$O(0,0), A(1,0), B(0,1)$ .

When plotted on the cartesian plane, these points make a right angled triangle

$OAB$ with

$OA=1$ ,

$OB=1$

Area of this right angled triangle

$=\dfrac12\times OA\times OB$

$=\cfrac12\times1\times1\ sq.unit$

$=\cfrac12\ sq.unit$

Distance between the points

$(2,-3)$ and

$(5,a)$ is

$5$ . Hence the value of

$a=$ ............

0%
$-1$
0%
$6$
0%
$1$
0%
$7$

Explanation

Let the two points be

$A(2,-3)$ and

$B(5,a)$

$|A-B|=5$

$\implies |A-B|^2=25$

$\implies {(2-5)}^{2}+{(-3-a)}^{2}=25$

$\implies 9+9+6a+{a}^{2}=25$

$\implies {a}^{2}+6a-7=0$

$\implies {a}^{2}+7a-a-7=0$

$\implies (a+7)(a-1)=0$

$\implies a=-7$ or

$a=1$
But distance can't be negative
Hence

$a=1$

The angle of inclination of a straight line parallel to x-axis is equal to

0%
${0}^{o}$
0%
${60}^{o}$
0%
${45}^{o}$
0%
${90}^{o}$

Explanation

The inclination of

$x-axis$ and every line parallel to

$x-axis$ is

$0^0$ .

i.e. the inclination of a line parallel to

$x-axis$ or the

$x-axis$ itself is

$0^0$ .

The points

$(5,1), (1,-1)$ and

$(11,4)$ are:

0%
Collinear
0%
Vertices of right angled triangle
0%
Vertices of equilateral triangle
0%
Vertices of isosceles triangle

Explanation

Let the points be

$A(5,1), B(1,-1),$ and

$C(11,4)$ .
Then,

$AB=2\sqrt { 5 } ,BC=5\sqrt { 5 }$ and

$CA=3\sqrt { 5 }$
Thus,

$AB+CA=BC$
Therefore,

$A, B,$ and

$C$ are collinear.

The graph of

$x = 5$ is perpendicular to _______.

0%
x-axis
0%
y-axis
0%
Line $y = x$
0%
Line $y = -x$

Explanation

General equation of a straight line is

$ax + by + c = 0$ where slope of line is

$\dfrac{-a}{b}$

Equation of line

$: x = 5$

Here slope is

$\dfrac{1}{0}$ which isn't defined, hence line makes angle of

$90^{\circ}$ with x-axis.

$\therefore$ Line is perpendicular to x-axis.

If a plane has X-intercept

$l$ , Y-intercept

$m$ and Z-intercept

$n$ , and perpendicular distance of plane from origin is

$k$ , then ?

0%
$\displaystyle l^2+m^2+n^2=\frac{1}{k^2}$
0%
$\displaystyle\frac{1}{l^2}+\frac{1}{m^2}+\frac{1}{n^2}=\frac{1}{k^2}$
0%
$l^2+m^2+n^2=k^2$
0%
$\displaystyle\frac{1}{l^2}+\frac{1}{m^2}+\frac{1}{n^2}=k^2$

Explanation

The equation of plane is

$\dfrac{x}{l}+\dfrac{y}{m}+\dfrac{z}{n}=1$

$\implies \dfrac{x}{l}+\dfrac{y}{m}+\dfrac{z}{n}-1=0$

Distance from origin

$=\dfrac{|\dfrac{0}{l}+\dfrac{0}{m}+\dfrac{0}{n}-1|}{\sqrt{{(\dfrac{1}{l})^2+(\dfrac{1}{m})^2+(\dfrac{1}{n})^2}}}$

$\implies k = \dfrac{1}{\sqrt{{(\dfrac{1}{l})^2+(\dfrac{1}{m})^2+(\dfrac{1}{n})^2}}}$

$\implies \dfrac{1}{l^2}+\dfrac{1}{m^2}+\dfrac{1}{n^2}=\dfrac{1}{k^2}$

Hence, the answer is option (B).

When the Y-coordinates of two points lying on a line are non-zero and equal, and the X-coordinates are unequal, then :

0%
The line is Parallel to the X-axis.
0%
The line is Parallel to the Y-axis.
0%
The line is inclined at an angle of $45^{\circ}$ with the positive direction of the X-axis.
0%
None of the above.

If the angle

$\theta$ gives the inclination of a line, then the slope of that line is given by :

0%
$\sin(\theta)$
0%
$\cos(\theta)$
0%
$\tan(\theta)$
0%
None of the above.

Point

$P(8,5)$ is the midpoint of segment

$AB$ , Co-ordinates of

$A$ are

$(5,y)$ and

$B$ are

$(x,-3)$ . What is the value of

$x$ ?

0%
$-11$
0%
$11$
0%
$7$
0%
$-7$

Explanation

$\cfrac { 5+x }{ 2 } =8\Rightarrow x=11\quad$

What is the maximum positive value that slope of a line can take?

0%
1.
0%
$\sqrt{2}$
0%
$\sqrt{3}$
0%
$\infty$

Statement-I : Inclination and Slope of a Line give an indication of the Length of that Line.
Statement-II : Inclination and Slope of a Line give an indication of the Direction of that Line.
Choose the correct Alternative.

0%
Only Statement I is True.
0%
Only Statement II is True.
0%
Both Statement I and Statement II are True.
0%
Both Statement I and Statement II are False.

If the projections of a line segment on the x, y and z axes in

$3$ -dimensional space are

$2, 3$ and

$6$ respectively, then the length of line segment is

0%
$6$
0%
$7$
0%
$9$
0%
$12$

Explanation

Given projections are 2, 3 and 6

$\therefore$ AB

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$\Rightarrow AB=\sqrt{4+9+36}=\sqrt{49}=7$ .

In which of these figures does the angle

$\theta$ represent the inclination of line L.

0%
0%
0%
0%
None of the above

Explanation

We know that inclination of a line is the angle it makes with the positive

$x-$ axis.

Hence, A will be correct.

The distance between the points

$(3,5)$ and

$(x,8)$ is

$5$ units. Then the value of

$x$

0%
$7$
0%
$2$
0%
$5$
0%
none of these

Explanation

Given the distance between the points

$(3,5)$ and

$(x,8)$ is

$5$ units.

Then we have,

$\sqrt{(x-3)^2+(8-5)^2}=5$

or,

$(x-3)^2+3^2=5^2$

or,

$(x-3)^2=5^2-3^2=4^2$

or,

$x=\pm 4+3$

or,

$x= 7,-1$

The point A

$(2a ,4a)$ , B

$(2a, 6a)$ and C

$(2a + \sqrt 3 a, 5a)$ (when

$a>0$ ) are vertices of

0%
an obtuse angled triangle
0%
an equilateral triangle
0%
an isosceles obtuse angled triangle
0%
a right angled triangle

Explanation

Distance AB = $\sqrt {{{(2a - 2a)}^2} + {{\left( {6a - 4a} \right)}^2}} = \sqrt {4{a^2}} = 2a$

Distance AC = $\sqrt {{{(2a + \sqrt {3a} - 2a)}^2} + {{\left( {5a - 4a} \right)}^2}}$

$\sqrt {3{a^2} + {a^2}} = 2a$

Distance BC= $\sqrt {{{\left( {2a + \sqrt {3a} - 2a} \right)}^2} + {{\left( {5a - 6a} \right)}^2}}$

$\sqrt {3{a^2} + {a^2}} = 2a$

AB = BC = CA

Equilateral Triangle

1113123_1046402_ans_d864cd9b3a69438b8cdd785302ea8858.PNG

Two adjacent sides of a rectangle are

$3{x^2} - 5{y^{2\,}}$ and

$7{x^2} - xy$ . Find its perimeter?

0%
$10x²-10y² -2xy$ .
0%
$20x²-10y² -2xy$ .
0%
$x²-10y² -2xy$ .
0%
$40x²-10y² -2xy$ .

Explanation

Adjacent sides in rectangle are lenght and base

Perimeter of rectangle =

$2(length + breadth)$

Perimeter =

$2(3x²-5y²+7x²-xy)$

Perimeter =

$2(10x² -5y²-xy)$

Perimeter =

$20x²-10y² -2xy$ .

A line passes through the point

$(3,4)$ and cut off intercepts from the coordinates axes such that their sum is

$14$ . The equation of the line is

0%
$4x-3y=24$
0%
$4x+3y=24$
0%
$3x-4y=24$
0%
$3x+4y=24$

Explanation

Let the intercept by the line on the axes are

$a$ and

$(14-a)$

Thus the equation,

$\dfrac{x}{a}+\dfrac{y}{14-a}=1$ and passes through

$(3,4)$ ----- (1)

$\dfrac{3}{a}+\dfrac{4}{(14-a)}=1$

Therefore,

$a=6,b=8$

So the equation is

Put in (1)

$4x+3y=24$

The distance between

$(2,7)$ and

$(-3,6)$ is

$\sqrt {26}$

0%
True
0%
False

Explanation

Given points

$(2,7)$ and

$(-3,6)$

The distacne between

$2$ points is known as

$\sqrt {(x_2-x_1)^2+(y_2-y_1^2)}\\\sqrt {(3+2)^2+(6-7)^2}\\\sqrt {25+1}=\sqrt {26}$

The points

$A(-4,1)$ ,

$B(-2,-2),C(4,0),D(2,3)$ are the vertices of

0%
parallelogram
0%
rectangle
0%
rhombus
0%
None of these

Explanation

$AB=\sqrt{(-4+2)^{2}+(1+2)^{2}}=\sqrt{13}$

$BC=\sqrt{(-2-4)^{2}+(-2-0)^{2}}=2\sqrt{10}$

$CD=\sqrt{(4-2)^{2}+(0-3)^{2}}=\sqrt{13}$

$DA=\sqrt{(-4-2)^{2}+(1-3)^{2}}=2\sqrt{10}$

Opposite sides are equal

$\implies$ it must be a parallelogram

The coordinates of a point on the line y=x where perpendicular from the line 3x+4y=12 is 4 units, are

0%
$(\dfrac{3}{7},\dfrac{5}{7})$
0%
$(\dfrac{3}{2},\dfrac{3}{2})$
0%
$(-\dfrac{8}{7},-\dfrac{8}{7})$
0%
$(\dfrac{32}{7},-\dfrac{32}{7})$

Explanation

Let any point on line

$y\, = x$

from point of P(a,a),

Distance from point,

$\begin{array}{l} \left| { \dfrac { { 3a+4a-12 } }{ { \sqrt { 9+16 } } } } \right| =4 \\ \left| { \dfrac { { 7a-12 } }{ 5 } } \right| \, \, =4,\, \, \, \dfrac { { 7a-12 } }{ 5 } =\pm 4 \\ we\, have\, two\, conditions, \\ 7a-12=20-----(1) \\ a=\dfrac { { 32 } }{ 7 } \\ 7a-12=-20------(2) \\ a=\dfrac { { -8 } }{ 7 } \\ Coordinate\, of\, po{ { int } }\, is\, a, \\ then, \\ P\left( { \dfrac { { 32 } }{ 7 } ,\dfrac { { 32 } }{ 7 } } \right) \, \, and\, \, P\left( { \dfrac { { -8 } }{ 7 } ,\dfrac { { -8 } }{ 7 } } \right) \\ So,the\, \, correct\, option\, is\, C. \end{array}$

Which of the following sets of points form an equilateral triangle

0%
$\left( {1,0} \right),\left( {4,0} \right),\left( {7, - 1} \right)$
0%
$\left( {\dfrac{2}{3},0} \right),\left( {0,\dfrac{2}{3}} \right),\left( {1,1} \right)$
0%
$\left( {0,0} \right),\left( {\dfrac{3}{2},\dfrac{4}{3}} \right),\left( {\dfrac{4}{3},\dfrac{3}{2}} \right)$
0%
$\left( {0,0} \right),\left( {3,\sqrt 3 } \right),\left( {0,2\sqrt 3 } \right)$

The point of y-axis equidistant from

$(2, 1)$ and

$(4, 5)$ is

0%
$(0, \dfrac{9}{2})$
0%
$(0, 9)$
0%
$(0, 2)$
0%
$(1, 0)$

Nature of the line

$y=7$ and

$y=9$ is

0%
Perpendicular
0%
Intersecting
0%
Parallel
0%
None

The mid points of three sides of a triangle are (0, 1) (0, 2) and (0, 3). Area of this triangle.

0%
3
0%
0
0%
1
0%
None of these

A rectangular hyperbola whose cente is C is cut by any circle of radius r in four point P, Q, R, S. The value of

$CP^{2}+CQ^{2}+CR^{2}+CS^{2}$ is equal to :

0%
$r^{2}$
0%
$2r^{2}$
0%
$3r^{2}$
0%
$4r^{2}$

Explanation

Let the equation of the rectangular hyperbola be

$xy=c^{2}$ ...(1)

And, equation of circle be

$x^{2}+y^{2}=r^{2}$ ....(2)

Put

$y=\dfrac{c^{2}}{x}$ in equation 2, we get,

$x^{2}+\dfrac{c^{4}}{x^{2}}=r^{2}$

$x^{4}-r^{2}x^{2}+c^{4}=0$ .....(3)

Now,

$CP^{2}+CQ^{2}+CR^{2}+CS^{2}$

$={x_{1}}^{2}+{y_{1}}^{2}+{x_{2}}^{2}+{y_{2}}^{2}+{x_{3}}^{2}+{y_{3}}^{2}+{x_{4}}^{2}+{y_{4}}^{2}$

$=2r^{2}+2r^{2}$

$=4r^{2}$

Slope of the line whose inclination is

$120^ {o}$

0%
$-\dfrac {1}{\sqrt {3}}$
0%
$\dfrac {1}{\sqrt {3}}$
0%
$-\sqrt {3}$
0%
$\dfrac {1}{2}$

Explanation

Given the inclination

$(\theta)$ of the line is

$120^o$ .

Then slope will be

$\tan \theta=\tan 120^o=-{\sqrt{3}}$ .

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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