MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 10

Point P(2,3) lines on the

$4x + 3y = 17$ . Then find the co-ordinated of points farthest from the line which are at 5 units distance from the P.

0%
$(6,6)$
0%
$(6 , -6)$
0%
$(2 , 0)$
0%
$(-2 , 0)$

Explanation

Let point

$(x,y)$ from the line.

$\left|\dfrac{4x+3y-17}{\sqrt{4^2+3^2}}=5\right |$

$\Rightarrow \boxed{4x+3y-17=\pm 25}$

but given point Only

$(A)$ option is satisfying line.

So, Answer should be

$(A)$ .

1086191_1114157_ans_653122a2190944cdbc218a19544257c8.png

If a circle

${ x }^{ 2 }+{ y }^{ 2 }=20$ meet the parabola

${ y }^{ 2 }=8x$ at

$P$ and

$Q$ . Then

$PQ=$ __

0%
$4$
0%
$8$
0%
$16$
0%
$\sqrt { 5 }$

The sum of the lengths of tangents and subtangent at a point of

$y=a\log{(x^{2}-a^{2})}$ ,

$(a>0)$ is porportional to

0%
$\left| x\right|$
0%
$\left| y\right|$
0%
$\left| xy\right|$
0%
$\left| x/y\right|$

Explanation

$y=a\log(x^2-a^2).a>0.$

$\dfrac{dy}{dx}=\dfrac{2ax}{x^2-a^2}$

Length of tangent

$=y\sqrt{1+\dfrac{1}{\left(\dfrac{dy}{dx}\right)^2}}$

$= \sqrt{1+\dfrac{(x^2-a^2)^2}{(2ax)^2}}$

$=\dfrac{|y(x^2+a^2)|}{|2ax|}$

Length of subtangent

$=\left|y\dfrac{dx}{dy}\right|=\left|y\times \dfrac{1}{\dfrac{dy}{dx}}\right|$

$=\dfrac{y(x^2-a^2)}{|2ax|}$

Sum

$=\dfrac{|x^2y|}{|2ax|}=\dfrac{|xy|}{2a}$

Proportional to

$|xy|$

Area of the triangle formed by

$(x_{1,}y_{1}),(x_{2},y_{2}), (3y_{2},(-2y_{1}))$

0%
$\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right]$
0%
$\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right]$
0%
$\dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right]$
0%
$=\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right]$

Explanation

We have,

$A\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},{{y}_{1}} \right)$

$B\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{x}_{2}},{{y}_{2}} \right)$

$C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3{{y}_{2}},-2{{y}_{1}} \right)$

We know that the area of triangle is

$Area\,of\,\Delta ABC=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

$=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}+2{{y}_{1}} \right)+{{x}_{2}}\left( -2{{y}_{1}}-{{y}_{1}} \right)+3{{y}_{2}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

$=\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right]$

Hence, the is the answer

Find the point on the x-axis which is equidistant from

$(2, -5)$ and

$(-2, 9)$ .

0%
$(1,2)$
0%
$(-7,0)$
0%
$(3,-5)$
0%
$(8,8)$

If the points (0,0),

$(3,\sqrt{3})$ ,(p,q) from an equilateral triangle and

$q_{1},q_{2}$ are the two values of q then

$q_{1}+q_{2}$ =

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

Explanation

Consider the given point

$A\left( {{x}_{1}},{{y}_{1}} \right)=\left( 0,0 \right)$

$B\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,\sqrt{3} \right)$

$C\left( {{x}_{3}},{{y}_{3}} \right)=\left( p,q \right)$

Then,

We know that it is an equilateral triangle,

$AB=BC$

$\Rightarrow \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}}$

$\Rightarrow \sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 0-\sqrt{3} \right)}^{2}}}=\sqrt{{{\left( 3-p \right)}^{2}}+{{\left( \sqrt{3}-q \right)}^{2}}}$

$\Rightarrow \sqrt{12}=\sqrt{9+{{p}^{2}}-6p+3+{{q}^{2}}-2\sqrt{3}q}$

$\text{Squaring both side and we get,}$

$\Rightarrow 12={{p}^{2}}+{{q}^{2}}-6p-2\sqrt{3}q+12$

$\Rightarrow {{p}^{2}}+{{q}^{2}}-6p-2\sqrt{3}q=0$ ....... (1)

Again,

$AB=CA$

$\Rightarrow \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{3}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{3}}-{{y}_{1}} \right)}^{2}}}$

$\Rightarrow \sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 0-\sqrt{3} \right)}^{2}}}=\sqrt{{{\left( p-0 \right)}^{2}}+{{\left( q-0 \right)}^{2}}}$

$\Rightarrow \sqrt{12}=\sqrt{{{p}^{2}}+{{q}^{2}}}$

$\text{squaring both side and we get,}$

${{p}^{2}}+{{q}^{2}}=12$ …… (2)

By equation (1) and (2) to, and we get,

$12-6p-2\sqrt{3}q=0$

$\Rightarrow 6p+2\sqrt{3}q=12$

$\Rightarrow 3p+\sqrt{3}q=6$

$\Rightarrow 3p=6-\sqrt{3}q$

$\Rightarrow p=\dfrac{6-\sqrt{3}q}{3}$

Put the value of p in equation (2) and we get,

${{p}^{2}}+{{q}^{2}}=12$

$\Rightarrow {{\left( \dfrac{6-\sqrt{3}q}{3} \right)}^{2}}+{{q}^{2}}=12$

$\Rightarrow 36+3{{q}^{2}}-12\sqrt{3}q+9{{q}^{2}}=108$

$\Rightarrow 12{{q}^{2}}-12\sqrt{3}q=72$

$$\begin{align}

$\Rightarrow 12\left( {{q}^{2}}-\sqrt{3}q-6 \right)=0$

$\Rightarrow {{q}^{2}}-\sqrt{3}q-6=0$

$\Rightarrow {{q}^{2}}-2\sqrt{3}q+\sqrt{3}q-6=0$

$\Rightarrow q\left( q-2\sqrt{3} \right)+\sqrt{3}\left( q-2\sqrt{3} \right)=0$

$\Rightarrow \left( q-2\sqrt{3} \right)\left( q+\sqrt{3} \right)=0$

Then, $q=2\sqrt{3},\,\,q=-\sqrt{3}$

Now, According to given question,

$q={{q}_{1}}+{{q}_{2}}$

$q=2\sqrt{3}-\sqrt{3}$

$q=\sqrt{3}$

Hence, this is the answer.

Three points

$\left( {0,0} \right),\left( {3,\sqrt 3 } \right),\left( {3,\lambda } \right)$ from an equilateral triangle, then

$\lambda$ is equal to

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$2$
0%
$-3$
0%
$-4$
0%
$- \sqrt 3$

Explanation

Given:An equilateral triangle

$ABC$ whose coordinates are

$A\left(0,0\right),\,B\left(3,\sqrt{3}\right)$ and

$C\left(3,\lambda\right)$

Since the triangle is equilateral, we have

${AB}^{2}={AC}^{2}$

So,

${\left(3-0\right)}^{2}+{\left(\sqrt{3}-0\right)}^{2}={\left(3-0\right)}^{2}+{\left(\lambda-0\right)}^{2}$

$\Rightarrow\,{\left(\sqrt{3}-0\right)}^{2}={\left(\lambda-0\right)}^{2}$

$\Rightarrow\,{\lambda}^{2}=3$

$\therefore\,\lambda=\pm \sqrt{3}$

Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are

$(2, 1)$ ,

$(-2, 3)$ ,

$(4, -3)$ .

0%
$1.5$ sq. units
0%
$3$ sq. units
0%
$6$ sq. units
0%
$20$ sq. units

Two points A

$\left ( x_{1},y_{1} \right )andB\left ( x_{2},y_{2} \right )$ are chosen on the graph of

$f(x)=inx with 0<x_{1}<x_{2}.$ the point C and D trisect line segment AB with AC<CB. through Ca horizontal line is drawn to cut the curve at

$E\left ( x_{3},y_{3} \right ).if x_{1}=1000$ then the value of

$x_{3}$ equals

0%
10
0%
$\sqrt{10}$
0%
$(10)^{2/3}$
0%
$(10)^{1/3}$

Explanation

We have,

Since, $C$ trisects $AB$

Now,

${{x}_{1}}=1,\,\,{{y}_{1}}=\ln 1=0$

${{x}_{2}}=1000,\,\,{{y}_{2}}=\ln 1000=3$

Now, Ratio ${{m}_{1}}:{{m}_{2}}=1:2$

$a=\dfrac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}}$

$=\dfrac{1\times 1000+2\times 1}{1+2}$

$=\dfrac{1002}{3}$

$a=334$

$b=\dfrac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}}$

$=\dfrac{1\times \ln 1000+2\times 0}{1+2}$

$b=\dfrac{\ln 1000}{3}\,\,\,\,.......\,\,\left( 1 \right)$

Given the curve

$y=\ln x\,\,\,......\,\,\left( 2 \right)$

From equation (1) and (2) to,

$\ln x=\dfrac{\ln 1000}{3}$

$3\ln x=\ln 1000$

$\ln x={{\left( \ln 1000 \right)}^{\dfrac{1}{3}}}$

$x={{\left( 1000 \right)}^{\dfrac{1}{3}}}$

$x=10$

Hence, ${{x}_{3}}=10$

This is the answer.

The points

$(-1,5),(-2,-3),(5,1),(6,9)$ taken on order are the vertices of a

0%
Parallelogram
0%
Rhombus
0%
Rectangle
0%
Suare

Explanation

$A(-1,5)$ ,

$B(-2,-3)$ ,

$C(5,1)$ and

$D(6,9)$

Slopes of sides of quadrilateral =>

${ m }_{ ab }=8\\ { m }_{ bc }=\cfrac { 4 }{ 7 } \\ { m }_{ cd }=8\\ { m }_{ da }=\cfrac { 4 }{ 7 }$

We see,

${ m }_{ ab }=8={ m }_{ cd }$ &

${ m }_{ bc }={ m }_{ da }=\cfrac { 4 }{ 7 }$ => ABCD is Parallelogram as opposite sides are Parallel.

The length of a line segment

$AB$ is

$10$ units. If the coordinates of one extremity are

$\left(2,-3\right)$ and the abscissa of the other extremity is

$10$ then the sum of all possible values of the ordinate of the other extremity is

0%
$3$
0%
$-4$
0%
$12$
0%
$-6$

Explanation

Let the ordinate be

$y$

$\therefore$ Point

$1(A)\equiv (2, -3)$

$\therefore$ Point

$2(B)\equiv (10, 4)$

$\therefore d(A, B)=\sqrt{(2-10)^2+(-3-4)^2}$

$\therefore \sqrt{64+(3+y)^2}=10$

$\therefore 36=(3+y)^2$

$\therefore 3+y=6$ OR

$3+y=-6$

$\therefore y=3$ OR

$y=-9$

$\therefore$ Sum

$=3+(-9)=-6$

If the line

$\left( \cfrac { x }{ 2 } +\cfrac { y }{ 3 } -1 \right) +\lambda(2x+y-1)=0$ is parallel to x-axis then

$\lambda=$

0%
$-\cfrac{1}{2}$
0%
$\cfrac{1}{2}$
0%
$-\cfrac{1}{4}$
0%
$\cfrac{1}{4}$

Explanation

$\left( \dfrac{x}{2}+ \dfrac{y}{3}-1 \right)+ \lambda (2x+y-1)=0$

$(3x+2y-6)+ 6 \lambda (2x+y-1)=0$

$(3+12 \lambda)x+ (2+6\lambda)y- (6+6\lambda)=0$

$m=0$

$-\left( \dfrac{3+12 \lambda}{2+6 \lambda} \right)=0$

$3+12 \lambda =0$

$\lambda=\dfrac{-3}{12}= \dfrac{-1}{4}$

The midpoints of the sides of a triangle are

$\left(1,1\right),\left(4,3\right)$ and

$\left(3,5\right)$ . The area of the triangle is ___ square units.

0%
$14$
0%
$16$
0%
$18$
0%
$20$

The angle between the lines

$x\cos{30}^{o}+y\sin{30}^{o}=3$

$x\cos{60}^{o}+y\sin{60}^{o}=5$ is

0%
${90}^{o}$
0%
${30}^{o}$
0%
${60}^{o}$
0%
None of these

Explanation

$\displaystyle x cos30^{\circ}+y sin 30^{\circ} = 3$

$y = -xcot30^{\circ}+3cosec 30^{\circ}$

$m_{1} = -cot 30^{\circ} = -\sqrt{3}$

$x cos60^{\circ}+y sin60^{\circ} = 5$

$y = -xcot60^{\circ}+5cosec60^{\circ}$

$m_{2} = -cot60^{\circ} = -1/\sqrt{3}$

$\displaystyle tan \theta = \left | \dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |$

$= \left | \dfrac{-3+1/\sqrt{3}}{1+(-1/\sqrt{3})(-1/\sqrt{3})} \right |$

$= \displaystyle \left | \dfrac{-3+1}{\sqrt{3}(1+1)} \right | = \left | \dfrac{1}{\sqrt{3}} \right |$

$\theta = 30^{\circ}$

1069943_1166827_ans_8865a91abee2462699f2dc5bcb2f1802.jpg

$B (1, 3)$ is equidistant from

$A (6, -1)$ and

$C (x, 8)$ , then

$x =$

0%
$3 \text { or } - 5$
0%
$- 3 \text { or } 5$
0%
$- 3 \text { or } - 5$
0%
$3 \text { or } 5$

Explanation

$\textbf{Step 1: Apply distance formula between the points.}$

$\text{Given that, B(1,3) is equidistant from the points }\mathrm{A(6,-1)\text{ and }C(x,8).}$

$\text{We know, distance between }\mathrm{(x_1,y_1)\ and\ (x_2,y_2)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\text{Distance between point B and point A,}$

$\Rightarrow\mathrm{BA=}\sqrt{(6-1)^2+(-1-3)^2}$

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

$=\sqrt{25+16}$

$=\sqrt{41}.......(1)$

$\text{Distance between point B and point C,}$

$\Rightarrow\mathrm{BC=}\sqrt{(x-1)^2+(8-3)^2}$

$=\sqrt{x^2+1-2x+25}$

$=\sqrt{x^2-2x+26}......(2)$

$\textbf{Step 2: Apply suitable condition.}$

$\text{Since, B(1,3) is equidistant from the points }\mathrm{A(6,-1)\text{ and }C(x,8).}$

$\mathrm{\therefore BC=BA}$

$\mathrm{\Rightarrow BC^2=BA^2}$

$\Rightarrow x^2-2x+26=41$

$\textbf{[From (1) and (2)]}$

$\Rightarrow x^2-2x-15=0$

$\Rightarrow x^2-5x+3x-15=0$

$\Rightarrow (x-5)(x+3)=0$

$\Rightarrow x=5\text{ or }x=-3$

$\textbf{Hence, the correct option is B.}$

$x+4y=7$ , where

$y\ \in\ N$ , then the minimum positive value of

$x+y$ equals

0%
$4$
0%
$-4$
0%
$3$
0%
$1$

The shortest distance of the point

$(h,k)$ from both the axes are

0%
$|h|,|k|$
0%
$(h,k)$
0%
$\sqrt{h^2+k^2},0$
0%
none of these

Explanation

Given the point is

$(h,k).$

Now foot of the perpendicular

from

$(h,k)$ on x-axis is

$(h,0)$ and

that on y-axis is

$(0,k)$

Now the shortest distance between

$(h,k)$ and

$(h,0) = \sqrt{(h-h)^{2}+(k-0)^{2}}$

$= |k|$ = the shortest distance of

$(h,k)$ from x- axis.

Again the shortest distance between

$(h,k)$ and

$(0,k)$ is

$\sqrt{(h-0)^{2}+(k-k)^{2}} = \left | h \right |$ = The shortest distance of

$(h,k)$ from y-axis.

Angle between

$x = 2$ and

$x - 3y = 6$ is

0%
$\infty$
0%
${\tan ^{ - 1}}\left( 3 \right)$
0%
${\tan ^{ - 1}}\left( {\frac{1}{3}} \right)$
0%
None of these

The points

$(0,1), (-2,3), (6,7), (8,3)$ form

0%
a parallelogram
0%
a rectangle
0%
a rhombus
0%
a quadrilateral

Explanation

$(0, 1)$

$(-2, 3)$

$(6, 7)$

$(5, 3)$

$A$

$B$

$C$

$D$

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

$BC=\sqrt{(6+2)^{2}+(7-3)^{2}}=4\sqrt{5}$

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

$DA=\sqrt{8^{2}+2^{2}}=2\sqrt{17}$

$AC=\sqrt{6^{2}+6^{2}}=2\sqrt{6}$

$BD=\sqrt{10^{2}}=10$

Neither of the sides are equal. then this is a general quadrilateral

One vertex of a square

$ABCD$ is

$A(-1, 1)$ and the equation of one diagonal

$BD$ is

$3x+y-8=0$ then

$C$ =

0%
$(-5, 3)$
0%
$(5, 3)$
0%
$(-5, -3)$
0%
$(2, 5)$

Explanation

Let point

$C (x,y)$ .

Slope of

$AC = \dfrac{{y - 1}}{{x + 1}}$

and slope of

$BD = 3x + y - 8 = 0$

$\begin{array}{l} y=-3x+8\to \left( i \right) \\ \Rightarrow Slope\, \, of\, \, BD\, \, is\, -3 \\ \Rightarrow AC\bot BD \\ \Rightarrow \dfrac { { y-1 } }{ { x+1 } } \times -3=-1 \\ \Rightarrow 3\left( { y-1 } \right) =x+1 \\ \Rightarrow 3y-3=x+1 \\ \Rightarrow 3y-x=4\to \left( { ii } \right) \end{array}$

Solving equation (i) and (ii)

$\begin{array}{l} \Rightarrow \dfrac { x }{ { 20 } } =\dfrac { y }{ { 20 } } =\dfrac { 1 }{ { 20 } } \\ \therefore x=2,y=2 \\ \Rightarrow then,\, \, by\, \, { { using } }\, \, mid\, formula,\dfrac { { x-1 } }{ 2 } =2,x=5 \\ \dfrac { { y+1 } }{ 2 } =2,y=3 \end{array}$

1216673_1206359_ans_a5cb2bcfc4064036af0ccc737fc4ff4a.PNG

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

0%
$\dfrac{3}{2} \lambda$
0%
$3\lambda$
0%
$4\lambda$
0%
$none\ of\ these$

Explanation

Area of

$\triangle \,= \dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$

Given area of triangle whose vertices are

$(2a,b), (a+b,2b+a), (2b,2a)$ is

$\lambda$

$\therefore \lambda = \dfrac{1}{2}\left | 2a(2b+a-2a)+(a+b)(2a-b)+2b(b-2b-a) \right |$

$\lambda = \dfrac{1}{2}\left | 4ab+2a^{2}-4a^{2}+2a^{2}-ab+2ab-b^{2}+2b^{2}-4b^{2}-2ab \right |$

$\lambda = \dfrac{1}{2}\left | 3ab - 3b^{2} \right |$ __ (1)

Now area of triangle whose vertices are

$(a+b,a-b), (3b-a,b+3a), (3a-b,3b-a)$ is given by

Area

$= \dfrac{1}{2}\left | (a+b)(b+3a-3b+a)+(3b-a)(3b-a-a+b)+(3a-b)(a-b-b-3a) \right |$

$= \dfrac{1}{2}\left | (a+b) (-2b+4a)+(3b-a)(4b-2a)+(3a-b)(-2a-2b) \right |$

$= \dfrac{1}{2}\left | -2ab+4a^{2}-2b^{2}+4ab+12b^{2}-6ab-4ab+2a^{2}-6a^{2}-6ab+2ab+2b^{2} \right |$

$= \dfrac{1}{2}\left | 12b^{2}-12ab \right |$

$= \dfrac{4}{2}\left | -(3ab-3b^{2}) \right |$

$= 4.\dfrac{1}{2}\left | 3ab-3b^{2} \right |$

$= 4 \lambda .$

If the points ( 5, 1), (1, p) & (4, 2) are collinear then the value of p will be ________.

0%
1
0%
5
0%
2
0%
-2

The value of m for which the lines

$3x = y - 8$ and

$6x + my + 16 = 0$ coincide is

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

$\left( \mathbf { a } \cos \theta _ { 1 } , \mathbf { a } \sin \theta _ { 1 } \right) , \left( \mathbf { a } \cos \theta _ { 2 } , \mathbf { a } \sin \theta _ { 2 } \right) \text { and } \left( \mathrm { a } \cos \theta _ { 3 } , \mathbf { a } \sin \theta _ { 3 } \right)$ represents the vertices of an equilateral triangle inscribed in a circle, then

0%
$\cos \theta _ { 1 } + \cos \theta _ { 2 } + \cos \theta _ { 3 } = 0$
0%
$\sin \theta _ { 1 } + \sin \theta _ { 2 } + \sin \theta _ { 3 } = 0$
0%
$\tan \theta _ { 1 } + \tan \theta _ { 2 } + \tan \theta _ { 3 } = 0$
0%
$\cot \theta _ { 1 } + \cot \theta _ { 2 } + \cot \theta _ { 3 } = 0$

A triangle with vertices

$(4,0),(-1,-1),(3,5)$ is:

0%
Isosceles and right angled
0%
Isosceles but not right angled
0%
right angled but not isosceles
0%
Neither right angled nor isosceles

Explanation

$\textbf{Step -1: Finding all the side lengths.}$

$\text{Let }(4,0)\text{ be point }A,\;(-1,-1)\text{ be point }B,\;(3,5)\text{ be point }C$

$AB=\sqrt{(4-(-1))^2+(0-(-1))^2}=\sqrt{5^2+1^2}=\sqrt{26}$

$BC=\sqrt{(-1-3)^2+(-1-5)^2}=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}$

$CA=\sqrt{(3-4)^2+(5-0)^2}=\sqrt{(-1)^2+(5)^2}=\sqrt{26}$

$\therefore AB=CA\text{ Hence, the triangle is isosceles.}$

$\textbf{Step -2: Checking whether the triangle is right angled.}$

$AB^2=26$

$BC^2=52$

$CA^2=26$

$\Rightarrow AB^2+CA^2=26+26=52=BC^2$

$\text{Hence, the triangle is right angled.}$

$\textbf{Hence, the triangle with vertices (4,0), (-1,-1), (3,5) is isosceles and right angled.}$

The area of the triangle vertices

$(1,0),(7,0)$ and

$(4,4)$ is ___ square units.

0%
$8$
0%
$10$
0%
$12$
0%
$14$

If the three distance points

$\left( { t }_{ i\quad }2{ at }_{ i }+{ { at }^{ 3 }_{ i } } \right) \quad for\quad i=1,2,3$ are collinear then the sum of the abscissae of the points is

0%
-1
0%
0
0%
1
0%
3

Coordinates of trisection of line joining points (-3, -3) and (6, 6) is -

0%
(0,0),(3,-3)
0%
(0,0),(3,3)
0%
(1,1),(3,3)
0%
(1,1), (-3,3)

Explanation

$C$ divide

$AB$ in ratio

$1:2$

$\Rightarrow$ By Section formula

$C=\left(\dfrac{6-6}{3},\dfrac{6-6}{3}\right)$

$\Rightarrow C(0,0),D$ is mid-point of

$CB$

$\Rightarrow D=\left(\dfrac{6}{2},\dfrac{6}{2}\right)=(3,3)\Rightarrow (B)$

1339172_1266752_ans_7f30eaa05a184ca9ba96615eaedcdb9f.PNG

Find the slope of the line

$y = {x^3} - 3x$ which is parallel to

$2x+18y=9$ .

0%
$2$
0%
$7$
0%
$9$
0%
$-2$

Explanation

$\begin{array}{l} y={ x^{ 3 } }-3x \\ \dfrac { { dy } }{ { dx } } =3{ x^{ 2 } }-3 \\ The\, normal\, is\, parallel\, \, to\, line\, 2x+18y=9 \\ solpe\, of\, the\, normal=\dfrac { { -1 } }{ { \dfrac { { dy } }{ { dx } } } } =\dfrac { { -1 } }{ 9 } \\ \dfrac { { dy } }{ { dx } } =9 \\ 3{ x^{ 2 } }-3=9 \\ \therefore x=\pm 2 \end{array}$

Number of lines that can be drawn through the point (4, -5) so that its distance from (-2, 3) will be equal to 12 is equal to _________.

0%
0
0%
1
0%
2
0%
3

Explanation

Distance between

$(4,-5)$ and

$(-2,3)$ is

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

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

$= \sqrt{36+64}$

$= 10$

$\therefore$ It can never be

$12$ and so line will

be There whose distance from

$(-2,3)$ and

passing The angle

$(4,-5)$ will be

$12$

1479796_1278582_ans_5e649725f6d3406890ad272b2aa08748.jpg

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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