MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 12

If the points

$A(3, 4)$ ,

$B(7, 12)$ and

$P(x, x)$ are such that

$(PA)^{2}> (PB)^{2}> (AB)^{2},$ then integral value of

$x$ can be

0%
$7$
0%
$12$
0%
$16$
0%
$20$

Explanation

consider first condition:

$(PA)^{2}> (PB)^{2}$

$\Rightarrow (x-3)^{2}+(x-4)^{2}> (x-7)^{2}+(x-12)^{2}$ ...(Hint: using distance formula)

$\Rightarrow 2x^{2}-14x+25> 2x^{2}-38x+193$

$\Rightarrow 24x-168> 0\Rightarrow x> 7$
and now consider the second condition:

$(PB)^{2}> (AB)^{2}$

$\Rightarrow (x-7)^{2}+(x-12)^{2}> (3-7)^{2}+(4-12)^{2}=80$ ..(similarly using distance formula).
which is satisfied by

$x=16$ or

$x=20$ .

The vertices of a triangle are

$A(3,4)$ ,

$B(7,2)$ and

$C(-2, -5)$ . Find the length of the median through the vertex A.

0%
$\dfrac { \sqrt { 111 } }{ 2 }$ units
0%
$\dfrac { \sqrt { 147 } }{ 2 }$ units
0%
$\dfrac { \sqrt { 137 } }{ 2 }$ units
0%
$\dfrac { \sqrt { 122 } }{ 2 }$ units

Explanation

A median of a triangle is a line segment that joins the

vertex of a triangle to the midpoint of the opposite side.

Mid point of two points ${ (x }_{ 1 },{ y }_{ 1 })$ and ${ (x }_{ 2 },{ y }_{ 2 })$ is calculated by the formula $\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right)$

Since AD is the median, this means, D is the mid point of BC.

Using this formula, mid point of BC $= \left( \frac { 7- 2 }{ 2 } ,\frac { 2 - 5 }{ 2 } \right) = (\dfrac {5}{2},\dfrac {-3}{2})$

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Hence, length of AD $= \sqrt { \left( \dfrac { 5 }{ 2 } -3 \right) ^{ 2 }+\left( \dfrac { -3 }{ 2 } -4 \right) ^{ 2 } } =\sqrt { \dfrac { 1 }{ 4 } +\dfrac { 121 }{ 4 } } =\sqrt { \dfrac { 122 }{ 4 } } =\dfrac { \sqrt { 122 } }{ 2 }$

The ends of a quadrant of a circle have the coordinates (1, 3) and (3, 1). Then the centre of such a circle is

0%
(2, 2)
0%
(1, 1)
0%
(4, 4)
0%
(2, 6)

Explanation

Let

$O(a,b)$ be the center of the circle.
AB is the chord of the circle. Draw

$OM \perp AB$

$\Rightarrow AM=MB$
So, M is the mid-point of AB.
Coordinates of mid-point M of AB is (2,2)
In right

$\triangle OMA$

$OA^2=OM^2+MA^2$

$\Rightarrow (a-3)^2+(b-1)^2=2+(a-2)^2+(b-2)^2$

$\Rightarrow a=b$ ....(1)
In right

$\triangle OAB$

$AB^2=OA^2+OB^2$

$\Rightarrow 8=2[(a-1)^2+(b-3)^2]$ (

$\because OA=OB$ )

$\Rightarrow a=1\ or\ a=3$
So, the center of circle is (1,1)

If four points are

$A(6,3),B(-3,5),C(4,-2)$ and

$P(x,y),$ then the ratio of the areas of

$\triangle PBC$ and

$\triangle ABC$ is:

0%
$\displaystyle \frac { x+y-2 }{ 7 }$
0%
$\displaystyle \frac { x-y-2 }{ 7 }$
0%
$\displaystyle \frac { x-y+2 }{ 2 }$
0%
$\displaystyle \frac { x+y+2 }{ 2 }$

Explanation

Given: Coordinates of points

$\displaystyle A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 6,3 \right) ,B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -3,5 \right) ,C\left( { x }_{ 3, }{ y }_{ 3 } \right) =\left( 4,-2 \right)$ and

$\displaystyle P\left( x,y \right) .$

We know that the area of:

$\displaystyle\triangle PBC=\frac { 1 }{ 2 } \left[ x\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 3 }\left( y-{ y }_{ 2 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-y \right) \right]$

$\displaystyle=\frac { 1 }{ 2 } \left[ x\left( 5+2 \right) +4\left( y-5 \right) -3\left( -2-y \right) \right]$

$\displaystyle=\frac { 1 }{ 2 } \left[ 7x+7y-14 \right]$

Similarly, the area of

$\displaystyle\triangle ABC=\frac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) \right] +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right)$

$\displaystyle=\dfrac{1}{2}[6\left( 5+2 \right) -3\left( -2-3 \right) +4\left( 3-5 \right)] =\dfrac{49}{2}$

Therefore, the ratio of the areas of

$\triangle PAB$ and

$\triangle ABC$

$\displaystyle=\frac { 7x+7y-14 }{ 49 } =\frac { 7\left( x+y-2 \right) }{ 49 } =\frac { x+y-2 }{ 7 }$

$\mathrm{P}_{1},\ \mathrm{P}_{2},\ldots\ldots.,\ \mathrm{P}_{\mathrm{n}}$ are points on the line

$y=x$ lying in the positive quadrant such that

$\mathrm{O}\mathrm{P}_{\mathrm{n}}=n\cdot\mathrm{O}\mathrm{P}_{\mathrm{n}-1}$ , where

$\mathrm{O}$ is the origin. If

$\mathrm{O}\mathrm{P}_1=1$ and the coordinates of

$\mathrm{P}_{\mathrm{n}}$ are

$(2520\sqrt{2},2520\sqrt{2})$ , then

$n$ is equal to

0%
$3$
0%
$5$
0%
$7$
0%
$9$

Explanation

$P_1,P_2.P_3....$ are points on line

$y=x$

$OP_1=1$

Let

$P_1=(x,x$ )

But

$OP_1=1\Rightarrow 2x^2=1\Rightarrow x^2=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{\sqrt2}$

$P_1=(\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2})$

$OP_2=2\times OP_1=2\Rightarrow 2x^2=4\Rightarrow x=\sqrt2\Rightarrow P_2=(\sqrt2,\sqrt2)$

$OP_3=2\times 3=6\Rightarrow2x^2=6^2\Rightarrow x=3\sqrt2 \Rightarrow P_3=(3\sqrt2,3\sqrt2)$

....coninuing this way get

$OP_7=7\times 720=5040\Rightarrow 2x^2=5040\times 5040\Rightarrow x^2=\Rightarrow x=2520\sqrt2\Rightarrow P_7=(2520\sqrt2,2520\sqrt2)$

Comparing with given point we get

$n=7$ .

Determine the area of the triangle whose vertices are

$\displaystyle \left( \frac { 1 }{ 2 } ,\frac { -1 }{ 2 } \right) ,\left( 2,\frac { -1 }{ 2 } \right)$ and

$\displaystyle \left( 2,\frac { \sqrt { 3 } -1 }{ 2 } \right)$ .

0%
$\displaystyle 3\sqrt { 3 }$
0%
$\displaystyle \frac { 3\sqrt { 3 } }{ 8 }$
0%
$\displaystyle 12\sqrt { 3 }$
0%
$\displaystyle \frac { 3\sqrt { 3 } }{ 4 }$

What is the angle between the straight lines

$\left( { m }^{ 2 }-mn \right) y=\left( mn+{ n }^{ 2 } \right) x+{ n }^{ 3 }$ and

$\left( mn+{ m }^{ 2 } \right) y=\left( mn-{ n }^{ 2 } \right) x+{ m }^{ 3 }$ , where

$m> n$ ?

0%
$\tan ^{ -1 }{ \left( \cfrac { 2mn }{ { m }^{ 2 }+{ n }^{ 2 } } \right) }$
0%
$\tan ^{ -1 }{ \left( \cfrac { 4{ m }^{ 2 }{ n }^{ 2 } }{ { m }^{ 4 }-{ n }^{ 4 } } \right) }$
0%
$\tan ^{ -1 }{ \left( \cfrac { 4{ m }^{ 2 }{ n }^{ 2 } }{ { m }^{ 4 }+{ n }^{ 4 } } \right) }$
0%
${ 45 }^{ o }$

Explanation

Given

$(m^{2} - mn)y = (mn + n^{2})x + n^{3}$

$(mn + m^{2})y = (mn - n^{2})x + m^{3}$

Slope of line

$(1) = \dfrac {mn + n^{2}}{m^{2} -mn}$

Slope of line

$(2) \rightarrow \dfrac {mn - n^{2}}{m^{2} + mn}$

By comparing with

$y = mx + c$

$\tan \theta = \pm \left (\dfrac {m_{1} - m_{2}}{1 + m_{1}m_{2}}\right ), m_{1}$ and

$m_{2}$ are slope

$\therefore \tan \theta = \dfrac {\left (\dfrac {mn + n^{2}}{m^{2} -mn} - \dfrac {mn - n^{2}}{m^{2} + mn}\right )}{\dfrac {1 + (mn + n^{2}(mn - n^{2})}{(m^{2} - mn)(m^{2} + mn)}}$

$\tan \theta = \dfrac {(mn + n^{2}) (m^{2} + mn) - (mn - n^{2})(m^{2} - mn)}{m^{4} - (mn)^{2} + (mn)^{2} - n^{4}}$

$\tan \theta = \dfrac {m^{3}n + (mn)^{2} + (mn)^{2} + mn^{3} - m^{3}n + (m^{2}n^{2}) + (mn)^{2} - mn^{3}}{m^{4} - n^{4}}$

$\tan \theta = \dfrac {4m^{2}n^{2}}{m^{4}- n^{4}}$

$\therefore \theta = \tan^{-1} \left (\dfrac {4m^{2} n^{2}}{m^{4} - n^{4}}\right )$ .

Three points (a, a), (-a, -a) and (

$\displaystyle -a,\sqrt { 3 } ,a\sqrt { 3 }$ ) on the co-ordinate plane are given. Which of the following is correct?

0%
The given points form the vertices of a right angled triangle.
0%
The given points form the vertices of an isosceles triangle.
0%
The given points form the vertices of an equilateral triangle.
0%
The given points are non-collinear.

A vertical line

$l$ passes through the point

$(2,3)$ . A horizontal line

$m$ passes through the point

$(-1,6)$ . Where do lines

$l$ and

$m$ intersect?

0%
$(0,5)$
0%
$(2,6)$
0%
$(6,2)$
0%
$(-1,3)$
0%
$(3,-1)$

Explanation

Given that the vertical line

$l$ passes through the point

$(2,3)$

Which implies that the equation of line

$l$ is

$x=2$

Given that the horizontal line

$m$ passes through the point

$(-1,6)$

Which implies that the equation of line

$l$ is

$y=6$

Therefore the point of intersection of lines

$l$ and

$m$ is

$(2,6)$

The line

$\displaystyle 3x+2y=24$ meets x-axis at A and y-axis at B. The perpendicular bisector of

$\displaystyle \overline { AB }$ meets the line through (0, -1) and parallel to x-axis at C. Find the area of

$\displaystyle \Delta ABC$ .

0%
85 $\displaystyle { units }^{ 2 }$
0%
87 $\displaystyle { units }^{ 2 }$
0%
90 $\displaystyle { units }^{ 2 }$
0%
91 $\displaystyle { units }^{ 2 }$

In which adjacent figures, the path is the shortest path?

Explanation

A. The total path length assuming each side of the square measures one unit is given by

$1 + \sqrt{5} + \sqrt{10} + \sqrt{8}$

B. Similarly, total length is

$1 + 2 + \sqrt{10} + \sqrt{8}$

C. Here, total length

$= \sqrt{5} + 2 + \sqrt{5} + \sqrt{5}$

D. Total length

$= 1 + 2 + \sqrt{13} + \sqrt{5}$

Clearly, Option A is greater than option B and so option A can be neglected for further comparison.

Comparing option B and option D,

$1 + 2 + \sqrt{10} + \sqrt{8}$ (inequality sign)

$1 + 2 + \sqrt{13} + \sqrt{5}$

i.e.

$\sqrt{10} + \sqrt{8}$ (inequality sign)

$\sqrt{13} + \sqrt{5}$

Squaring both sides, we get

$18 + 2\sqrt{80}$ (inequality sign)

$18 + 2\sqrt{65}$

Since

$80 > 65$ , option B is greater than option D.

Comparing C and D,

$2\sqrt{5}$ (inequality sign)

$1 + \sqrt{13}$

Squaring both sides,

$20$ (inequality sign)

$14 + 2\sqrt{13}$

i.e.

$3$ (inequality sign)

$\sqrt{13}$

Since

$3 < \sqrt{13}$ , option C is the shortest path

The points A

$(2a, 4a)$ , B

$(2a, 6a)$ and C

$(2a+\sqrt{3a}, 5a)$ (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

$a$ and

$b$ are real numbers between

$0$ and

$1$ such that the points

$(a,1),(1,b)$ and

$(0,0)$ from an equilateral triangle then the values of

$a$ and

$b$ respectively ?

0%
$2-\sqrt{3},\ 2-\sqrt{3}$
0%
$-2+\sqrt{3},\ -2+\sqrt{3}$
0%
$2\pm \sqrt{3},\ 2\pm \sqrt{3}$
0%
$None\ of\ these$

An equilateral triangle has one vertex at

$(3, 4)$ and another at

$(-2, 3)$ . Find the coordinates of the third vertex.

0%
$(\frac{1+\sqrt3}{2} , \frac{7+\sqrt3}{2})$
0%
$(\frac{1-\sqrt3}{2} , \frac{7-\sqrt3}{2})$
0%
$(\frac{1+\sqrt2}{3} , \frac{7+5\sqrt3}{2})$
0%
both $A$ and $B$

$x$ co-ordinates of two points

$B$ and

$C$ are the roots of equation

$x^{2}+4x+3=0$ and their

$y$ co-ordinate are the roots of equation

$x^{2}-x-6=0$ . If

$x$ co-ordinates of

$B$ is less than

$x$ co-ordinate of

$C$ and

$y$ co-ordinate of

$B$ is greater than the

$y$ co-ordinate of

$C$ and co-ordinate of a third point

$A$ be

$(3,-5)$ , find the length of the bisector of the interior angle at

$A$ ?

0%
$\dfrac{7\sqrt{2}}{3}$
0%
$\dfrac{14\sqrt{2}}{3}$
0%
$\dfrac{5\sqrt{2}}{3}$
0%
$None\ of\ these$

The point whose abscissa is equal to its ordinate and which is equidistant from

$A(5,0)$ and

$B(0,3)$ is

0%
$(1,1)$
0%
$(2,2)$
0%
$(3,3)$
0%
$(4,4)$

Explanation

Let

$P(x,y)$ be the point which is equidistant from

$A(5,0)$ and

$B(0,3)$

Given that in

$P$ , abscissa is equal to ordinate i.e.,

$x=y$ . Also it is given that

$AP=BP$

$\therefore \sqrt{(x-5)^2+(x-0)^2}=\sqrt{(x-0)^2+(x-3)^2}$

Squaring on both sides

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

$\Rightarrow x^2+25-10x+x^2=x^2+x^2+9-6x$

$\Rightarrow 25-10x=9-6x$

$\Rightarrow 4x=16$

$\Rightarrow x=4$

Therefore the point is

$(4,4)$

The area of the triangle whose co-ordinates are

$(2012, 7), (2014, 7)$ and

$(2014, a)$ is

$1 \,sq$ unit. The sum of possible values of

$a$ is

0%
$6$
0%
$8$
0%
$14$
0%
$16$

If a line L is perpendicular to the line 5x - y = 1, and the area of the triangle formed by the line L and the coordinates axes is 5, then the distance of line L from the line x + 5y = 0 is

0%
$\dfrac{7}{\sqrt 5}$
0%
$\dfrac{5}{\sqrt 13}$
0%
$\dfrac{7}{\sqrt 13}$
0%
$\dfrac{5}{\sqrt 7}$

Explanation

Slope of line

$5x-y=1,m_1=5$

$\therefore$ slope of line

$L,m_2=-\dfrac{1}{m_1}=\dfrac{-1}{5}$

$\therefore$ Equation of line

$L :y=\dfrac{-1}{5}x+C$

Y-intercept

$=C$ , then x-intercept

$=5C$

$\therefore$ Area=

$\dfrac{1}{2}(5C)(C)=5$

$\implies C^2=2$

$C=\pm \sqrt{2}$

$\therefore$ Equation of line

$L$ is

$y=-\dfrac{x}{5}\pm \sqrt{2}$

$\implies x+5y=\pm 5\sqrt{2}$

Distance of line

$L$ from

$x+5y=\pm 5\sqrt{2}$

$d=\dfrac{\pm 5\sqrt{2}}{\sqrt{(1)^2+(5)^2}}$

$d=\dfrac{5\sqrt{2}}{\sqrt{26}}=\dfrac{5}{\sqrt{13}}$

Let

$\triangle ABC$ be the right triangle with vertices of

$A\left(0,2\right)$ ,

$B\left(1,0\right)$ and

$C\left(0,0\right)$ . If D is the point on AB such that the segment CD bisects angle C, then the length of CD is ?

0%
$\dfrac{1}{\sqrt{2}}$
0%
$\dfrac{\sqrt{5}}{2}$
0%
$\dfrac{\sqrt{3}}{2}$
0%
$\dfrac{2\sqrt{2}}{3}$

Explanation

Given:

$\triangle{ABC}$ is a Rightangled triangle

From the figure

$C\left(0,0\right)$ is at the origin and

$A\left(1,0\right)$ is at

$x-$ axis and

$B\left(1,0\right)$ is at

$y-$ axis

$\therefore \angle{C}={90}^{0}$ and

$AB$ are joined.

Distance

$BC=1$ unit and

$CA=2$ units

Given that

$CD$ bisects

$angle{C}={90}^{\circ}$

$\therefore \angle{BCD}={45}^{\circ}$

$\sin{{45}^{\circ}}=\dfrac{CD}{AB}$

$\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{CD}{1}$

$\therefore$ Length of

$CD=\dfrac{1}{\sqrt{2}}$ units

If the straight line drawn through the point

$p\left( \sqrt { 3 } ,2 \right)$ making an angle

$\dfrac { \pi }{ 6 }$ with x-axis meet the line

$\sqrt { 3 } x-4y+8=0$ at Q. Then PQ is :-

0%
4
0%
5
0%
6
0%
9

The extremities of the base of an isoscelestriangle ABC are the points

$\mathrm { A } ( 2,0 )$ and

$\mathrm { B } ( 0,1 )$ .If the equation of the side

$\mathrm { AC }$ is

$\mathrm { x } = 2$ then theslope of the side BC is -

0%
$\frac { 3 } { 4 }$
0%
$\frac { 4 } { 3 }$
0%
$\frac { 3 } { 2 }$
0%
$\sqrt { 3 }$

If the line

$y= \sqrt{3x}$ cuts the curve

$x^{4}+ax^{2}y+bxy+cx+dy+6=0$ at

$A,B,C$ and

$D$ , then

$OA.OB.OC.OD$ is equal to (

$O$ being origin)

0%
$a-4b+c$
0%
$2cd$
0%
$96$
0%
$36$

The lines

$\dfrac { x+4 }{ 3 }$ =

$\dfrac { y+6 }{ 5 }$ =

$\dfrac { z-1 }{ -2 }$ and 3x-2y+z+5=0=2x+3y+4z-k are coplanar for _______.

0%
4
0%
0
0%
6
0%
5

A circle touches the y-axis at the point (0,4) and cuts the x-axis in a chord of length 6 units . The radius of the circle is

0%
3
0%
4
0%
5
0%
6

If the tangent at

$\theta =\frac { \pi }{ 4 }$ to the curve

$x=a\cos ^{ 3 }{ \theta } ,y=a\sin ^{ 3 }{ \theta }$ meets the x and y axes in A and B then the area of the triangle OAB is

0%
$\frac { { a }^{ 2 } }{ 4 } sq.units$
0%
$\frac { { a }^{ 2 } }{ 2 } sq.units$
0%
$\frac { { 3a }^{ 2 } }{ 4 } sq.units$
0%
$\frac { { 5a }^{ 2 } }{ 4 } sq.units$

$A = (-3,4) , B =(-1,-2) , C=(5,6) D= (x,-4)$ are the vertices of a quadrilateral such that area triangle

$ABD= 2 \times$ (area of a triangle

$ACD$ ), then

$x =$

0%
6
0%
9
0%
69
0%
96

A line L passes through the points

$(1,1)$ and

$(2,0)$ and another line

$L^{\prime}$ passes through

$\left(\frac{1}{2}, 0\right)$ and perpendicular to L.Then the area of the triangle formed by the lines

$L, L^{\prime}$ and

$y-$ axis, is

0%
$15 / 8$
0%
$25 / 4$
0%
$25 / 8$
0%
$25 / 16$

Explanation

$\textbf{Hint: Product of slopes of two perpendicular lines is -1.}$

$\textbf{Step 1: Find the equation of two lines.}$

$\text{Using two points form, equation of line L is given by,}$

$\Rightarrow y-0=\dfrac{0-1}{2-1}(x-2)$

$\Rightarrow y=-(x-2)$

$\Rightarrow x+y=2.......(1)$

$\text{Slope of L = - 1}$

$\text{L and }$

$\text{L' are perpendicular to each other}$

$\text{Slope of L' =}$

$\dfrac{-1}{-1}=1$

$\text{Using point slope form, equation of line L' is given by,}$

$\Rightarrow y-0=1(x-\dfrac{1}{2})$

$\Rightarrow 2x-2y=1 ......(2)$

$\textbf{Step 2: Find the required area.}$

$\text{Let, L and L' intersect each other at B.}$

$\text{Solving eq(1) and eq(2) , we get,}$

$\Rightarrow A(x,y)=\left(\dfrac{5}{4},\dfrac{3}{4}\right)$

$\text{The intersection points of lines L and L' with y-axis are B(0,2) and C}$

$\left(0,-\dfrac{1}{2}\right)$

$\text{respectively.}$

$\text{Using distance formula,}$

$AB=\sqrt{\left(0-\dfrac{5}{4}\right)^2+\left(2-\dfrac{3}{4}\right)^2}=\dfrac{5}{4}\sqrt2$

$\text{And,}$

$AC=\sqrt{\left(\dfrac{5}{4}-0\right)^2+\left(\dfrac{3}{4}+\dfrac{1}{2}\right)^2}=\dfrac{5}{4}\sqrt2$

$\text{Area of the triangle}$

$=\dfrac{1}{2}\times AB\times AC=\dfrac{1}{2}\times\dfrac{5}{4}\sqrt2\times\dfrac{5}{4}\sqrt2\ sq.unit$

$=\dfrac{25}{16}\ sq.unit$

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

The area of the triangle formed by the lines

$x=0;y=0$ and

$x\sin { { 18 }^{ 0 } } +y\cos { { 36 }^{ 0 } } +1=0$ is

0%
1
0%
2
0%
3
0%
4

The line

$( k + 1 ) ^ { 2 } x + k y - 2 k ^ { 2 } - 2 = 0$ passes through a fixed point. Distance of this point (-2, -1) is

0%
3
0%
4
0%
5
0%
6

the points on the curve

$y={ x }^{ 3 }$ , the tangents at which are inclined at an angle

${ 60 }^{ 0 }$ to x-axis is

0%
$\quad \quad \quad \left( { 3 }^{ -1/4 },{ 3 }^{ -3/4 } \right)$
0%
$\quad \left( 3^{ -1/2 },{ 3 }^{ -2/5 } \right) ,\quad \left( -3^{ 1/3 },{ -3 }^{ -2/3 } \right)$
0%
$\quad \left( { 2 }^{ 1/4 },{ 2 }^{ -2/5 } \right) , \quad \left( -3^{ 1/2 },{ -3 }^{ -1/2 } \right)$
0%
none

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 12 - 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 12 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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