MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 5

$M (x, y)$ is equidistant from

$A (a + b, b - a)$ and

$B (a - b, a + b)$ , then

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$bx + ay =0$
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$bx - ay = 0$
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$ax + by = 0$
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$ax - by = 0$

Explanation

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

$\text{Let, M(x,y) is equidistant from the points }\mathrm{A(a+b,b-a)\text{ and }B(a-b,a+b).}$

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

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

$\Rightarrow\mathrm{MA=}\sqrt{(a+b-x)^2+(b-a-y)^2}$

$=\sqrt{a^2+b^2+x^2+2ab-2ax-2bx+b^2+a^2+y^2-2ab-2by+2ay}$

$=\sqrt{a^2+b^2+x^2-2ax-2bx+b^2+a^2+y^2-2by+2ay}.....(1)$

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

$\Rightarrow\mathrm{MB=}\sqrt{(a-b-x)^2+(a+b-y)^2}$

$=\sqrt{a^2+b^2+x^2-2ab-2ax+2bx+a^2+b^2+y^2+2ab-2by-2ay}$

$=\sqrt{a^2+b^2+x^2-2ax+2bx+a^2+b^2+y^2-2by-2ay}.....(2)$

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

$\text{Point M(x,y) is equidistant from the points }\mathrm{A(a+b,b-a)\text{ and }B(a-b,a+b).}$

$\mathrm{\therefore MA=MB}$

$\mathrm{\Rightarrow MA^2=MB^2}$

$\Rightarrow -2ax-2bx-2by+2ay=-2ax+2bx-2by-2ay$

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

$\Rightarrow -2bx+2ay=2bx-2ay$

$\Rightarrow 4bx-4ay=0$

$\Rightarrow bx-ay=0$

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

$(3, 1), (-3, 2)$ and

$\displaystyle (0,2-\sqrt{3})$ are the vertices of __________ triangle of area ___________.

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an isosceles, $81$ sq. units
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a scalene, $\cfrac {-3+6\sqrt{3}}{2}$ sq. units
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an equilateral, $\displaystyle 9\sqrt{3}$ sq. units
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a right angled, $81$ sq. units

Explanation

Distance between two points $= \sqrt { \left( { x }_{ 2 }-{x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$
Distance between the points $A (3,1)$ and $B (-3,2) = \sqrt { \left( -3-3 \right) ^{ 2 }+\left( 2 - 1 \right) ^{ 2 } } = \sqrt { 36 + 1 } = \sqrt { 37 }$

Distance between the points $B(-3,2)$ and $C (0,2-\sqrt {3}) = \sqrt { \left( 0 + 3 \right) ^{ 2 }+\left(2 - \sqrt {3} - 2\right) ^{ 2 } } = \sqrt { 9 + 3 } = \sqrt { 12 }$

Distance between the points $A(3,1)$ and $C(0,2-\sqrt {3})$

$= \sqrt { \left( 0-3\right) ^{ 2 }+\left( 2-\sqrt {3} - 1\right) ^{ 2 } } = \sqrt { 9 + 1 + 3 -2\sqrt {3} } = \sqrt { 13 - 2\sqrt {3} }$

Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.

Area of a triangle $= \left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Area of $\triangle ABC= \left| \cfrac { (3)(2-2+\sqrt {3})+(-3)(2-\sqrt {3} - 1)+0(1-2) }{ 2 } \right|$

$= \left| \cfrac { 3\sqrt {3} -3 + 3\sqrt {3} }{ 2 } \right|$

$= \cfrac{-3 + 6 \sqrt {3}}{2}$ sq. units

The angle between the lines

$y -x + 5 = 0$ and

$\sqrt 3x - y + 7= 0$ is/are

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$15^o$
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$60^o$
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$165^o$
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$75^o$

Explanation

Slope of line

$y-x+5=0$ is

${ m }_{ 1 }=1$
And slope of line

$\sqrt { 3 } x-y+7=0$ is

${ m }_{ 2 }=\sqrt { 3 }$
Then angle between them is

$\displaystyle \tan { \theta } =\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { 1-\sqrt { 3 } }{ 1+\sqrt { 3 } } \right|$
Hence

$\theta ={ 15 }^{ 0 },{ 165 }^{ 0 }$

The points

$A(a, b + c), B(b, c + a)$ and

$C(c, a + b)$ are:

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collinear
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doesn't lie in the same plane
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doesn't lie on the same line
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nothing can be said

Explanation

Slope of

$AB=\dfrac{(c+a)-(b+c)}{b-a}=\dfrac{a-b}{b-a}=-1$
Slope of

$BC=\dfrac{(b+a)-(a+c)}{c-b}=\dfrac{b-c}{c-b}=-1$

This is only possible if three points lie on same line.
Therefore collinear.

Find the area of the triangle whose vertices are

$(a, b + c), (a, b - c)$ and

$(-a, c)$ .

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$2ac$
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$2bc$
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$b(a + c)$
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$c (a - b)$

Explanation

Area of a triangle

$(A) =\left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (a,b+c)$ ,

$({ x }_{ 2 },{ y }_{ 2 }) = (a,b-c)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (-a,c)$

$A=\left| \cfrac { a(b-c-c)+a(c-b-c)-a(b+c-b+c) }{ 2 } \right|$

$=\left| \cfrac { a(b-2c)+a(-b)-a(2c) }{ 2 } \right|$

$=\left| \cfrac { ab-2ac-ab-2ac }{ 2 } \right|$

$=\left| \cfrac { -4ac }{ 2 } \right|$

$=2ac$ square units

The triangle with vertices A(4, 4), B(-2, -6) and C(4, -1) is shown in the diagram. The area of

$\Delta$ ABC is _______

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5 sq. units
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12 sq. units
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15 sq. units
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20 sq. units

Explanation

We know that the area of the triangle whose vertices are

$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and

$(x_{3},y_{3})$ is

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

$=\dfrac{1}{2}|4(-6-(-1))-2(-1-4)+4(4-(-6))|$

$=\dfrac{1}{2}|4(-5)-2(-5)+4(10)|$

$=\dfrac{1}{2}|30|$

$=15$ sq. units

$\triangle OAB$ is an equilateral triangle where

$\displaystyle O\equiv \left( 0,0 \right) ,A\equiv \left( 1,\frac { 1 }{ \sqrt {3}}\right)$ . The co-ordinates of point

$B$ can be :-

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$\displaystyle \left( \frac { 2 }{ \sqrt { 3 } } ,0 \right)$
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$\displaystyle \left( 0,-\frac { 1 }{ \sqrt { 3 } } \right)$
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$\displaystyle \left( 0,\frac { 1 }{ \sqrt { 3 } } \right)$
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$\displaystyle \left( 1,-\frac { 1 }{ \sqrt { 3 } } \right)$

Explanation

Lenth of sides is

$\displaystyle \sqrt { { \left( 1-0 \right) }^{ 2 }+{ \left( \frac { 1 }{ \sqrt { 3 } } -0 \right) }^{ 2 } } =\sqrt { 1+\frac { 1 }{ 3 } } =\sqrt { \frac { 4 }{ 3 } }$
Then from options only point

$\displaystyle \left( 1,-\frac { 1 }{ \sqrt { 3 } } \right) \quad$
gives the same length

Supriya is standing in the Sun close to a lamp post. Supriya is 5 feet tall and her shadow is 2 feet long. The shadow of the lamp post is 8 feet long. How tall is the lamp post ?

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25ft
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18ft
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16ft
0%
14ft

Explanation

R.E.F image

$\Delta ABE$ and

$\Delta CDE$

$\angle B = \angle D = 90^{\circ}$

$\angle E = \angle E$

$\therefore \Delta ABC \sim \Delta CDE$

Therefore,

$\dfrac{AB}{CD} = \dfrac{BE}{DE}$

$\therefore \dfrac{AB}{5} = \dfrac{10}{2}$

$AB = 25\,fts$

$\boxed {Height\,of\,lamp\,post = 25\,ft}$

1172196_315415_ans_f87aefd75b4e4b92b3ffce3023318c69.jpg

Three points A, B and C have coordinates

$(a, b + c), \ (b, c + a)$ and

$(c, a + b)$ , respectively. The area of the triangle ABC will be:

0%
$\displaystyle a^{2}+b^{2}+c^{2}$
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$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$
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$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$
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$0$

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 the points:

$A(a, b+c) , B(b, c+a) , C(c, a+b)$

Therefore, area:

$A =\dfrac{ 1 }{ 2 }\left[ a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a) \right]$

$A =\dfrac{ 1 }{ 2}{\left[ a(c-b)+b(a-c)+c(b-a) \right] }$

$A=\dfrac{ 1 }{ 2}{\left[ ac-ab+ba-bc+cb-ac \right] }=0$

A special fully automatic car is designed by the Indian scientist in the Hindustan Automobiles Ltd The car follows only the following instructions

$\displaystyle G_{1}$ (x) : The car shall move forward to x meters

$\displaystyle G_{2}$ (x) : The car shall turn in right direction and move x meters

$\displaystyle G_{3}$ (x) : The car shall turn in left direction and move x meters

$\displaystyle G_{4}$ (x) : The car shall move left y meetrs

The car is given instruction

$\displaystyle G_{1}$ (100),

$\displaystyle G_{3}$ (50),

$\displaystyle G_{4}$ (10). Assume that car was initially at origin Find the shortest distance of the car from the original position

0%
$\displaystyle 10\sqrt{105}$
0%
$\displaystyle 10\sqrt{106}$
0%
$\displaystyle 8\sqrt{106}$
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$\displaystyle 5\sqrt{106}$

Explanation

Suppose the car is at origin

$(0,0)$ at the start.

After instruction G1, the car will move

$100 m$ forward, and be at the position

$(0,100 )$
After instruction G3, the car will move

$50 m$ left, and be at the position

$(-50,100 )$
After instruction G4, the car will move

$10 m$ left, and be at the position

$(-50,90 )$

Distance between the points

$(0,0)$ and D

$(-50,90) = \sqrt { \left( -50-0 \right) ^{ 2 }+\left( 90-0 \right) ^{ 2 } } = \sqrt { 2500 + 8100 } = \sqrt { 10600 } = 10 \sqrt {106}$

If the coordinates of two points A and B are

$(3, 4)$ and

$(5, -2)$ , respectively, then the coordinates of any point P if

$PA = PB$ and area of

$\displaystyle \Delta PAB=10$ is

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

Explanation

Let P

$= (x, y)$
Given,

$PA = PB$

$=> {PA}^{2} = {PB}^{2}$

$=> {(x-3)}^{2} + {(y-4)}^{2} = {(x-5)}^{2} + {(y+2)}^{2}$

$=> {x} ^{2} -6x + 9 + {y} ^{2} -8y + 16 = {x}^{2} -10x + 25 + {y}^{2} + 4y + 4$

$=> 4x - 12y = 4$

$=> x - 3y = 1$ .......(i)

Also, Area of

$\Delta PAB=10$

$\left| \cfrac {x (4+2)+3(-2-y)+5(y-4)}{ 2 } \right|=10$

$\left| \cfrac { 6x -6-2y +5y -20 }{ 2 } \right| = 10$

$\cfrac {6x+3y -26}{2} = \pm 10$

$6x+2y-26= \pm 20$

$6x+2y = 46$ ........(ii)
or

$6x + 2y = 6$ .......(iii)
Solving equation (i) and (ii), we get

$x = 7, y = 2$

Solving equation (i) and (iii), we get

$x = 1, y = 0$ .
So, the co-ordinates of P are

$(7, 2)$ or

$(1, 0)$ .

If the slope of a line passing through the point A(3, 2) be

$\displaystyle \frac{3}{4}$ then the points on the line which are 5 units away from A are

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$(5, 5), (-1, -1)$
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$(7, 5), (-1, -1)$
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$(5, 7), (-1, -1)$
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$(7, 5), (1, 1)$

Explanation

Given a point A(3,2) and slope

$\dfrac{3}{4}$ .

The equation of slope is given by ,

$y-{ y }_{ 1 }=m(x-{ x }_{ 1 })$

$\\ \Rightarrow y-2=\frac { 3 }{ 4 } (x-3)\\ \Rightarrow 3x-4y-1=0$

The points which are 5 units away from A are when

$x=7$

$3\times7-4y-1=0$

$\therefore y=5 \rightarrow(7,5)$

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

$(7, 6)$ and

$(-3, 4)$

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

Explanation

Let the point on the x-axis be $(x,0)$ .
Distance between $(x,0)$ and $(7,6) = \sqrt { \left( 7-x \right) ^{ 2}+\left( 6 - 0 \right) ^{ 2 } } = \sqrt { { 7}^{ 2 }+ { x }^{ 2 } - 14x + 36 }= \sqrt { { x }^{ 2 } - 14x + 85 }$
Distance between $(x,0)$ and $(-3,4) = \sqrt { \left( -3-x \right) ^{ 2}+\left( 4 - 0 \right) ^{ 2 } } = \sqrt { 9 + { x }^{ 2 } + 6x + 16 } =\sqrt { { x }^{ 2 } + 6x + 25 }$
As the point $(x,0)$ is equidistant from the two points, both the distances calculated are equal.
$\sqrt { { x }^{ 2 } - 14x + 85 } = \sqrt { { x }^{ 2 } + 6x + 25}$
$\Rightarrow { x }^{ 2 } - 14x + 85 = { x }^{ 2 } + 6x + 25$
$\Rightarrow 85 - 25 = 6x + 14x$
$\Rightarrow 60 = 20x$

$\Rightarrow x = 3$
Thus, the point is $(3,0)$ .

If the co-ordinates of two points A and B are (3, 4) and (5, -2) respectively then the co-ordinates of any point P if PA = PB and area of

$\displaystyle \Delta PAB=10$ is

0%
(7, 2) or (1, 0)
0%
(-7, 2) or (3, 0)
0%
(7, -2) or (5, 0)
0%
(7, -2) or (-1, 0)

Explanation

$4x - 12y = 4$
$x - 3y = 1$ .......(i)
Also Area of $PAB=10$
$| \dfrac {(x (4+2)+3(-2-y)+5(y-4)}{ 2 } =10$
$\left| \dfrac { 6x -6-2y +5y -20 }{ 2 } \right| = 10$
$\dfrac {6x+3y -26}{2} = \pm 10$
$$ 6x+2y26=
\pm 20 $$
$$ 6x+2y =46 $........(ii) or$ 6x + 2y = 6 $.......(iii) Solving\ equation\ (i)\ and\ (ii)$ x = 7, y = 2 $$ and solving equation (i) and (iii) $x = 1, y = 0$

So, the co-ordinates of P are $(7, 2) or (1, 0)$

Find the point P(x, y) if its distance from (-3, 0) & (3, 0) is 4 units individually

0%
$\displaystyle \left ( 0,\sqrt{5} \right )$
0%
$\displaystyle \left ( 0,-\sqrt{5} \right )$
0%
$\displaystyle \left ( 0,-\sqrt{7} \right )$
0%
$(1, 0)$

Explanation

Distance

between two points

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

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula

$\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$
Given,
Distance

between the points

$(x,y) ; (-3,0) =$ 4

between

$(x,y) ; (3,0)= 4$

Solving,we get

$=> 12x = 0$

$=> x = 0$

Now,

Distance

between the points

$(x,y) ; (-3,0) = 4$

$=> \sqrt { \left( -3-x \right) ^{ 2 }+\left( 0 - y \right) ^{ 2 } } = 4$

But $x = 0$

$=> \sqrt { 9 + { y }^{ 2 } } = 4$

$=> 9 + {y}^{2} = 16$

${y}^{2} = 7$

$=> y = \sqrt {7}$ or $- \sqrt {7}$

$A$ &

$B$ are the points

$(-3, 4)$ and

$(2, 1)$ , then the coordinates of the point

$C$ on produced

$AB$ such that

$AC = 2BC$ are

0%
$(2, 4)$
0%
$(3, 7)$
0%
$(7, -2)$
0%
$\displaystyle \left ( \frac{1}{2},\frac{5}{2} \right )$

Explanation

Using the section formula, if a point $(x,y)$ divides the

line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ externally in the ratio $m:n$ , then $(x,y) = \left( \dfrac { m{ x }_{ 2 }-n{ x }_{ 1 } }{ m-n } ,\dfrac { m{ y }_{ 2 }-n{ y }_{ 1 } }{ m-n } \right)$

Since, $AC = 2BC => \dfrac { AC }{ BC } =\dfrac { 2 }{ 1 }$
Substituting $({ x }_{ 1 },{ y }_{ 1 }) = (-3,4)$ and $({x }_{ 2 },{ y }_{ 2 }) = (2,1)$ and $m = 2, n = 1$ in the section formula, we get

$C = \left( \dfrac { 2(2)-1(-3) }{ 2-1 } ,\dfrac { 2(1)-1(4) }{ 2-1 } \right) =\left( 7,-2 \right)$

AOBC is a rectangle whose three vertices are A (0, 3) O (O, 0) and B (5, 0). The length of its diagonal is

0%
5
0%
3
0%
4
0%
$\displaystyle \sqrt{34}$

Explanation

Length of diagonal is the distance between the points A and B.

Distance

between two points

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

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula

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

Distance between the points $(0,3)$ and $(5,0) = \sqrt { \left( 5-0\right) ^{ 2 }+\left( 0-3\right) ^{ 2 } } = \sqrt { 25 + 9 } = \sqrt { 34 }$

The coordinates of

$A$ for which area of triangle, whose vertices are

$A(a, 2a),\ B(-2, 6)$ and

$C(3, 1)$ is

$10$ square units, are:

0%
$(0,\:3)$
0%
$(5,\:8)$
0%
$(\displaystyle 3, \frac{8}{3})$
0%
None of these

Explanation

Given: Vertices of the triangle are

$A(a, 2a),\ B(-2, 6),\ C(3,1)$

The area of the triangle is

$10$ square units.

$\Rightarrow \dfrac{1}{2}\bigg | a(6-1)+(-2)(1-2a)+3(2a-6) \bigg |=10$

$\Rightarrow 5a-2+4a+6a-18=20$

$\Rightarrow 15a=40$

$\Rightarrow a=\dfrac{40}{15}=\dfrac{8}{3}$

$\therefore$ The coordinates of vertex

$A$ are

$\displaystyle \left ( \dfrac{8}{3}, \frac{16}{3} \right )$

Find a point on the X-axis which is equidistant from the points

$(5, 4)$ and

$(-2, 3)$ .

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

Explanation

Let the point on the x-axis be $(x,0)$
Distance between $(x,0)$ and $(5,4) = \sqrt { \left( 5-x \right) ^{ 2}+\left( 4 - 0 \right) ^{ 2 } } = \sqrt { { 5}^{ 2 }+ { x }^{ 2 } - 10x + 16 }= \sqrt { { x }^{ 2 } - 10x + 41 }$
Distance between $(x,0)$ and $(-2,3) = \sqrt { \left( -2-x \right) ^{ 2}+\left( 3 - 0 \right) ^{ 2 } } = \sqrt { { 2}^{ 2 }+ { x }^{ 2 } + 4x + 9 } =\sqrt { { x }^{ 2 } + 4x + 13 }$

As the point $(x,0)$ is equidistant from the two points, both the distances calculated are equal.

$\sqrt { { x }^{ 2 } - 10x + 41 } = \sqrt { { x }^{ 2 } + 4x + 13 }$

$=> { x }^{ 2 } - 10x + 41 = { x }^{ 2 } + 4x + 13$

$41 - 13 = 10x + 4x$

$28 = 14x$
$x = 2$ . Thus, the point is $(2,0)$ .

If A(2, 2), B(-4, -4), C(5, -8) are the vertices of any triangle the length of median passes through C will be

0%
$\displaystyle \sqrt{65}$
0%
$\displaystyle \sqrt{117}$
0%
$\displaystyle \sqrt{85}$
0%
$\displaystyle \sqrt{113}$

Explanation

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

the midpoint of the opposite side.

Median through C will pass through midpoint of AB.

Mid point of two points

${ (x }_{ 1 },{ y }_{ 1 })$ and

${ (x }_{ 2 },{ y }_{ 2 })$ is calculated by the formula

$\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right)$

Using this formula,

mid point of AB $= D = \left( \dfrac { 2-4 }{ 2 } ,\dfrac { 2-4 }{ 2 } \right) \quad =\quad (-1,-1)$

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

Distance between the points C $(5,-8)$ and D $(-1,-1) = \sqrt { \left( -1-5 \right) ^{ 2 }+\left( -1 +8 \right) ^{ 2 } } = \sqrt { 36 + 49} = \sqrt {85 }$

If the area of a triangle formed by the points (k, 2k) (-2, 6) and (3, 1) is 20 square units. Find the value of k.

0%
$5$
0%
$4$
0%
$\displaystyle \frac{3}{5}$
0%
$\displaystyle \frac{2}{3}$

Explanation

Area of a triangle with vertices

$({ x }_{ 1 },{ y }_{ 1 })$ ;

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is

$\left| \frac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Hence, area of the triangle with given vertices

$\left| \frac { k(6-1)-2(1-2k)+3(2k-6) }{ 2 } \right| = 20$

$\Longrightarrow \left| \frac { 6k-k-2+4k+6k-18 }{ 2 } \right| = 20$

$\left| \frac { 15k-20 }{ 2 } \right| = 20$

$\left| 15k-20 \right| = 40$

$15k - 20 = 40$

$15k = 60$

$k = 4$

The distance between

$(x+y, x-y), (x-y, x+y)$ is

0%
$0$
0%
$\sqrt2$ xy
0%
$2\sqrt 2$ y
0%
$2$ xy

Explanation

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated

using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Distance between the points $(x+y,x-y)$ and $(x-y,x+y) = \sqrt { \left(x-y-x-y\right) ^{ 2 }+\left( x+y-x+y \right) ^{ 2 } } = \sqrt { { (2y) }^{ 2 }+{ (2y) }^{ 2 } } = 2\sqrt {2} y$

A(1, 1) and B(2, -3) are two points and D is a point on AB produced such that AD = 3 AB Then the co-ordinates of D is

0%
(4, 11)
0%
(4, -11)
0%
(-2, 5)
0%
(-4, -11)

Explanation

Since, $AD = 3AB => \dfrac { AD }{ BD} =\dfrac { 3 }{2 }$

Using the section formula, if a point $(x,y)$ divides the

line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ externally in the ratio $m:n$ , then $(x,y) = \left( \dfrac { m{ x }_{ 2 }-n{ x }_{ 1 } }{ m-n } ,\dfrac { m{ y }_{ 2 }-n{ y }_{ 1 } }{ m-n } \right)$

Substituting $({ x }_{ 1 },{ y }_{ 1 }) = (1,1)$ and $({x }_{ 2 },{ y }_{ 2 }) = (2,-3)$ and $m = 3, n = 2$ in the section formula, we get

$C = \left( \dfrac { 3(2)-2(1) }{ 3-2 } ,\dfrac { 3(-3)-2(1) }{ 3-2 } \right) =\left( 4,-11 \right)$

In the diagram PQR is an isosceles triangle and QR = 5 units
The coordinates of Q are

0%
(1, 5)
0%
(3, 4)
0%
(2, 4)
0%
(1, 4)

Explanation

Given triangle PQR is an isosceles triangle having PQ = QR.

Let the co-ordinates of Q be (x, y).

Therefore,

${ PQ }^{ 2 }={ QR }^{ 2 }$ .

${ (x+2) }^{ 2 }+{ (y-0) }^{ 2 }={ (x-4) }^{ 2 }+{ (y-0) }^{ 2 }\\ \Longrightarrow { x }^{ 2 }+4+4x={ x }^{ 2 }+16-8x\\ \Longrightarrow 12x=12\\ \Longrightarrow x=1$

Again, QR = 5 units.

$\therefore { (x-4) }^{ 2 }+{ (y-0) }^{ 2 }={ 5 }^{ 2 }\\ \Longrightarrow { (1-4) }^{ 2 }+{ (y) }^{ 2 }={ 5 }^{ 2 }\\ \Longrightarrow { (y) }^{ 2 }=25-9=16\\ \Longrightarrow y=\pm 4$

Now, y can't be negative.

Therefore, y = 1.

So, coordinates of Q = (x, y) = (1, 4).

Value of a when the distance between the points

$(3, a)$ and

$(4, 1)$ is

$\displaystyle \sqrt{10}$ is

0%
$4\ or -2$
0%
$-2\ or\ 4$
0%
$6\ or\ 2$
0%
None

Explanation

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the

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

Distance between the points $(3,a)$ and $(4,1) = \sqrt {10}$

$=> \sqrt { \left( 4-3 \right) ^{ 2 }+\left( 1 - a \right) ^{ 2 } } = \sqrt {10}$

$\sqrt { 1 + { (1 - a) }^{ 2 } } = \sqrt {10}$

$1 + { (1 - a) }^{ 2 } = 10$

${ (1-a) }^{ 2 } = 9$

$1-a = 3$ or $- 3$

$=> a = -2$ or $4$

Do the points

$(-2, 5), (3, -4)$ and

$(7, 10)$ represent the vertices of a right triangle?

0%
Yes
0%
No
0%
Cannot be determined
0%
None

Explanation

Let $A (2,5), B (3,-4)$ and $C (7,10)$

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

Distance between the points A and B $AB = \sqrt { \left( 3+2 \right) ^{ 2 }+\left( -4-5\right) ^{ 2 } } = \sqrt { 25 + 81 } = \sqrt { 106 }$

Distance between the points B and C $BC = \sqrt { \left(7-3 \right) ^{ 2 }+\left( 10+4\right) ^{ 2 } } = \sqrt { 16 + 196 } = \sqrt { 212 }$

Distance between the points A and C $AC = \sqrt { \left(7+2 \right) ^{ 2 }+\left(10-5\right) ^{ 2 } } = \sqrt { 81 + 25 } = \sqrt { 106 }$

Since, ${(\sqrt { 212}) }^{2} = {(\sqrt { 106 }) }^{2} + {(\sqrt { 106 })}^{2}$ , the triangle is a right angled triangle.

If the distance between the points (a, 2) and (3, 4) be 8 then a =

0%
$\displaystyle 2+3\sqrt{15}$
0%
$\displaystyle 2-3\sqrt{15}$
0%
$\displaystyle 2\pm 3\sqrt{15}$
0%
$\displaystyle 3\pm 2\sqrt{15}$

Explanation

Distance between $(a,2)$ and $(3,4) = 8$

$\sqrt { \left( 3-a \right) ^{ 2 }+\left( 4-2 \right) ^{ 2 } } = 8$

$\sqrt { 9 + { a }^{ 2 } - 6a + 4 } = 8$

$\sqrt { { a }^{ 2 } - 6a + 13 } = 8$

${ a }^{ 2 }- 6a + 13 = 64$

${ a }^{ 2 } - 6a - 51 = 0$

$a= \dfrac { -(-6)\pm \sqrt { \left( { -6 }^{ 2 } \right) -4(1)(-51) } }{ 2(1) }$

$=> a = \dfrac {6 \pm \sqrt{36 + 204}}{2}$

$=> a = 3 \pm 2\sqrt {15}$

If the area of a triangle is

$68$ sq. units and the vertices are

$(6, 7), (-4, 1)$ and

$(a, -9)$ then the value of

$a$ is

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

Explanation

Area of a triangle with vertices

$({ x }_{ 1 },{ y }_{ 1 })$ ;

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is:

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

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (6,7)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (-4,1)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (a,-9)$ in the formula for area, we get:

Area of triangle

$= \left| \dfrac { (6)(1+9)+(-4)(-9-7) + a(7-1) }{ 2 } \right| = 68$

$\left| \dfrac { 60 + 64 + 6a }{ 2 } \right| = 68$

$\dfrac{124 + 6a}{2} = 68$

$124 + 6a = 136$

$6a = 12$

$\implies a = 2$

The value of k when the distance between the points

$(3,k)$ and

$(4,1)$ is

$\displaystyle \sqrt{10}$ is

0%
$3 or 4$
0%
$-4 or -2$
0%
$-4 or 2$
0%
$4 o r-2$

Explanation

Consider given, distance between the points $(3,k)$ and $(4,1)$ is $\sqrt{10}$ .

Now,

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

${{\left( -1 \right)}^{2}}+{{\left( k-1 \right)}^{2}}=10$

${{\left( k-1 \right)}^{2}}=9$

$k-1=\pm 3$

If,

$k-1=-3$

$k=-2$

if,

$k-1=+3$

$k=4$

Hence, this is the answer.

If P is (-3, 4) and My Mx (P) shows the reflection ofthe point P in the x-axis and then the reflection of the image in the y-axis, then

$M_y M_x (P)$ is

0%
(3,4)
0%
(-3,-4)
0%
(-3,4)
0%
(3,-4)

Explanation

$\because$ P is (-3, 4)

$\therefore$ Reflection of P in the X-axis is

$M_x P = (-3, -4)$
Hence, the reflection of the image in the Y-axis is

$M_y M_x$ (P) is (3, - 4).

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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