Straight Lines Test 32

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Straight Lines Test 32

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Question 1
1 / -0

If $$a$$ 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 ?

A
$$2-\sqrt{3},\ 2-\sqrt{3}$$

B
$$-2+\sqrt{3},\ -2+\sqrt{3}$$

C
$$2\pm \sqrt{3},\ 2\pm \sqrt{3}$$

D
$$None\ of\ these$$

Solution
Question 2
1 / -0

$$x$$ 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 $$A$$ ?

A
$$\dfrac{7\sqrt{2}}{3}$$

B
$$\dfrac{14\sqrt{2}}{3}$$

C
$$\dfrac{5\sqrt{2}}{3}$$

D
$$None\ of\ these$$

Solution
Question 3
1 / -0

The point whose abscissa is equal to its ordinate and which is equidistant from $$A(5,0)$$ and $$B(0,3)$$ is

A
$$(1,1)$$

B
$$(2,2)$$

C
$$(3,3)$$

D
$$(4,4)$$

Solution

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 $$(4,4)$$
Question 4
1 / -0

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

A
$$6$$

B
$$8$$

C
$$14$$

D
$$16$$

Solution
Question 5
1 / -0

If the points $$\left( {0,0} \right),\left( {3,\sqrt 3 } \right),\left( {p,q} \right)$$ form an equilateral triangle $$q_1, q_2$$ are the two values of $$q$$ then $$q_1 + q_2 = $$

A
$$2 \sqrt 3$$

B
$$\sqrt 3$$

C
$$ - \sqrt 3$$

D
$$0$$

Solution

REF.Image.
Given ABC is a equilateral
triangle
so, $$ AB^{2}=AC^{2} = BC^{2}$$
$$ AB^{2}=(3-0)^{2}+(\sqrt{3}-0)^{2} = 9+3 = 12 $$
$$ AC^{2} = (p-0)^{2}+(q-0)^{2}=p^{2}+q^{2}$$
$$ BC^{2} = (p-3)^{2}+(q-\sqrt{3})^{2}$$
$$ BC^{2}=p^{2}+q^{2}-6p-2\sqrt{3}2+12$$
Now do$$ \rightarrow AC^{2} = BC^{2}$$
$$ p^{2}+q^{2} = p^{2}+q^{2}-6p-2\sqrt{3}q+12$$
$$ \Rightarrow 6p+2\sqrt{3}q = 12 $$ __ (i)
Now do $$ \rightarrow AB^{2} = AC^{2}\Rightarrow 12 = p^{2}+q^{2}$$ __ (ii)
From equation (i) $$ \Rightarrow 6p = 12-2\sqrt{3}q$$
$$ p = \dfrac{12}{6}-\dfrac{2\sqrt{3}}{6}q = (2-\dfrac{2}{\sqrt{3}})$$
put the value of 'p' in equation (ii)
$$ 12 = (2-\dfrac{q}{\sqrt{3}})^{2}+q^{2} = 4+\dfrac{q^{2}}{3}-\dfrac{4q}{\sqrt{3}}+q^{2}$$
$$ 12 = \dfrac{12+q^{2}-4\sqrt{3}q+3q^{2}}{3} \Rightarrow 12\times 3 = 12+4q^{2}-4\sqrt{3}q$$
$$ \Rightarrow 4q^{2}-4\sqrt{3}q = 36-12 = 24 \Rightarrow q^{2}-\sqrt{3}q = 6 $$
$$ q^{2}-\sqrt{3}q -6 = 0$$ $$(q_{1} q^{_{2}})\rightarrow $$ This equation will have two roots.
so, sum of roots $$ = q_{1}+q_{2} = \dfrac{(\sqrt{3})}{1}$$
$$ q_{1}+q_{2} = \sqrt{3}$$
Question 6
1 / -0

The equation ot the line passing through the point $$( 1 , - 2,3 )$$ and parallel to the line$$x - y + 2 z = 5$$ and $$3 x + y + z = 6$$ is

A
$$\frac { x - 1 } { - 3 } = \frac { y + 2 } { 5 } = \frac { z - 3 } { 4 }$$

B
$$\frac { x - 1 } { 1 } = \frac { y + 2 } { -1 } = \frac { z - 3 } { 2 }$$

C
$$\frac { x - 1 } { 3 } = \frac { y + 2 } { 1 } = \frac { z - 3 } { 6 }$$

D
$$\frac { x - 1 } { 3 } = \frac { y + 2 } { -1 } = \frac { z - 3 } { 2 }$$

Solution

$$\begin{array}{l} \vec { r } =\vec { a } +\lambda \vec { b } \\ \Rightarrow \vec{r} = \hat { i } -2\hat { j } +3\hat { k } +\lambda \left( { \hat { i } -j+2\hat { k } } \right) \\ \Rightarrow x\hat { i } +y\hat { j } +z\hat { k } =\left( { 1+\lambda } \right) \hat { i } +\left( { 2-\lambda } \right) \hat { j } +\left( { 3+2\lambda } \right) \hat { k } \\ \Rightarrow x=1+\lambda \\ \Rightarrow y=2-\lambda \\ \Rightarrow z=3+2\lambda \\ Then,\, \, cartesion\, \, equation\, \, of\, \, the\, \, line \\ \frac { { x-1 } }{ 1 } =\frac { { y+2 } }{ { -1 } } =\frac { { z-3 } }{ 2 } \end{array}$$
Question 7
1 / -0

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)

A
$$a-4b+c$$

B
$$2cd$$

C
$$96$$

D
$$36$$

Solution
Question 8
1 / -0

ABC is an isosceles triangle with AB = AC, If the coordinates of the vertices of the base are B (1, 3) nd C(-2, 7), then the coordinates of the vertex A can be

A
$$( - 5 / 6,6 )$$

B
$$( - 7 , - 1 / 8 )$$

C
$$( 5 / 6,6 )$$

D
none of these

Solution
Question 9
1 / -0

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

A
6

B
9

C
69

D
96

Solution
Question 10
1 / -0

the points on the cirve y={ x }^{ 3 }, the tangents at which are inclined at an angle $${ 60 }^{ 0 }$$ to x-axis is

A
$$\quad \quad \quad \left( { 3 }^{ -1/4 },{ 3 }^{ -3/4 } \right) $$

B
$$\quad \left( 3^{ -1/2 },{ 3 }^{ -2/5 } \right) ,\quad \left( -3^{ 1/3 },{ -3 }^{ -2/3 } \right) $$

C
$$\quad \left( { 2 }^{ 1/4 },{ 2 }^{ -2/5 } \right) , \quad \left( -3^{ 1/2 },{ -3 }^{ -1/2 } \right) $$

D
none