Straight Lines Test 22

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Straight Lines Test 22

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Weekly Quiz Competition

Question 1
1 / -0

The area of the triangle with vertices at $$(-4, 1), (1, 2)(4, -3)$$ is

A
$$14$$

B
$$16$$

C
$$15$$

D
None of these

Solution
Question 2
1 / -0

Given the family of lines $$a(2x+y+4)+b(x-2y-3)=0$$. The numbers of lines belonging to the family at a distance $$\sqrt { 10 } $$ from any point $$(2,-3)$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution
Question 3
1 / -0

The coordinates of the point where the line joining $$P(3, 4, 1)$$ and $$Q(5, 1, 6)$$ crosses the xy-plane are:

A
$$\left( { - {{13} \over 5}, - {{23} \over 5},0} \right)$$

B
$$\left( {{{13} \over 5},{{23} \over 5},0} \right)$$

C
$$\left( {{{13} \over 5}, - {{23} \over 5},0} \right)$$

D
$$\left( { - {{13} \over 5},{{23} \over 5},0} \right)$$

Solution

$$P(3,4,1)$$
$$Q(5,1,6)$$
Equation of PQ
$$\cfrac { x-3 }{ 5-3 } =\cfrac { y-4 }{ 1-4 } =\cfrac { z-1 }{ 6-1 } \\ \cfrac { x-3 }{ 2 } =\cfrac { y-4 }{ -3 } =\cfrac { z-1 }{ 5 } \\ In\quad xy\quad plane\quad z=0\\ \therefore \cfrac { z-1 }{ 5 } =\cfrac { -1 }{ 5 } \\ \cfrac { x-3 }{ 2 } =\cfrac { -1 }{ 5 } \\ x=3-\cfrac { 2 }{ 5 } =\cfrac { 13 }{ 5 } \\ \cfrac { y-4 }{ -3 } =\cfrac { -1 }{ 5 } \\ y=4+\cfrac { 3 }{ 5 } =\cfrac { 23 }{ 5 } \\ \therefore Coordinates\quad are\left( \cfrac { 13 }{ 5 } ,\cfrac { 23 }{ 5 } ,0 \right) \\ \\ $$
Question 4
1 / -0

The area of the triangle whose vertices are (3,8), (-4,2) and (5,-1) is

A
$$75 \ sq. units$$

B
$$37.5 \ sq. units$$

C
$$45 \ sq. units$$

D
$$22.5 \ sq. units$$

Solution

Let $$A(3,8), B(-4,2),C(5,-1)$$ be the vertices of the given $$\triangle ABC$$.

Then,

$$(x_{1}=3,y_{1}=8),(x_{2}=-4,y_{2}=2),(x_{3}=5,y_{3}=-1)$$

Area of $$\triangle ABC$$ = $$\dfrac{1}{2}|[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]|$$

$$=\dfrac{1}{2}|3[2-(-1)]-4(-1-8)+5(8-2)|$$

$$=\dfrac{1}{2}|9+36+30|=\dfrac{75}{2}=37.5 \ sq. units$$
Question 5
1 / -0

For points $$A(1, -1, 1), B(1, 3, 1), C(4, 3, 1)$$ and $$D(4, -1, 1)$$ taken in order are the vertices of

A
a parallelogram which is neither a square nor a rhombus

B
rhombus

C
as isosceles trapezium

D
a cyclic quadrilateral

Solution

$$\dfrac{\bar A+\bar C}{2}=\dfrac{\bar B+\bar D}{2}\Rightarrow $$ it should be $$\parallel \ gm$$ May be square Rectangle or Rhombus
$$AB=\sqrt{(1-1)^2+(3+1)^2+(1-1)^2}=4$$
$$CD=\sqrt{(4-4)^2+(3+1)^2+(1-1)^2}=4$$
$$AD=\sqrt{(4-1)^2+(-1+1)^2+(1-1)^2}=3$$
$$BC=\sqrt{(4-1)^2+(3-3)^2+(1-1)^2}=3$$
As All sides are not equal it can be Rectangle or $$\parallel el\ gm$$ not but cant be square and Rhombus But $$\overline {AB}.\overline {AD}\neq 0$$ hence it should be Parallelogram
Question 6
1 / -0

The sides $$AB,BC,CD$$ and $$DA$$ of quadrilateral are $$x+2y=3,x-3y=4,\ 5x+y+12=0$$ respectively. The angle between diagonals $$AC$$ and $$BD$$ is

A
$$45^{o}$$

B
$$60^{o}$$

C
$$90^{o}$$

D
$$30^{o}$$

Solution
Question 7
1 / -0

If the distance between the points$$\left( {x,2} \right)$$and$$\left( {3,4} \right)$$ is $$2$$,then the value of $$x$$ is

A
$$2$$

B
$$1$$

C
$$3$$

D
$$4$$

Solution

$$(x-3)^{ 2 }+(2-4)^{ 2 }=2^{ 2 }\\ (x-3)^{ 2 }=2^{ 2 }-2^{ 2 }\\ x-3=0\\ x=3\\ $$
Question 8
1 / -0

If the axes are transformed from origin to the point $$(-2,1)$$, then new coordinates of $$(4,-5)$$ are

A
$$(2,6)$$

B
$$(6,4)$$

C
$$(6,-6)$$

D
$$(2,-4)$$

Solution
Question 9
1 / -0

If $$a, b, c$$ and $$d$$ are points on a number line such that $$a < b < c < d, b$$ is twice as far from $$c$$ as from $$a,$$ and $$c$$ is twice as far from $$b$$ as from $$d,$$ then what is the value of $$\dfrac{c-a}{d-b}$$ ?

A
$$\dfrac{1}{3}$$

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

C
$$\dfrac{1}{2}$$

D
$$1$$

Solution

Let distance between ab be $$x$$
$$\therefore $$ distance between $$bc = 2x$$
Let distance between $$cd$$ be $$y$$
$$\therefore $$ distance between $$bc=2x$$
$$\therefore 2x = 2y$$
$$\therefore x = y$$
$$\cfrac { c-a }{ d-b } =\cfrac { 2x+x }{ y+2x } =\cfrac { 3x }{ 3x } =1$$
Question 10
1 / -0

The curve $$y=ax^3+bx^2+cx+5$$ touches the x-axis at $$P(-2,0)$$ and cuts the y-axis at a point $$Q$$ where its gradient is $$3$$. Then the value of $${a,b,c} $$ is

A
$$A=\dfrac{9}{2}$$, $$B=\dfrac{1}{2}$$, $$C=\dfrac{3}{5}$$

B
$$A=\dfrac{-1}{2}$$, $$B=\dfrac{-3}{4}$$, $$C=3$$

C
$$A=\dfrac{9}{2}$$, $$B=\dfrac{-3}{4}$$, $$C=3$$

D
$$A=\dfrac{-9}{2}$$, $$B=\dfrac{1}{2}$$, $$C=\dfrac{-3}{5}$$

Solution

$$y=ax^{3}+bx^{2}+cx+5$$ $$P(-2, 0)$$ at $$y$$ axis
differentiate $$y$$ wrt $$x$$ $$y=5 @\phi (0, 5) m=3$$
$$\left(\dfrac{dy}{dx}\right)_{@\phi}=3ax^{2}+2bx+c$$
$$\left(\dfrac{dy}{dx}\right)_{@(0, 5)}=3a(0)^{2}+2b(0)+c=3$$
$$c=3$$
$$\left(\dfrac{dy}{dx}\right)_{@ P=(-2, 0)}=3a(-2)^{2}+2b(-2)+3=0$$
$$12a-4b+3=0 .... (1)$$
$$-8a+4b-1=0 ..... (2)$$
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ $$
$$a=-1/2 b=-3/4$$