Straight Lines Test 20

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Straight Lines Test 20

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Weekly Quiz Competition

Question 1
1 / -0

Let $$ABCD$$ be a square of a side length $$1$$. Let $$P,Q,R,S$$ be points in the interiors of the sides $$AD,BC,AB,CD$$ respectively, such that $$PQ$$ and $$RS$$ intersect at right angles. If $$PQ=\cfrac { 3\sqrt { 3 } }{ 4 } $$ then $$RS$$ equals

A
$$\cfrac { 2 }{ \sqrt { 3 } } $$

B
$$\cfrac { 3\sqrt { 3 } }{ 4 } $$

C
$$\cfrac { \sqrt { 2 } +1 }{ 2 } $$

D
$$4-2\sqrt { 2 } $$

Solution

$$PQ\bot RS\Rightarrow $$
$$c-a=b-d.......(1)$$
$$PQ=\cfrac { 3\sqrt { 3 } }{ 4 } $$
$${ PQ }^{ 2 }=\cfrac { 27 }{ 16 } $$
$$1+{ (a-c) }^{ 2 }=\cfrac { 27 }{ 16 } .....(2)$$
$$RS=\sqrt { { (b-d) }^{ 2 }+1 } ....(3)$$
By equation (1),(2),(3)
$$RS=\cfrac { 3\sqrt { 3 } }{ 4 } $$
Question 2
1 / -0

A frog is presently located at the origin $$\left( 0,0 \right)$$ in the $$xy$$-plane. It always jumps from a point with integer coordinates to a point with integer coordinates moving a distance of $$5$$ units in each jump. What is the minimum number of jumps required for the frog to go from $$\left( 0,0 \right)$$ to $$\left( 0,1 \right) $$?

A
$$2$$

B
$$3$$

C
$$4$$

D
$$9$$

Solution

The frog can jump to only integral coordinates and have to move $$5$$ units in each jump
So, he must jumps from $$(0,0)$$ to $$(4,3)$$
Then from $$(4,3)$$ to $$(0,6)$$
Finally from$$(0,6)$$ to $$(0,1)$$ covering a distance of $$5$$ unit in each jump
So, minimum of three jumps are required
Hence, option $$B$$ is correct
Question 3
1 / -0

Find the area (in square units) of the triangle whose vertices are $$(a, b+c), (a, b-c) $$ and $$(-a, c). $$

A
$$2ac$$

B
$$2bc$$

C
$$b(a+c)$$

D
$$c(a-h)$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times |[ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] |$$
Given the vertices of triangle,
$$A (a,b+c) ,B(a,b-c) ,C(-a,c)$$
Therefore, area is given by
$$=\dfrac { 1 }{ 2 } |\left[ a\left[ b-c-c \right] +a\left[ c-b-c \right] +(-a)\left[ b+c-b+c \right] \right]| \\ =\dfrac { 1 }{ 2 } |\left[ a(b-2c)+a(-b)-a(2c) \right]| \\ =\dfrac { 1 }{ 2 } |\left[ ab-2ac-ab-2ac \right]| =\left| \dfrac { -4ac }{ 2 } \right| =2ac$$
Question 4
1 / -0

The angle between the lines $$ 2x+11y-7=0$$
and $$ x+3y+5=0$$ is equal to :

A
$$ \tan^{-1} \left (\dfrac {17}{31}\right) $$

B
$$ \tan^{-1} \left (\dfrac {11}{35} \right)$$

C
$$ \tan^{-1} \left ( \dfrac {1}{7} \right)$$

D
$$ \tan^{-1} \left (\dfrac {33}{35} \right)$$

E
$$ \tan^{-1} \left ( \dfrac {7}{33} \right)$$

Solution

Given lines are $$ 2x+11y-7=0$$ and $$ x+3y+5=0$$
$$\Rightarrow y = - \dfrac {2}{11} x + \dfrac {7}{11} $$
and $$ y = - \dfrac {1}{3} x - \dfrac {5}{3} $$
Their, slopes are $$ m_1 = - \dfrac {2}{11} $$ and $$ m_2 = - \dfrac {1}{3} $$
Therefore, angle between lines is
$$ \tan \theta = \dfrac {m_1 - m_2 }{1+m_1m_2} $$
$$ = \dfrac {- \dfrac {2}{11} + \dfrac {1}{3} } {1 + \dfrac {2}{11} \times \dfrac {1}{3} } = \dfrac {5}{35} $$
Thus $$ \theta = \tan^{-1} \left( \dfrac {1}{7} \right) $$
Question 5
1 / -0

The slope of a straight line passing through A( -2, 3) is -4/3. The points on the line that are 10 units away from A are

A
(-8, 11), (4, -5)

B
(- 7, 9), (17,-1)

C
(7, 5) (- 1, -1)

D
(6, 10), (3, 5)

Solution

The equation line will be

$$y-3=\frac{-4}{3}(x+2)$$

$$4x+3y=1$$

Now, let the point $$10$$ unit away is $$(h,k)$$

Then,

$$4h+3k=1$$ ------ $$(1)$$

And

$$\sqrt {(h+2)^2+(k-3)^2}=10$$

$$ {(h+2)^2+(k-3)^2}=10$$ ------ $$(2)$$

Now, from $$(1)$$

$$h=\frac{1-3k}{4}$$

Put in equation $$(2)$$, we get

$$25k^2-150k-1663=0$$

On solving quadric equation

$$K=11$$ and $$k=-5$$

Put $$k=11$$ in equation $$(1)$$

$$h=-8$$

And put $$k=-5$$ in equation $$(1)$$

$$h=4$$

Hence the points are
$$(-8,11),(4,-5)$$

Hence, the correct answer is $$A$$.
Question 6
1 / -0

If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of $$\Delta DEF$$, the area of triangle DEF is sq. units.

A
$$11$$

B
$$22$$

C
$$48$$

D
$$50$$

Solution

Area of triangle = $$\dfrac{1}{2} {x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$$

Area of triangle $$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$$

$$=|\dfrac{1}{2}(-3+16-35)|$$

$$=11$$ sq. Units.
Question 7
1 / -0

Let $$P$$ be $$(5, 3)$$ and a point $$R$$ on $$y = x$$ and $$Q$$ on the x-axis be such that $$PQ + QR + RP$$ is minimum. Then the coordinates of $$Q$$ are

A
$$(17/4, 0)$$

B
$$(17, 0)$$

C
$$(17/2, 0)$$

D
$$None\ of\ these$$

Solution

REF.Image.
Given
$$ P= (5,3) $$
on x -axis
$$Q = (x_1 ,0)$$
R on $$ y = x $$
$$ R = (x_{2},x_{2})$$
Now,
$$ PQ + QR + RP $$ is minimum
$$ PQ = (5-x_{1}, 3-0) ; QR = (x_{1}- x_{2}, 0-x_{2})$$
$$ RP = (x_{2}-5, x_{2}-3)$$
& $$ (x_{1}, x_{2}) = |PQ |+ |QR| + |RP|$$
$$ = \sqrt{(5-x_{1})^2 + 3^2 }+ \sqrt{(x_{1}-x_{2})^2 + (-x_{2})^2} + \sqrt{(x_{2}-5)^2 + (x_{2}-3)^2}$$
min of $$ f (x_{1},x_{2})$$
$$ x_{2} = \dfrac{17}{5}, x_{1} = \dfrac{17}{4}$$
$$ \therefore Q = \left ( \dfrac{17}{4} ,0\right )$$
Question 8
1 / -0

The points $$(-1,0)$$ and $$(-2,1)$$ are the two extremities of a diagonal of a parallelogram. If $$(-6,5)$$ is the third vertex, then the fourth vertex of the parallelogram is

A
$$(2,-6)$$

B
$$(2,-5)$$

C
$$(3,-4)$$

D
$$(-3,4)$$

E
$$(3,-5)$$

Solution

Let the points of parallelogram are $$A(-2,1),B(-6,5),C(-1,0)$$ and $$D(x,y)$$
We know that, diagonals of parallelogram bisect each other
$$\left( \cfrac { -2-1 }{ 2 } ,\cfrac { 1+0 }{ 2 } \right) =\left( \cfrac { x-6 }{ 2 } ,\cfrac { y+5 }{ 2 } \right) $$
$$\quad \Rightarrow \cfrac { -3 }{ 2 } =\cfrac { x-6 }{ 2 } ;\cfrac { 1 }{ 2 } =\cfrac { y+5 }{ 2 } $$
$$\Rightarrow x=3;y=-4$$
Thus the coordinates of fourth vertex of parallelogram are
$$(3,-4)$$
Question 9
1 / -0

The point on the line $$4x+3y=5$$, which is equidistant from $$\left( 1,2 \right)$$ and $$\left( 3,4 \right) $$, is

A
$$\left( 7,-4 \right) $$

B
$$\left( -10,15 \right) $$

C
$$\left( \dfrac { 1 }{ 7 } ,\dfrac { 8 }{ 7 } \right) $$

D
$$\left( 0,\dfrac { 5 }{ 4 } \right) $$

Solution

Let the point $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ be on the line $$4x+3y=5$$.
$$\therefore 4{ x }_{ 1 }+3{ y }_{ 1 }=5$$ ....(i)
Also, we have
$${ \left( { x }_{ 1 }-1 \right) }^{ 2 }+{ \left( { y }_{ 1 }-2 \right) }^{ 2 }={ \left( { x }_{ 1 }-3 \right) }^{ 3 }+{ \left( { y }_{ 1 }-4 \right) }^{ 2 }$$
$$\Rightarrow { x }_{ 1 }^{ 2 }+1-2{ x }_{ 1 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }={ x }_{ 1 }^{ 2 }+9-6{ x }_{ 1 }+{ y }_{ 1 }^{ 2 }+16-8{ y }_{ 1 }$$
$$\Rightarrow 4{ x }_{ 1 }+4{ y }_{ 1 }=20$$
$$\Rightarrow { x }_{ 1 }+{ y }_{ 1 }=5$$ ....(ii)
From equations (i) and (ii), we get
$${ y }_{ 1 }=15$$ and $${ x }_{ 1 }=-10$$
Question 10
1 / -0

$$A=(4, 2)$$ and $$B=(2, 4)$$ are two given points and a point $$P$$ on the line $$3x + 2y + 10 = 0 $$ is given then. which of the following is true.

A
$$(PA + PB)$$ is minimum when $$P (\dfrac {-14}{5}, \dfrac {-4}{5}) $$

B
$$(PA + PB)$$ is maximum when $$P (\dfrac {-14}{5}, \dfrac {-4}{5}) $$

C
$$(PA - PB)$$ is maximum when $$ (-22, 28) $$

D
$$(PA - PB)$$ is minimum when $$ P (-22, 28) $$

Solution

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 20

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Straight Lines Test 20

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SHARING IS CARING

Question 1

$$\cfrac { 2 }{ \sqrt { 3 } } $$

$$\cfrac { 3\sqrt { 3 } }{ 4 } $$

$$\cfrac { \sqrt { 2 } +1 }{ 2 } $$

$$4-2\sqrt { 2 } $$

Solution

Question 2

$$2$$

$$3$$

$$4$$

$$9$$

Solution

Question 3

$$2ac$$

$$2bc$$

$$b(a+c)$$

$$c(a-h)$$

Solution

Question 4

$$ \tan^{-1} \left (\dfrac {17}{31}\right) $$

$$ \tan^{-1} \left (\dfrac {11}{35} \right)$$

$$ \tan^{-1} \left ( \dfrac {1}{7} \right)$$

$$ \tan^{-1} \left (\dfrac {33}{35} \right)$$

$$ \tan^{-1} \left ( \dfrac {7}{33} \right)$$

Solution

Question 5

(-8, 11), (4, -5)

(- 7, 9), (17,-1)

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

(6, 10), (3, 5)

Solution

Question 6

$$11$$

$$22$$

$$48$$

$$50$$

Solution

Question 7

$$(17/4, 0)$$

$$(17, 0)$$

$$(17/2, 0)$$

$$None\ of\ these$$

Solution

Question 8

$$(2,-6)$$

$$(2,-5)$$

$$(3,-4)$$

$$(-3,4)$$

$$(3,-5)$$

Solution

Question 9

$$\left( 7,-4 \right) $$

$$\left( -10,15 \right) $$

$$\left( \dfrac { 1 }{ 7 } ,\dfrac { 8 }{ 7 } \right) $$

$$\left( 0,\dfrac { 5 }{ 4 } \right) $$

Solution

Question 10

$$(PA + PB)$$ is minimum when $$P (\dfrac {-14}{5}, \dfrac {-4}{5}) $$

$$(PA + PB)$$ is maximum when $$P (\dfrac {-14}{5}, \dfrac {-4}{5}) $$

$$(PA - PB)$$ is maximum when $$ (-22, 28) $$

$$(PA - PB)$$ is minimum when $$ P (-22, 28) $$

Solution

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