Straight Lines Test 25

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Straight Lines Test 25

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Weekly Quiz Competition

Question 1
1 / -0

An aeroplane flying horizontally $$1$$, above the ground is observed at an elevation of $$60$$ and after $$10$$ seconds the elevation is observed to be $$30$$. The uniform speed of the aeroplane in $$/h$$ is-

A
$$240$$

B
$$240\sqrt { 3 } $$

C
$$60\sqrt { 3 } $$

D
None of these

Solution

At $$t=0$$ plane is at A and at $$t=10$$, the plane is at B. Distance covered$$=AB$$
In $$\triangle AHO$$
$$\tan { 60° } =\cfrac { OH }{ AH } $$
$$AH=\cfrac { (1000) }{ \sqrt { 3 } } m$$
In $$\triangle BHO$$
$$\tan { 30° } =\cfrac { OH }{ BH } $$
$$BH=\sqrt { 3 } \left( 1000 \right) m$$
$$\therefore AB=BH-AH=\left( \sqrt { 3 } -\cfrac { 1 }{ \sqrt { 3 } } \right) \left( 1000 \right) m$$
$$=\cfrac { 2 }{ \sqrt { 3 } } .1000m$$
Speed$$=\cfrac { Distance }{ Time } $$
Speed$$=\cfrac { 2 }{ \sqrt { 3 } } \times \cfrac { 1000 }{ 10 } ㎧$$
$$=\cfrac { 200 }{ \sqrt { 3 } } ㎧$$
$$=\cfrac { 200 }{ \sqrt { 3 } } \times \cfrac { 18 }{ 5 } ㎞{ h }^{ -1 }$$
$$=40\times 6\sqrt { 3 }$$
Speed$$=240\sqrt { 3 } ㎞{ h }^{ -1 }$$
Question 2
1 / -0

Points $$A(\sqrt{2}, 4)$$, $$B(6, \sqrt{3})$$ and C are collinear. If B is the midpoint of line segment AC, approximately calculate the (x, y) coordinates of point C.

A
$$(3.71, 1.13)$$

B
$$(3.71, 5.73)$$

C
$$(7.41, -7.46)$$

D
$$(10.59, -7.46)$$

E
$$(10.59,-0.54)$$

Solution

We are given point $$C(x, y)$$
Point $$B$$ is the midpoint of line segment $$AC$$
i.e $$B(6,\sqrt{3})=B\left(\dfrac{\sqrt{2}+x}{2}, \dfrac{4+y}{2}\right)$$
So, $$\dfrac{\sqrt{2}+x}{2}=6$$ & $$\dfrac{4+y}{2}=\sqrt{3}$$
$$\sqrt{2}+x=12$$ & $$4+y=2\sqrt{3}$$
$$x=12-\sqrt{2}$$ & $$y=2\sqrt{3}-4$$
$$x=12-1.41$$ & $$y=2(1.73)-4$$
$$x=10.59$$ $$y=-0.54$$
$$C(x,y)=C(10.59, -0.54)$$
Question 3
1 / -0

A straight line L through the point $$(3,-2)$$ is inclined at an angle $$60$$ to the line $$\sqrt { 3 } x+y=1$$. If L also intersects the x-axis, the equation of L is-

A
$$y+\sqrt { 3 }x+2-3\sqrt { 3 }=0$$

B
$$y-\sqrt { 3 }x+2+3\sqrt { 3 }=0$$

C
$$\sqrt { 3 } y-x+3+2\sqrt { 3 } =0$$

D
$$\sqrt { 3 } y+x-3+2\sqrt { 3 } =0$$

Solution

$$\sqrt{3}x+y=1\implies m=\tan \theta =-\sqrt{3}\implies $$ this line makes an angle of $$120^{\circ}$$ with x-axis.

Since $$ L$$ makes an angle $$60^{\circ}$$ with this line.

$$\implies L$$ must be either x- axis $$($$ which is not correct because given $$L$$ intersect the x-axis$$)$$ or a line making an angle $$60^{\circ}$$ with the x-axis.

$$\implies m=\tan 60^{\circ}=\sqrt{3}$$

$$L:y=m{x}+c$$

$$L:y=\sqrt{3}x+c$$

$$L$$ passes through $$(3,-2)\implies -2=3\sqrt{3}+c\implies c=-2-3\sqrt{3}$$

$$L:y-\sqrt{3}x+2+3\sqrt{3}=0$$
Question 4
1 / -0

A straight line is drawn through the point $$p\ (2,3)$$ and is inclined at an angle of $$30^{o}$$ with the $$x-$$axis, the co-ordinates of two points on it at a distance of $$4$$ from $$p$$ is/are

A
$$\left(2+2\sqrt{3},5\right)$$, $$\left(2-2\sqrt{3},1\right)$$

B
$$\left(2+2\sqrt{3},5\right)$$, $$\left(2+2\sqrt{3},1\right)$$

C
$$\left(2-2\sqrt{3},5\right)$$, $$\left(2-2\sqrt{3},1\right)$$

D
$$none\ of\ these$$

Solution

The parametric points of a line is given as $$(x_1\pm r\cos \theta,y_1\pm r\sin \theta)\\(2\pm 4\cos 30,3\pm 4 \sin 30)\\\left(2\pm 4\dfrac{\sqrt3}{2},3\pm 4\dfrac 12\right)\\(2\pm2\sqrt3,3\pm 2)\\(2+2\sqrt3,5),(2-2\sqrt3,1) $$
Question 5
1 / -0

If the vertices of a triangle are $$(1,2),(4,-6)$$ and $$(3,5)$$, then its area is

A
$$\cfrac{23}{2}$$ sq. units

B
$$\cfrac{25}{2}$$ sq. units

C
$$12$$ sq.unit

D
none of these

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore, area of required triangle is given by
Area $$ = \dfrac{1}{2} \times [ 1 (-6 - 5) + 4(5 - 2) + 3(2 + 6) ] $$
$$=\dfrac{1}{2}[-11+12+24]$$
$$=\dfrac{1}{2}\times25$$
$$=\dfrac{25}{2}\ sq.unit$$
Question 6
1 / -0

The points given are $$(1, 1)$$, $$(-2, 7)$$ and $$(3, 3)$$.Find distance between the points.

A
$$\sqrt45,$$$$7,$$$$\sqrt{8}$$

B
$$\sqrt{10},7,8$$

C
$$\sqrt{3},1,2$$

D
None of the above

Solution

Let the points be
$$A(1,1)$$, $$B(-2,7)$$, $$C(3,3)$$
Distance of $$AB=\sqrt { { \left( 2-1 \right) }^{ 2 }+{ \left( 7-1 \right) }^{ 2 } } $$
$$=\sqrt { 9+36 } =\sqrt { 45 } $$
Distance of $$BC=\sqrt { { \left( 3+2 \right) }^{ 2 }+{ \left( 3-7 \right) }^{ 2 } } $$
$$=\sqrt { 25+16 } =\sqrt { 49 } =7$$
Distance of $$AC=\sqrt { { \left( 3-1 \right) }^{ 2 }+{ \left( 3-1 \right) }^{ 2 } } =\sqrt { 8 } $$
Question 7
1 / -0

The points $$(a, b), (-a, -b), (b\sqrt{3}, -a\sqrt{3})$$ are the vertices of a triangle which is

A
Isosceles

B
Equilateral

C
Right angled

D
Scalene

Solution
Question 8
1 / -0

A line passing through $$(3,4)$$ meets the axis $$\overline {OX} $$ and $$\overline {OY} $$ at $$A$$ and $$B$$ respectively. Find the minimum area of triangle OAB.

A
$$12$$

B
$$10$$

C
$$24$$

D
$$6$$

Solution
Question 9
1 / -0

The intercepts on the straight $$y=mx$$ by the line $$y=2$$ and $$y=6$$ is less than $$5$$, then $$m$$ belongs to

A
$$\left( { - \infty , - \frac{4}{3}} \right) \cup \left( {\frac{4}{3},\infty } \right)$$

B
$$\left( {\frac{4}{3},\frac{3}{8}} \right)$$

C
$$\left( { - \frac{4}{3},\frac{4}{3}} \right)$$

D
$$\left( {\frac{4}{3},\infty } \right)$$

Solution
Question 10
1 / -0

$$ABC$$ is an isosceles triangle. If the coordinates of the base are $$B(1,3)$$ and $$C(-2,7)$$, the coordinates of vertex $$A$$ can be

A
$$(1,6)$$

B
$$(-1/2,5)$$

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

D
none of these

Solution

$$AB=AC$$
$$\sqrt{(x-1)^2+(y-3)^2}=\sqrt{(x-(-2))^2+(y-7)^2}$$
$$(x-1)^2+(y-3)^2=(x+2)^2+(y-7)^2$$
$$=2x+1-6y+9=4x+4-14y+49$$
$$\Rightarrow 6x-8y+43=0$$
By substituting $$\left(\dfrac{5}{6}, 6\right)$$
$$\Rightarrow \not{6}\times \dfrac{5}{\not{6}}-8\times 6+43$$
$$\Rightarrow 0$$
Hence $$\left(\dfrac{5}{6}, 6\right)$$ is answer.

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 25

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Straight Lines Test 25

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SHARING IS CARING

Question 1

$$240$$

$$240\sqrt { 3 } $$

$$60\sqrt { 3 } $$

None of these

Solution

Question 2

$$(3.71, 1.13)$$

$$(3.71, 5.73)$$

$$(7.41, -7.46)$$

$$(10.59, -7.46)$$

$$(10.59,-0.54)$$

Solution

Question 3

$$y+\sqrt { 3 }x+2-3\sqrt { 3 }=0$$

$$y-\sqrt { 3 }x+2+3\sqrt { 3 }=0$$

$$\sqrt { 3 } y-x+3+2\sqrt { 3 } =0$$

$$\sqrt { 3 } y+x-3+2\sqrt { 3 } =0$$

Solution

Question 4

$$\left(2+2\sqrt{3},5\right)$$, $$\left(2-2\sqrt{3},1\right)$$

$$\left(2+2\sqrt{3},5\right)$$, $$\left(2+2\sqrt{3},1\right)$$

$$\left(2-2\sqrt{3},5\right)$$, $$\left(2-2\sqrt{3},1\right)$$

$$none\ of\ these$$

Solution

Question 5

$$\cfrac{23}{2}$$ sq. units

$$\cfrac{25}{2}$$ sq. units

$$12$$ sq.unit

none of these

Solution

Question 6

$$\sqrt45,$$$$7,$$$$\sqrt{8}$$

$$\sqrt{10},7,8$$

$$\sqrt{3},1,2$$

None of the above

Solution

Question 7

Isosceles

Equilateral

Right angled

Scalene

Solution

Question 8

$$12$$

$$10$$

$$24$$

$$6$$

Solution

Question 9

$$\left( { - \infty , - \frac{4}{3}} \right) \cup \left( {\frac{4}{3},\infty } \right)$$

$$\left( {\frac{4}{3},\frac{3}{8}} \right)$$

$$\left( { - \frac{4}{3},\frac{4}{3}} \right)$$

$$\left( {\frac{4}{3},\infty } \right)$$

Solution

Question 10

$$(1,6)$$

$$(-1/2,5)$$

$$(5/6,6)$$

none of these

Solution

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