Straight Lines Test 24

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Straight Lines Test 24

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Weekly Quiz Competition

Question 1
1 / -0

The length of the segment of the straight line passing through $$(3, 3)$$ and $$(7, 6)$$ cut off by the coordinate axes is?

A
$$\dfrac{4}{5}$$

B
$$\dfrac{5}{4}$$

C
$$\dfrac{7}{4}$$

D
$$\dfrac{4}{8}$$

Solution
Question 2
1 / -0

If $$Q(0, 1)$$ is equidistant from $$P(5, -3)$$ and $$R(x, 6)$$, find the values of x. Also find the distances QR and PQ.

A
PQ=QR=$$\sqrt { 41} $$

B
PQ=20, QR=$$\sqrt { 29 } $$

C
PQ= $$\sqrt { 29 } $$, QR= 20

D
PQ$$=\sqrt { 19 } $$, QR = 21

Solution

Q is equidistant from P and R
=>PQ=RQ
=>$$P{ Q }^{ 2 }=R{ Q }^{ 2 }$$
=>$$(5)^{ 2 }+(-4)^{ 2 }={ x }^{ 2 }+{ (5) }^{ 2 }\\ =>{ x }^{ 2 }=16\\ =>x=\pm 4$$
Now PQ=QR=$$\sqrt { 16+25 } =\sqrt { 41 } $$
Question 3
1 / -0

The angle between the lines $$3x + y - 7 = 0$$ and $$x + 2y + 9 = 0$$ is

A
$$\dfrac{\pi}{3}$$

B
$$\dfrac{\pi}{6}$$

C
$$\dfrac{\pi}{2}$$

D
$$\dfrac{\pi}{4}$$

Solution

$$3x+y-7=0 $$ and $$ x+2y+9=0$$

write in the form $$ y = mx+c$$

$$i.e. y = -3x+7, y= \dfrac{-x}{2}-\frac{9}{2}$$

$$\therefore m_{1}=-3$$ and $$m_{2}\mp -\dfrac{1}{2}$$

$$tan ( A-B)= \left | \dfrac{\tan A-\tan B}{1+\tan A \tan B} \right |$$

$$\left | \dfrac{-3-(-y_{2})}{1+(-3)(-y_{2})} \right |$$

$$=\left | \dfrac{-5/2}{5/2} \right |= 1$$

$$\therefore $$ Angle b/w then is $$\tan^{-1}(1)=45^{\circ}$$

$$=\dfrac{\pi }{4}$$
Question 4
1 / -0

Find the area of the triangle whose vertices are $$(-5, -1), (3, -5), (5, 2)$$

A
$$31 sq$$ $$units$$

B
$$32sq$$ $$units$$

C
$$33 sq$$ $$units$$

D
$$\text{no triangle can be formed}$$

Solution

Area of a triangle whose vertices are $$A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$$ is given as,
$$Area=(\dfrac{1}{2})[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] $$
Let $$A$$ be the required area
$$A=(\dfrac{1}{2})[(-5)(-5-2)+3(2+1)+5(-1+5)]\\ (\dfrac{1}{2})[35+9+20]\\=32sq\>unit$$
So, the correct option is (B)
Question 5
1 / -0

Find the coordinates of the point equidistant from the points $$A(-2, -3), B(-1, 0)$$ and $$C(7, -6)$$.

A
$$(3, -3)$$

B
$$(-3, 3)$$

C
$$(-2, -3)$$

D
$$(-3, -3)$$

Solution

Let us consider the co-ordinate $$(x,y)$$ which is the point equidistant from the points $$A(-2,-3), B(-1,0)$$ and $$C(7,-6).$$
So, $$d_{1}= \sqrt{(x-(-2))^{2}+ (y-(-3))^{2}} = \sqrt{(x+2)^{2}+ (y+3)^{2}}$$
$$d_{2}= \sqrt{(x-(-1))^{2}+ (y-0)^{2}} = \sqrt{(X+1)^{2} +y^{2}}$$ &
$$d_{3}= \sqrt{(x-7)^{2}+ (y-(-6))^{2}}^{2} = \sqrt{(x-7)^{2}+ (y+6)^{2}}$$
here $$d_{1}= d_{2}= d_{3}$$
Let compare $$d_{1}$$ & $$ d_{2}$$. we get,
$$\sqrt{(x+2)^{2} + (y+3)^{2}}= \sqrt{(x+1)^{2}}+y^{2}$$
$$\therefore (x+2)^{2}+ (y+3)^{2}= (x+1)^{2}+ y^{2}$$
$$\therefore x^{2}+4x+4+ y^{2}+ 6y+ 9 = x^{2}+ 2x+1+y^{2}$$
$$\therefore 2x+ 12+ 6y= 0$$
$$\therefore x+ 3y+ 6 = 0 \rightarrow (1)$$
Now, compare $$d_{2}$$ & $$d_{3}$$. we get.
$$\sqrt{(x+1)^{2}+y^{2}}= \sqrt{(x-7)^{2}+ (y+6)^{2}}$$
$$x^{2}+2x+1+y^{2}= x^{2}-14x+49+y^{2}+12y+36$$
$$16x-12y= 84$$
$$4 (4x-3y)= 84$$
$$4x-3y- 21= 0 \rightarrow (2)$$
From $$(1) $$ & $$(2)\ 5x- 15 = 0 \Rightarrow x=3$$
From $$(1) \Rightarrow y= \dfrac{-x-6}{3}= \dfrac{-3-6}{3}=\dfrac{-9}{3}= -3$$
So, $$(x,y)= (3,-3)$$
Question 6
1 / -0

Find a point on x-axis which is equidistant from the points $$(5, 4)$$ and $$(-2, 3)$$.

A
$$(2, 0)$$

B
$$(-2, 0)$$

C
$$(3, 0)$$

D
$$(-3, 0)$$

Solution

$$AC=BC$$ (Equidistance)

$$AC^2=BC^2$$

$$(x-5)^2+(0-4)^2=(x+2)^2+(0-3)^2$$

$$x^2-10x+25+16=x^2+4+4x+9$$

$$-14x+41-13=0$$

$$-14x+28=0$$

$$=14x=-28$$

$$x=28/14\\ \boxed {x=2}$$

These the coordinates are $$(2, 0)$$
Question 7
1 / -0

An equilibrium triangle $$ABC$$ has its centroid at the origin and the base $$BC$$ lies along the line $$x+y=1$$. Then the area of $$\Delta ABC$$ is

A
$$\frac {3 \sqrt 8}{2}$$

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

C
$$\frac {3 \sqrt 2}{2}$$

D
$$\frac {3 \sqrt 3}{2}$$

Solution
Question 8
1 / -0

Distance between A(x, y) and B(-4, 7) is $$\sqrt{41}.$$ Find x and y, if A's ordinate is thrice of its abscissa.

A
$$(\frac{-12}{5},\frac{36}{5})$$

B
$$(1,\dfrac{12}{5})$$

C
$$\text{both a and b}$$

D
none

Solution

$$\\(x+4)^2(y-7)^2=41\\and y=3x \\\therefore (x+4)^2+(3x-7)^2=41\\x^2+8x+16+9x^2+49-42x=41\\10x^2-34x+24=0\\5x^2-17x+12\\5x^2-5x-12x+12=0\\5x(x-1)-12(x-1)=0\\or (x-1)(5x-12)=0\\\therefore x=1 and x=(\frac{12}{5})$$
Question 9
1 / -0

The ends of the base of an isosceles triangle are at $$(2a, 0)$$ and $$(0, a)$$ and one side is parallel to Y-axis. The equation of the other side is?

A
$$x+2y-a=0$$

B
$$x+2y=2a$$

C
$$3x+4y-4a=0$$

D
$$3x-4y+4a=0$$

Solution
Question 10
1 / -0

If $$A(-2,1),B(2,3)$$ and $$C(-2,-4)$$ are three points, then the angle between $$BA$$ and $$BC$$ is:

A
$$\tan ^{ -1 }{ \left( \cfrac { 3 }{ 2 } \right) } $$

B
$$\tan ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) } $$

C
$$\tan ^{ -1 }{ \left( \cfrac { 7 }{ 4 } \right) } $$

D
$$none\ of\ these$$

Solution

Slope of line joining $$(x_1,y_1), (x_2,y_2)$$ is $$m=\dfrac{y_2-y_1}{x_2-x_1}$$

First Consider the side $$BA$$. Let $$m_1$$ be the slope of $$BA$$

$$\Rightarrow m_1=\dfrac{3-1}{2+2}$$

$$\Rightarrow m_1=\dfrac{2}{4}$$

First Consider the side $$BC$$. Let $$m_2$$ be the slope of $$BC$$

$$\Rightarrow m_2=\dfrac{3+4}{2+2}$$

$$\Rightarrow m_2=\dfrac{7}{4}$$

Angle between $$BA, BC$$ is given by $$\tan\theta=\dfrac{m_1-m_2}{1+m_1m_2}$$

Substituting the values of $$m_1, m_2$$ we get

$$\Rightarrow \tan\theta=\dfrac{\dfrac{5}{4}}{1+\dfrac{14}{16}}$$

On solving we get

$$\Rightarrow \tan\theta=\dfrac{2}{3}$$

$$\Rightarrow \theta=\tan^{-1}(\dfrac{2}{3})$$

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Straight Lines Test 24

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Straight Lines Test 24

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SHARING IS CARING

Question 1

$$\dfrac{4}{5}$$

$$\dfrac{5}{4}$$

$$\dfrac{7}{4}$$

$$\dfrac{4}{8}$$

Solution

Question 2

PQ=QR=$$\sqrt { 41} $$

PQ=20, QR=$$\sqrt { 29 } $$

PQ= $$\sqrt { 29 } $$, QR= 20

PQ$$=\sqrt { 19 } $$, QR = 21

Solution

Question 3

$$\dfrac{\pi}{3}$$

$$\dfrac{\pi}{6}$$

$$\dfrac{\pi}{2}$$

$$\dfrac{\pi}{4}$$

Solution

Question 4

$$31 sq$$ $$units$$

$$32sq$$ $$units$$

$$33 sq$$ $$units$$

$$\text{no triangle can be formed}$$

Solution

Question 5

$$(3, -3)$$

$$(-3, 3)$$

$$(-2, -3)$$

$$(-3, -3)$$

Solution

Question 6

$$(2, 0)$$

$$(-2, 0)$$

$$(3, 0)$$

$$(-3, 0)$$

Solution

Question 7

$$\frac {3 \sqrt 8}{2}$$

$$\frac {3 \sqrt 3}{4}$$

$$\frac {3 \sqrt 2}{2}$$

$$\frac {3 \sqrt 3}{2}$$

Solution

Question 8

$$(\frac{-12}{5},\frac{36}{5})$$

$$(1,\dfrac{12}{5})$$

$$\text{both a and b}$$

none

Solution

Question 9

$$x+2y-a=0$$

$$x+2y=2a$$

$$3x+4y-4a=0$$

$$3x-4y+4a=0$$

Solution

Question 10

$$\tan ^{ -1 }{ \left( \cfrac { 3 }{ 2 } \right) } $$

$$\tan ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) } $$

$$\tan ^{ -1 }{ \left( \cfrac { 7 }{ 4 } \right) } $$

$$none\ of\ these$$

Solution

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