Straight Lines Test 30

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Straight Lines Test 30

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Weekly Quiz Competition

Question 1
1 / -0

The coordinates of two consecutive vertices $$A$$ and $$B$$ of a regular hexagon $$ABCDEF$$ are $$(1, 0)$$ and $$(2, 0)$$, respectively. The equation of the diagonal $$CE$$ is

A
$$\sqrt{3} x + y =4$$

B
$$x + \sqrt{3} y + 4 =0$$

C
$$x + \sqrt{3} y = 4$$

D
None of these

Solution

Refer to the figure attached. $$ABCDEF$$ is a regular hexagon with side length $$1$$ with other sides above $$x-$$ axis. The coordinates of $$A$$ and $$B$$ are given.
Length of $$AE$$ can be found using sine rule in triangle $$FEA$$
$$ \dfrac{\sin {120^{o}}}{AE}=\dfrac{\sin {30^{o}}}{1}$$
$$ \Rightarrow AE = \sqrt{3}$$
Since for the given hexagon $$ AE \perp AB$$, we get the coordinates of $$E$$ as $$(1, \sqrt{3})$$....(1)
Now, $$FC || AB$$ and $$\angle {ECF} = 30^{o}$$
Hence, slope of $$CE$$ is $$ \tan {120^{o}}$$....(2)
From (1) and (2) equation of $$CE$$ is $$y- \sqrt 3 = \tan {120^{o}}(x-1)$$
Simplifying we get
$$ \sqrt{3}y + x =4$$
Another hexagon can also be formed with vertices other than $$A$$ and $$B$$ below x-axis. From symmetry, the slope of that line will be $$ \tan {30^{o}}$$
But options A and B have negative slopes. Hence, only option C is correct.
Question 2
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One diagonal of a square is along the line $$8x - 15 y =0$$ and one of its vertices is $$(1, 2)$$. Then the equations of the sides of the square passing through this vertex are

A
$$23 x + 7y = 9, 7x+23 y = 53$$

B
$$23 x - 7y + 9 = 0, 7x + 23 y + 53 = 0$$

C
$$23x - 7y - 9 = 0, 7x + 23 y - 53 =0$$

D
None of these

Solution

Let the square be $$ABCD$$ and vertex $$C$$ be $$(1, 2) $$ . Slope of $$BD$$ is $$\dfrac8{15}$$ and angle made by $$BD$$ with $$DC$$ and $$BC$$ is $$45$$$$^{\circ}$$. So let slope of $$DC$$ be $$m$$. Then,
$$\tan 45^{\circ} = \pm \displaystyle \dfrac{m - \dfrac{8}{15}}{ 1 + \dfrac{8}{15}m}$$
$$\Rightarrow (15 + 8m) = \pm (15 m -8)$$
$$\Rightarrow m = \dfrac{23}{7}$$ and $$ -\dfrac{7}{23}$$
Hence, the equations of $$DC$$ and $$BC$$ are
$$\Rightarrow y -2 = \dfrac{23}{7} (x-1)$$
$$\Rightarrow 23x - 7y - 9 =0$$
and $$y - 2 = - \dfrac{7}{23} (x-1)$$
$$\Rightarrow 7x + 23 y - 53 = 0$$
Question 3
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The diagonals of a parallelogram $$PQRS$$ are along the lines $$x + 3y = 4$$ and $$6x - 2y = 7$$. Then $$PQRS$$ must be a

A
Rectangle

B
Square

C
Cyclic quadrilateral

D
Rhombus

Solution

$$\textbf{Step 1: Check the slopes of given lines}$$

$$\text{We have,}$$

$$x + 3y = 4$$ $$\text{and}$$ $$6x - 2y = 7$$

$$\Rightarrow y = \dfrac{-x}{3} + \dfrac{4}{3}$$ $$\text{and}$$ $$\Rightarrow 6x - 7 = 2y$$
$$\text{and}$$ $$\Rightarrow y = 3x - \dfrac{7}{2}$$

$$\therefore$$ $$\text{Slope}$$ $$\left(m_1\right) = \dfrac{-1}{3}$$ $$\therefore$$ $$\text{Slope}$$ $$\left(m_2\right) = 3$$

$$\therefore m_1\times m_2 = \dfrac{1}{3}\times -3 = -1$$

$$\textbf{Step 2: Use the above results and get the required unknown}$$

$$\because$$ $$\text{the two lines are perpendicular as the multiplication of slope is -1.}$$

$$\therefore$$ $$\text{Diagonals are perpendicular.}$$

$$\textbf{Hence, PQRS is a rhombus.}$$
Question 4
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The ends of a quadrant of a circle have the coordinates (1, 3) and (3, 1). Then the centre of such a circle is

A
(2, 2)

B
(1, 1)

C
(4, 4)

D
(2, 6)

Solution

Let $$O(a,b)$$ be the center of the circle.
AB is the chord of the circle. Draw $$OM \perp AB$$
$$\Rightarrow AM=MB$$
So, M is the mid-point of AB.
Coordinates of mid-point M of AB is (2,2)
In right $$\triangle OMA$$
$$OA^2=OM^2+MA^2$$
$$\Rightarrow (a-3)^2+(b-1)^2=2+(a-2)^2+(b-2)^2$$
$$\Rightarrow a=b$$ ....(1)
In right $$\triangle OAB$$
$$AB^2=OA^2+OB^2$$
$$\Rightarrow 8=2[(a-1)^2+(b-3)^2]$$ ($$\because OA=OB$$)
$$\Rightarrow a=1\ or\ a=3$$
So, the center of circle is (1,1)
Question 5
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If four points are $$A(6,3),B(-3,5),C(4,-2)$$ and $$P(x,y),$$ then the ratio of the areas of $$\triangle PBC$$ and $$\triangle ABC$$ is:

A
$$\displaystyle \frac { x+y-2 }{ 7 } $$

B
$$\displaystyle \frac { x-y-2 }{ 7 } $$

C
$$\displaystyle \frac { x-y+2 }{ 2 } $$

D
$$\displaystyle \frac { x+y+2 }{ 2 } $$

Solution

Given: Coordinates of points $$\displaystyle A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 6,3 \right) ,B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -3,5 \right) ,C\left( { x }_{ 3, }{ y }_{ 3 } \right) =\left( 4,-2 \right) $$ and $$\displaystyle P\left( x,y \right) .$$

We know that the area of:
$$\displaystyle\triangle PBC=\frac { 1 }{ 2 } \left[ x\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 3 }\left( y-{ y }_{ 2 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-y \right) \right] $$

$$\displaystyle=\frac { 1 }{ 2 } \left[ x\left( 5+2 \right) +4\left( y-5 \right) -3\left( -2-y \right) \right] $$ $$\displaystyle=\frac { 1 }{ 2 } \left[ 7x+7y-14 \right] $$

Similarly, the area of

$$\displaystyle\triangle ABC=\frac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) \right] +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) $$

$$\displaystyle=\dfrac{1}{2}[6\left( 5+2 \right) -3\left( -2-3 \right) +4\left( 3-5 \right)] =\dfrac{49}{2}$$

Therefore, the ratio of the areas of $$\triangle PAB$$ and $$\triangle ABC$$

$$\displaystyle=\frac { 7x+7y-14 }{ 49 } =\frac { 7\left( x+y-2 \right) }{ 49 } =\frac { x+y-2 }{ 7 } $$
Question 6
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$$\mathrm{P}_{1},\ \mathrm{P}_{2},\ldots\ldots.,\ \mathrm{P}_{\mathrm{n}}$$ are points on the line $$y=x$$ lying in the positive quadrant such that $$\mathrm{O}\mathrm{P}_{\mathrm{n}}=n\cdot\mathrm{O}\mathrm{P}_{\mathrm{n}-1}$$, where $$\mathrm{O}$$ is the origin. If $$\mathrm{O}\mathrm{P}_1=1$$ and the coordinates of $$\mathrm{P}_{\mathrm{n}}$$ are $$(2520\sqrt{2},2520\sqrt{2})$$, then $$n$$ is equal to

A
$$3$$

B
$$5$$

C
$$7$$

D
$$9$$

Solution

$$P_1,P_2.P_3....$$ are points on line $$y=x$$
$$OP_1=1$$
Let $$P_1=(x,x$$)
But $$ OP_1=1\Rightarrow 2x^2=1\Rightarrow x^2=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{\sqrt2}$$
So $$P_1=(\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2})$$
$$OP_2=2\times OP_1=2\Rightarrow 2x^2=4\Rightarrow x=\sqrt2\Rightarrow P_2=(\sqrt2,\sqrt2)$$
$$OP_3=2\times 3=6\Rightarrow2x^2=6^2\Rightarrow x=3\sqrt2 \Rightarrow P_3=(3\sqrt2,3\sqrt2)$$
....coninuing this way get
$$OP_7=7\times 720=5040\Rightarrow 2x^2=5040\times 5040\Rightarrow x^2=\Rightarrow x=2520\sqrt2\Rightarrow P_7=(2520\sqrt2,2520\sqrt2)$$
Comparing with given point we get $$n=7$$.
Question 7
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If $$A(-2,4)$$, $$B(0,0)$$ and $$C(4,2)$$ are the vertices of a $$\Delta ABC$$, then find the length of median through the vertex A.

A

2

B
4

C
3

D
5

Solution

A median of a triangle is a line segment that joins the vertex of a triangle to the midpoint of the opposite side.

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{

2 }) $$ is calculated by the formula $$ \left( \frac { { x }_{ 1 }+{ x

}_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) $$

Using this formula,

mid point of BC $$ D = \left( \frac { 0 + 4}{ 2 } ,\frac { 0 + 2 }{ 2 } \right)

\quad =\quad (2,1) $$
Distance between two points $$

\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 }

\right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{

x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$

Length of the median A$$ (-2,4) $$ and D $$ (2,1) = \sqrt { \left( 2 + 2

\right) ^{ 2 }+\left( 1 - 4 \right) ^{ 2 } } = \sqrt { 16 + 9 } = \sqrt { 25 }

= 5 $$
Question 8
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The vertices of a triangle are $$A(3,4)$$, $$B(7,2)$$ and $$C(-2, -5)$$. Find the length of the median through the vertex A.

A
$$\dfrac { \sqrt { 111 } }{ 2 } $$units

B
$$\dfrac { \sqrt { 147 } }{ 2 } $$units

C
$$\dfrac { \sqrt { 137 } }{ 2 } $$units

D
$$\dfrac { \sqrt { 122 } }{ 2 } $$units

Solution

A median of a triangle is a line segment that joins the

vertex of a triangle to the midpoint of the opposite side.

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y

}_{ 2 }) $$ is calculated by the formula $$ \left( \frac { { x }_{ 1 }+{

x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) $$

Since AD is the median, this means, D is the mid point of BC.

Using this formula, mid point of BC $$= \left( \frac { 7- 2 }{ 2 } ,\frac { 2 - 5 }{ 2 }

\right) = (\dfrac {5}{2},\dfrac {-3}{2}) $$
Distance between two points $$

\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 }

\right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{

x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Hence, length of AD $$ = \sqrt { \left( \dfrac { 5 }{ 2 } -3 \right) ^{ 2 }+\left( \dfrac { -3 }{ 2 } -4 \right) ^{ 2 } } =\sqrt { \dfrac { 1 }{ 4 } +\dfrac { 121 }{ 4 } } =\sqrt { \dfrac { 122 }{ 4 } } =\dfrac { \sqrt { 122 } }{ 2 } $$
Question 9
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Directions For Questions

The area of a triangle whose vertices are $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ and $$\displaystyle \left( { x }_{ 3 },{ y }_{ 3 } \right) $$ is given by $$\displaystyle \Delta =\frac { 1 }{ 2 } \left| { x }_{ 1 }\left( { y }_{ 2 },{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 },{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 },{ y }_{ 2 } \right) \right| $$. The points $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ and $$\displaystyle \left( { x }_{ 3 },{ y }_{ 3 } \right) $$ are collinear of $$\displaystyle \Delta =0$$

...view full instructions

Determine the area of the triangle whose vertices are $$\displaystyle \left( \frac { 1 }{ 2 } ,\frac { -1 }{ 2 } \right) ,\left( 2,\frac { -1 }{ 2 } \right) $$ and $$\displaystyle \left( 2,\frac { \sqrt { 3 } -1 }{ 2 } \right) $$.

A
$$\displaystyle 3\sqrt { 3 } $$

B
$$\displaystyle \frac { 3\sqrt { 3 } }{ 8 } $$

C
$$\displaystyle 12\sqrt { 3 } $$

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

Solution
Question 10
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Find the equation of the line that passes through the points $$(-1,0)$$ and $$(-4,12)$$

A
$$y+4x=-1$$

B
$$y+4x=-4$$

C
$$-y+4x=-8$$

D
$$y-8x=-12$$

Solution

Since slope of line passing through two points $$\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ is $${ m }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$
We now find the slope of the line passing through the points $$(-1,0)$$ and $$(-4,12)$$ as shown below:
$${ m }=\dfrac { 12-0 }{ -4-(-1) } =\dfrac { 12 }{ -3 } =-4$$
Therefore, the slope of the line is $$-4$$.
Now use the slope and either of the two points to find the $$y$$-intercept.
$$y=mx+b$$
$$0=(-4)(-1)+b$$
$$0=4+b$$
$$b=-4$$
Write the equation in slope intercept form as:
$$y=mx+b$$
$$y=(-4)x-4$$
$$y=-4x-4$$
$$y+4x=-4$$
Hence, the equation of the line is $$y+4x=-4$$.

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Straight Lines Test 30

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Straight Lines Test 30

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SHARING IS CARING

Question 1

$$\sqrt{3} x + y =4$$

$$x + \sqrt{3} y + 4 =0$$

$$x + \sqrt{3} y = 4$$

None of these

Solution

Question 2

$$23 x + 7y = 9, 7x+23 y = 53$$

$$23 x - 7y + 9 = 0, 7x + 23 y + 53 = 0$$

$$23x - 7y - 9 = 0, 7x + 23 y - 53 =0$$

None of these

Solution

Question 3

Rectangle

Square

Cyclic quadrilateral

Rhombus

Solution

Question 4

(2, 2)

(1, 1)

(4, 4)

(2, 6)

Solution

Question 5

$$\displaystyle \frac { x+y-2 }{ 7 } $$

$$\displaystyle \frac { x-y-2 }{ 7 } $$

$$\displaystyle \frac { x-y+2 }{ 2 } $$

$$\displaystyle \frac { x+y+2 }{ 2 } $$

Solution

Question 6

$$3$$

$$5$$

$$7$$

$$9$$

Solution

Question 7

2

4

3

5

Solution

Question 8

$$\dfrac { \sqrt { 111 } }{ 2 } $$units

$$\dfrac { \sqrt { 147 } }{ 2 } $$units

$$\dfrac { \sqrt { 137 } }{ 2 } $$units

$$\dfrac { \sqrt { 122 } }{ 2 } $$units

Solution

Question 9

$$\displaystyle 3\sqrt { 3 } $$

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

$$\displaystyle 12\sqrt { 3 } $$

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

Solution

Question 10

$$y+4x=-1$$

$$y+4x=-4$$

$$-y+4x=-8$$

$$y-8x=-12$$

Solution

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