Straight Lines Test 15

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Straight Lines Test 15

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Weekly Quiz Competition

Question 1
1 / -0

The points $$(-4,-4), (-1,-2)$$ and $$(x,-8)$$ are the vertices of a right triangle with the right angle at $$(-1,-2)$$. Find the value of $$x$$.

A
$$0$$

B
$$2$$

C
$$3$$

D
$$-1$$

E
$$-4$$

Solution

Since the right angle is at the point $$B$$, $$BA$$ and $$BC$$ are the legs of the right triangle and $$AC$$ is the hypotenuse.
So we find the squares of distances $$AC$$, $$BC$$ and $$AB$$:
$${ \left( AB \right) }^{ 2 }={ \left( -1+4 \right) }^{ 2 }+{ \left( -2+4 \right) }^{ 2 }=13$$
$${ \left( BC \right) }^{ 2 }={ \left( x+1 \right) }^{ 2 }+{ \left( -8+2 \right) }^{ 2 }={ \left( x+1 \right) }^{ 2 }+36$$
$${ \left( AC \right) }^{ 2 }={ \left( x+4 \right) }^{ 2 }+{ \left( -8+4 \right) }^{ 2 }={ \left( x+4 \right) }^{ 2 }+16$$
Apply pythagoras theorem:
$$\Rightarrow { \left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ \left( AC \right) }^{ 2 }$$
$$\Rightarrow 13+{ \left( x+1 \right) }^{ 2 }+36={ \left( x+4 \right) }^{ 2 }+16$$
$$\Rightarrow 13+x^{ 2 }+1+2x+36=x^{ 2 }+16+8x+16$$
$$\Rightarrow -6x=-18$$
$$\Rightarrow x=3$$
Hence, option C is correct.
Question 2
1 / -0

The perimeter of the triangle with vertices $$(1,3), (1,7)$$ and $$(4,4)$$ is

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

B
$$3\sqrt {2}$$

C
$$6 + 3\sqrt {2}$$

D
$$9 + \sqrt {2}$$

Solution

Assume $$A(1, 3), B(1, 7), C(4, 4)$$
Formula of distance $$=$$ $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2)}$$
Distance of $$AB =$$ $$\sqrt{(1-1)^2+(7-3)^2}$$
$$=$$ $$\sqrt{0+4^2}$$
$$= 4$$
Likewise calculate the distance for $$BC$$ and $$CA$$.
Distance of $$BC =$$ $$\sqrt{(4-1)^2+(4-7)^2} = 3\sqrt{2}$$
Distance of $$CA =$$ $$\sqrt{(1-4)^2+(3-4)^2} = 2$$
Perimeter $$=$$ distance $$AB$$ $$+$$ distance $$BC$$ $$+$$ distance $$CA$$
$$=$$ $$4+ 3\sqrt{2}+2$$
$$=$$ $$6+3\sqrt{2}$$
Question 3
1 / -0

From the above figure, calculate the length of $$AG$$, if point $$G$$ is the center of rectangle $$BCEF$$.

A
$$\sqrt {10}$$

B
$$\sqrt {13}$$

C
$$\sqrt {85}$$

D
$$\sqrt {97}$$

E
$$11$$

Solution

$$G$$ will be the midpoint of $$BE$$.
Hence, $$G=\dfrac{6+12}{2},\dfrac{4+0}{2}=9,2$$
Hence, $$AG=\sqrt{(9-0)^{2}+(2-0)^{2}}$$
$$=\sqrt{81+4}$$
$$=\sqrt{85}$$
Question 4
1 / -0

Find the possible co-ordinates of the fourth corner of a parallelogram if its three corners are located at $$(3, 3), (4, 4)$$, and $$(2, 1)$$.

A
$$(8, 0)$$

B
$$(8, -1)$$

C
$$(-1, 9)$$

D
$$(1, 0)$$

E
$$(4,-3)$$

Solution

Let the corners be $$A (3,3) ; B (4,4) ; C (2,1)$$
Let the fourth vertex D $$ = (x,y) $$
We know that the diagonals of a parallelogram bisect each other. So, the midpoint of AC is same as the midpoint of BD.
Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{ 2 }) $$ is calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) $$.
So, midpoint of $$ AC = $$ Mid point of $$ BD $$
$$ \Rightarrow \left( \dfrac { 3+2 }{ 2 } ,\dfrac { 3 + 1 }{ 2 } \right) = \left( \dfrac { 4+x }{ 2 } ,\dfrac { 4 +y }{ 2 } \right) $$
$$\Rightarrow \left( \dfrac { 5 }{ 2 } ,\dfrac { 4 }{ 2 } \right) = \left( \dfrac { 4+x }{ 2 } ,\dfrac { 4 +y }{ 2 } \right)$$
$$\Rightarrow 4+x=5 ; 4 + y = 4 $$
$$\Rightarrow x = 1 ; y = 0 $$
Hence, $$ D = (1,0) $$.
Question 5
1 / -0

In the XY-coordinate plane, point P is a distance of $$4$$ from the point $$(1, 1)$$. Which of the following could be P?

A
$$(1, 2)$$

B
$$(1, 5)$$

C
$$(3, 1)$$

D
$$(4, 1)$$

Solution

It may help a lot to make a figure so you can visualize what is going on:
The figure above shows the point $$(1, 1)$$ along with the five possible answers for point P.
Only answer B, point $$(1, 5)$$, is a distance of $$4$$ from point $$(1, 1)$$: the y value increases by $$4$$ units to go from $$(1, 1)$$ to $$(1, 5)$$, and the x-value doesn't change.
Question 6
1 / -0

The distance between the points $$A(3,4)$$ and $$B(5,8)$$

A
$$2\sqrt 5$$

B
$$2\sqrt 3$$

C
$$2\sqrt 7$$

D
$$\sqrt 5$$

Solution

To find the value of $$x$$, we’ll utilize the distance formula which can be used to calculate the distance $$d$$ between any two points in a coordinate plane.
This formula is given as follows:
$$d = [(x_2 – x_1)^2 + (y_2 – y_1)^2]^(½)$$
$$d = [(5-3 )^2 + (8-4)^2]^(½)$$
$$d = [(2 )^2 + (4)^2]^(½)$$
$$d = [4 + 16]^(½)$$
$$d = [20]^(½)$$
$$d=2\sqrt { 5 }$$
Question 7
1 / -0

The value of $$x$$ if the distance between the points $$(x,4)$$ and $$(-5,3)$$ is equal to $$5$$.

A
$$x=-5\pm2\sqrt{6}$$

B
$$x=-5\pm2\sqrt{5}$$

C
$$x=-5\pm2\sqrt{23}$$

D
$$x=-5\pm2\sqrt{7}$$

Solution

Let $$A(x,4)$$ and $$B(-5,3)$$. Then it is given that $$AB=5units$$. Therefore, $$\sqrt{(x+5)^{2}+(4-3)^{2}}=5$$ or $$(x+5)^{2}+1=25$$ or $$(x+5)^{2}=24$$ or $$x+5=\pm2\sqrt{6}$$ or $$x=-5\pm2\sqrt{6}$$. Thus we get two values of x satisfying the above condition.
Question 8
1 / -0

In Figure 1, calculate the distance from the midpoint of segment $$AC$$ to the midpoint of segment $$BD$$.

A
1.118

B
1.414

C
1.803

D
2.236

E
2.828

Solution

Midpoint of $$AC$$ is $$\left (0 , \dfrac {3}{2}\right)$$.......[hint: using mid-point formula i.e, $$ x = (x_1 + x_2)/2\ ,\ y = (y_1 + y_2)/2$$ ]
Midpoint of $$BD$$ is $$(-1,1/2)$$
Distance between those two midpoints is $$\sqrt { { (1) }^{ 2 }+{ \left (\dfrac {3}{2} - \dfrac {1}{2}\right) }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } = 1.414$$
Question 9
1 / -0

Find the length of the hypotenuse of the right triangle whose vertices are given by the points $$(-2,1),(1,1)$$ and $$(1,2)$$

A
$$\sqrt {10}$$

B
$$\sqrt 5$$

C
$$10$$

D
$$2 \sqrt 5$$

Solution

Let the points be $$A(-2,1)$$, $$B(1,1)$$ and $$C(1,2)$$
Given: $$\triangle ABC$$ is right-angled.
By distance formula,,
$$d(A,B) = \sqrt {(1+2)^2 + (1-1)^2} = \sqrt {9} = 3$$
$$d(A,C) = \sqrt {(1+2)^2 + (2-1)^2} = \sqrt {10} $$
$$d(B,C) = \sqrt {(1-1)^2 + (2-1)^2} = \sqrt {1} = 1$$
Hence, the longest side i.e. hypotenuse of the right angled $$\triangle ABC$$ is $$AC = \sqrt {10}$$ units.
Question 10
1 / -0

Find the distance between the points $$(2,3)$$ and $$(0,6)$$.

A
$$\sqrt 3$$

B
$$\sqrt {13}$$

C
$$\sqrt {14}$$

D
$$\sqrt 5$$

Solution

We’ll use the distance formula which can be used to calculate the distance d between any two points in a coordinate plane.
This formula is given as follows:
$$d = \sqrt {(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
$$d =\sqrt {(0-2 )^2 + (6-3)^2}$$
$$d = \sqrt{(-2 )^2 + (3)^2}$$
$$d = \sqrt{4 + 9}$$
$$d=\sqrt {13}$$.

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Straight Lines Test 15

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Straight Lines Test 15

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SHARING IS CARING

Question 1

$$0$$

$$2$$

$$3$$

$$-1$$

$$-4$$

Solution

Question 2

$$3 + \sqrt {2}$$

$$3\sqrt {2}$$

$$6 + 3\sqrt {2}$$

$$9 + \sqrt {2}$$

Solution

Question 3

$$\sqrt {10}$$

$$\sqrt {13}$$

$$\sqrt {85}$$

$$\sqrt {97}$$

$$11$$

Solution

Question 4

$$(8, 0)$$

$$(8, -1)$$

$$(-1, 9)$$

$$(1, 0)$$

$$(4,-3)$$

Solution

Question 5

$$(1, 2)$$

$$(1, 5)$$

$$(3, 1)$$

$$(4, 1)$$

Solution

Question 6

$$2\sqrt 5$$

$$2\sqrt 3$$

$$2\sqrt 7$$

$$\sqrt 5$$

Solution

Question 7

$$x=-5\pm2\sqrt{6}$$

$$x=-5\pm2\sqrt{5}$$

$$x=-5\pm2\sqrt{23}$$

$$x=-5\pm2\sqrt{7}$$

Solution

Question 8

1.118

1.414

1.803

2.236

2.828

Solution

Question 9

$$\sqrt {10}$$

$$\sqrt 5$$

$$10$$

$$2 \sqrt 5$$

Solution

Question 10

$$\sqrt 3$$

$$\sqrt {13}$$

$$\sqrt {14}$$

$$\sqrt 5$$

Solution

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