Straight Lines Test 11

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Straight Lines Test 11

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Weekly Quiz Competition

Question 1
1 / -0

If mid point of the line segment joining (2a,4) and (-2,3b) is (1,2a+1), then the values of a and b are given by

A
a=2, b=-2

B
a=b=2

C
a=1=b

D
a=-2, b=2

Solution

Since $$(1,2a+1)$$ is the midpoint of line segment that means it divides line in ratio $$1:1$$
Now using section formula, we have
$$1=\frac{2a-2}{2}$$ and $$(2a+1) =\frac{4+3b}{2}$$
On solving we get
$$a=2\;\;and \;\; b=2$$
Question 2
1 / -0

$$(0, 0), (0, 3)$$ and $$(3, 0)$$ form ___________.

A
a straight line

B
an equilateral triangle

C
a scalene triangle

D
a right triangle

Solution

Let the vertices be $$A(0,0)$$, $$B(0,3)$$ and $$C(3,0)$$. Now the measure of sides are
$$AB=3$$ units
$$BC=\sqrt{(0-3)^{2}+(3-0)^{2}}=\sqrt{3^{2}+3^{2}}=3\sqrt{2}units$$.
$$AC=3$$ units.
Hence, we can observe that $$AB^{2}+AC^{2}=3^2+3^{2}=2(3^{2})=BC^{2}$$.
Thus, it follows Pythagoras theorem with BC being the hyptenuse and AC and AB being the perpendicular sides.
Hence, the above vertices form a right angled triangle.
Question 3
1 / -0

Find the point on the y-axis which is equidistant from $$A(3, -6)$$ and $$B(-2, 5)$$.

A
$$\displaystyle \left ( \frac{9}{11}, 0 \right )$$

B
$$\displaystyle \left ( 0, \frac{8}{11} \right )$$

C
$$\displaystyle \left ( 0, \frac{-8}{11} \right )$$

D
$$\displaystyle \left ( 0, \frac{9}{11} \right )$$

Solution

Let the point be $$(x,y)$$.
Distance formula: $$d^2=(x_1-x_2)^2+(y_1-y_2)^2$$
Now equate the distances of the points $$A(3,-6)$$ and $$B(-2,5)$$ from the point $$(x,y)$$.
$$(x-3)^2+(y+6)^2=(x+2)^2+(y-5)^2$$
The $$x-$$coordinate of a point on $$y-$$axis is $$0$$.
$$(0-3)^2+(y+6)^2=(0+2)^2+(y-5)^2$$
$$9+y^2+12y+36=4+y^2-10y+25$$
$$12y+10y=29-45$$
$$22y=-16$$
$$y=-\dfrac { 8 }{ 11 }$$
Hence, the point on the y-axis is $$\left( 0,-\dfrac { 8 }{ 11 } \right)$$.
Question 4
1 / -0

$$L, M$$ and $$N$$ are the midpoints of the sides $$BC, CA$$ and $$AB$$ respectively of triangle $$ABC$$. If the vertices are $$A(3,-4), B(5,-2)$$ and $$C(1,3)$$ the area of $$\displaystyle \triangle LMN$$ is ____ square units.

A
2.25

B
2.5

C
2.75

D
3

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$

Let $$(x_1,y_1)$$ $$=(3,-4)$$, $$(x_2,y_2)$$ $$=(5,-2)$$ and $$(x_3,y_3)$$ $$=(1,3)$$

Area of $$\displaystyle \Delta ABC=\dfrac{1}{2}\left [ 3\left ( -2-3 \right )+5\left ( 3+4 \right )+1\left ( -4+2 \right ) \right ]$$

$$=\dfrac{1}{2}\left ( -15+35-2 \right )$$

$$=9$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta LMN=\frac{1}{4}\times 9=2.25$$ square units.
Question 5
1 / -0

Find the perimeter of the triangle formed by the points $$(3, 5), (4, 8)$$ and $$(5, 6)$$.

A
$$\sqrt{5}(2 + \sqrt{2})$$

B
$$\sqrt{3}(2 + \sqrt{2})$$

C
$$\sqrt{2}(5 + \sqrt{3})$$

D
$$\sqrt{5}(2 + \sqrt{3})$$

Solution

To find the perimeter of the triangle formed by the points $$A(3,5)$$, $$B(4,8)$$ and $$C(5,6)$$, we must find the distance between these three points that can be determined as follows:

$$AB=\sqrt { { (4-3) }^{ 2 }+{ (8-5) }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 1+9 } =\sqrt { 10 }$$

$$BC=\sqrt { { (5-4) }^{ 2 }+{ (6-8) }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 1+4 } =\sqrt { 5 }$$

$$AC=\sqrt { { (5-3) }^{ 2 }+{ (6-5) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 4+1 } =\sqrt { 5 }$$

Now the perimeter is $$AB+BC+AC$$ that is

$$AB+BC+AC=\sqrt { 10 } +\sqrt { 5 } +\sqrt { 5 } =\sqrt { 10 } +2\sqrt { 5 } =\sqrt { 5 } (2+\sqrt { 2 } )$$

Hence the perimeter is $$\sqrt { 5 } (2+\sqrt { 2 } )$$.
Question 6
1 / -0

Which of the following points is equidistant from $$(3,2)$$ and $$(-5,-2)$$?

A
$$(0,2)$$

B
$$(0,-2)$$

C
$$(2,0)$$

D
$$(2,-2)$$

Solution

Let the coordinates of the required point be $$x$$ and $$y$$.
Then application of distance formula gives us
$$(x-3)^{2}+(y-2)^{2}=(x+5)^{2}+(y+2)^{2}$$
$$(x+5)^{2}-(x-3)^{2}+(y+2)^{2}-(y-2)^{2}=0$$
$$(2x+2)(8)+(2y)(4)=0$$
Or $$(x+1)(16)+8y=0$$ or $$2x+y+2=0$$.
Hence, the locus of all those points which are equidistant from the points $$(3,2)$$ and $$(-5,-2)$$ is $$2x+y+2=0$$
Now the point lying on the $$y-axis$$ and is equidistant from the above two points will be $$(2x+y+2=0)_{x=0}$$ or $$y+2=0$$ or $$y=-2$$.
Hence the point $$(0,-2)$$ is point which lies on the y-axis and is equidistant from the given two points.
Conversely, from the above given options, only option B satisfies the locus $$2x+y+2=0$$.
Question 7
1 / -0

The midpoints of the sides of triangle $$ABC$$ are $$ (-1,-2), (6,1)$$ and $$(3,5) $$. The area of $$\displaystyle \triangle ABC$$ is ____ square units.

A
$$72$$

B
$$73$$

C
$$74$$

D
$$75$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle formed with midpoints
$$\displaystyle =\frac{1}{2}\left [ -1\left ( 1-5 \right )+6\left ( 5+2 \right )+3\left ( -2-1 \right ) \right ]$$
$$\displaystyle =\frac{37}{2}$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta ABC=4\times \frac{37}{2}=74$$ square units.
$$[\because$$ Area of triangle formed by joining mid-points of the sides of given triangle is $$\left(\dfrac{1}{4}\right)^{th}$$ of the area of original triangle $$]$$
Question 8
1 / -0

Which of the following points are the vertices of an equilateral triangle?

A
$$(a,a), (-a,-a), (2a,a)$$

B
$$(a,a), (-a,-a), \displaystyle \left ( -a\sqrt{3},a\sqrt{3} \right )$$

C
$$\displaystyle \left ( \sqrt{2}a,-a \right ),\left ( a,\sqrt{2}a \right ),\left ( a,-a \right )$$

D
$$(0,0), (a,-a), \displaystyle \left ( a,\sqrt{2}a \right )$$

Solution

Consider option A;
Let $$A=(a,a)$$ ,$$B=(-a,-a)$$ and $$C=(2a,a)$$. Then the distance between A and B is $$AB=2a\sqrt{2}$$. Similarly $$BC=\sqrt{13}a$$. Hence we can see that the sides are not equal in length. Hence these are not the vertices of an equilateral triangle.

Consider B
Let $$A=(a,a)$$ ,$$B=(-a,-a)$$ and $$C=(-a\sqrt{3},a\sqrt{3})$$. Then the distance between A and B is $$AB=2a\sqrt{2}$$. Similarly $$BC=2\sqrt{2}a$$ and $$AC=2\sqrt{2}a$$. Hence we can see that the sides are equal in length. Also $$(a,a)$$ and $$(-a,-a)$$ lie on $$y=x$$ where as $$(-a\sqrt{3},a\sqrt{3})$$ lies on $$y=-x$$. Hence the points are non-collinear. Hence these are the vertices of an equilateral triangle.
Question 9
1 / -0

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

A
$$(-2,0)$$

B
$$(2,0)$$

C
$$(0,2)$$

D
$$(2,2)$$

Solution

Let the point of x-axis be $$P(x, 0)\equiv(x_1,y_1)$$

Given: $$A(5,4)\equiv(x_2,y_2)$$ and $$B(-2, 3)$$ are equidistant from $$P$$

That is $$PA = PB$$

Hence $$PA^2 = PB^2$$ ................. (1)

Distance between two points is $$\sqrt { { ({ x }_{ 2 } }-{ x }_{ 1 })^{ 2 }+{ ({ y }_{ 2 } }-{ y }_{ 1 })^{ 2 } }$$

$$PA=\sqrt { (5-x)^{ 2 }+(4-0)^{ 2 } } =\sqrt { x^{ 2 }+25-10x+16 } =\sqrt { x^{ 2 }-10x+41 }$$

Therefore, $$PA^{ 2 }=x^{ 2 }-10x+41$$

Here $$P(x, 0)\equiv(x_1,y_1)$$ $$B(-2, 3)\equiv(x_2,y_2)$$

Similarly, $$PB=\sqrt { (-2-x)^{ 2 }+(3-0)^{ 2 } } =\sqrt { x^{ 2 }+4+4x+9 } =\sqrt { x^{ 2 }+4x+13 }$$

Therefore, $$PB^{ 2 }=x^{ 2 }+4x+13$$

Equation (1) becomes

$$x^2-10x+41=x^2+4x+13$$

$$-10x-4x=13-41$$

$$-14x = -28$$

$$x =2$$

Hence the point on x-axis is $$(2, 0)$$.
Question 10
1 / -0

If the distances of $$P(x,y)$$ from $$A(-1,5)$$ and $$B(5,1)$$ are equal, then

A
$$2x=y$$

B
$$3x=2y$$

C
$$3x=y$$

D
$$2x=3y$$

Solution

Let the points be $$P(x, y)$$, $$A(-1,5)$$ and $$B(5,1)$$

Distance between two points is $$\sqrt { { ({ x }_{ 2 } }-{ x }_{ 1 })^{ 2 }+{ ({ y }_{ 2 } }-{ y }_{ 1 })^{ 2 } }$$

$$PA=\sqrt { (-1-x)^{ 2 }+(5-y)^{ 2 } } =\sqrt { x^{ 2 }+1+2x+25+y^2-10y } =\sqrt { x^2+y^2+2x-10y+26 }$$

Therefore, $$PA^{ 2 }=x^2+y^2+2x-10y+26$$

Similarly,

$$PB=\sqrt { (5-x)^{ 2 }+(1-y)^{ 2 } } =\sqrt { 25+x^2-10x+1+y^2-2y } =\sqrt { x^2+y^2-10x-2y+26 }$$

Therefore, $$PB^{ 2 }=x^2+y^2-10x-2y+26$$

Since it is given that the distances are equal therefore $$PA^2=PB^2$$ that is

$$x^2+y^2+2x-10y+26=x^2+y^2-10x-2y+26$$
$$2x-10y=-10x-2y$$
$$2x+10x=-2y+10y$$
$$12x=8y$$
$$3x=2y$$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 11

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Straight Lines Test 11

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SHARING IS CARING

Question 1

a=2, b=-2

a=b=2

a=1=b

a=-2, b=2

Solution

Question 2

a straight line

an equilateral triangle

a scalene triangle

a right triangle

Solution

Question 3

$$\displaystyle \left ( \frac{9}{11}, 0 \right )$$

$$\displaystyle \left ( 0, \frac{8}{11} \right )$$

$$\displaystyle \left ( 0, \frac{-8}{11} \right )$$

$$\displaystyle \left ( 0, \frac{9}{11} \right )$$

Solution

Question 4

2.25

2.5

2.75

3

Solution

Question 5

$$\sqrt{5}(2 + \sqrt{2})$$

$$\sqrt{3}(2 + \sqrt{2})$$

$$\sqrt{2}(5 + \sqrt{3})$$

$$\sqrt{5}(2 + \sqrt{3})$$

Solution

Question 6

$$(0,2)$$

$$(0,-2)$$

$$(2,0)$$

$$(2,-2)$$

Solution

Question 7

$$72$$

$$73$$

$$74$$

$$75$$

Solution

Question 8

$$(a,a), (-a,-a), (2a,a)$$

$$(a,a), (-a,-a), \displaystyle \left ( -a\sqrt{3},a\sqrt{3} \right )$$

$$\displaystyle \left ( \sqrt{2}a,-a \right ),\left ( a,\sqrt{2}a \right ),\left ( a,-a \right )$$

$$(0,0), (a,-a), \displaystyle \left ( a,\sqrt{2}a \right )$$

Solution

Question 9

$$(-2,0)$$

$$(2,0)$$

$$(0,2)$$

$$(2,2)$$

Solution

Question 10

$$2x=y$$

$$3x=2y$$

$$3x=y$$

$$2x=3y$$

Solution

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