Coordinate Geometry Test

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Coordinate Geometry Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The mid-point of line segment joining the points (3, 0) and (-1, 4) is :

A
1, 2

B
2, 3

C
-1, 2

D
None of these

Solution

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) $$

Using this formula, mid point of given points $$= \left( \dfrac { 3+(-1) }{ 2 } ,\dfrac { 0+4 }{ 2 }

\right) \quad =\quad (1,2) $$
Question 2
1 / -0

There are two point $$P(1,-4)$$ and $$Q(4,2)$$. Find a point X dividing the line PQ in the ratio $$1:2$$

A
$$2, -2$$

B
$$2,-3$$

C
$$1,-3$$

D
$$1,-2$$

Solution

Using the section formula, if a point $$(x,y)$$ divides the

line joining the points $$({ x }_{ 1 },{ y }_{ 1 })$$ and $$({ x }_{ 2 },{ y

}_{ 2 })$$ internally in the ratio $$ m:n $$, then $$(x,y) = \left( \frac { m{

x }_{ 2 } + n{ x }_{ 1 } }{ m + n } ,\frac { m{ y }_{ 2 } + n{ y }_{ 1 }

}{ m + n } \right) $$

Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (1,-4) $$ and $$({x }_{ 2 },{ y }_{

2 }) = (4,2) $$ and $$ m = 1, n = 2 $$ in the section formula, we get

the point $$ \left( \frac { 1(4) + 2(1) }{ 2 +1 } ,\frac { 1(2) + 2(-4)

}{ 2 + 1 } \right) =\left( 2,-2 \right) $$
Question 3
1 / -0

The co-ordinates of a point (x,y) which divides the straight line joining two points $$\displaystyle \left ( x_{1},y_{1} \right )$$ and $$\displaystyle \left ( x_{2} ,y_{2}\right )$$ internally in the ratio $$\displaystyle m_{1}$$ and $$\displaystyle m_{2}$$ are

A
$$\displaystyle x=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}},y=\frac{m_{1}y_{1}+m_{2}y_{2}}{m_{1}+m_{2}}$$

B
$$\displaystyle x=\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},y=\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$$

C
x=0,y=0

D
$$\displaystyle x=\frac{m_{1}x_{1}-m_{2}x_{2}}{m_{1}-m_{2}},y=\frac{m_{1}y_{1}-m_{2}y_{2}}{m_{1}-m_{2}}$$

Solution

This is the Section formula, i.e. If a point divides the line segment $$AB$$ joining given two points in given m:n ratio internally, then by section formula, the co-ordinates of a point are given as

$$=\left (\dfrac {mx_{2}+nx_{1}}{m+n},\dfrac {my_{2}+ny_{1}}{m+n}\right) $$
Question 4
1 / -0

The mid-point of a line segment is $$(-4,-2)$$ and one end of the line segement is $$(-6,4)$$. The co-ordinates of the other end are

A
$$(2,-8)$$

B
$$(-2,8)$$

C
$$(-2,-8)$$

D
$$(2,8)$$

Solution

The co-ordinates of the mid point of a line segment whose end points are $$(x_1,y_1) $$ and $$(x_2,y_2)$$ are given by $$(\cfrac{x_1+x_2}{2},\cfrac{y_1+y_2}{2})$$

Substituting $$(x_1,y_1)=(-6,4)$$ and mid-point $$(-4,-2)$$.

$$\Rightarrow (-4,-2)=(\cfrac{-6+x_2}{2},\cfrac{4+y_2}{2})$$

Equating the $$x-$$coordinates,

$$\Rightarrow -4=\cfrac{-6+x_2}{2}$$

$$\Rightarrow -4\times 2=-6+x_2$$

$$\Rightarrow x_2=6-8$$

$$\Rightarrow x_2=-2$$

Equating the $$y-$$coordinates

$$\Rightarrow -2=\cfrac{4+y_2}{2}$$

$$\Rightarrow -2\times 2=4+y_2$$

$$\Rightarrow -y_2=4+4$$

$$\Rightarrow y_2=-8$$

The coordinates of the other end are $$(-2,-8)$$.
Question 5
1 / -0

Find the point, which bisects line segment joining the points $$P(-2,0)$$ and $$Q(2,5)$$

A
$$\left (0,\dfrac{5}{2}\right )$$

B
$$(0,2)$$

C
$$(2,0)$$

D
$$(-2,0)$$

Solution

Given points are $$P(-2,0) = (x_1,y_1)$$ and $$Q(2,5)=(x_2,y_2)$$

Formula for Midpoint $$=( \dfrac { { x }_{ 2 }+{ x }_{ 1 } }{ 2 } ,\dfrac { { y }_{ 2 }+{ y }_{ 1 } }{ 2 })$$

Midpoint of $$PQ=\left (\dfrac {-2+2}{2},\dfrac{0+5}{2}\right)=\left (0,\dfrac {5}{2}\right) $$
Question 6
1 / -0

If the area of a triangle is $$68 $$ sq. units and the vertices are $$(6, 7), (-4, 1)$$ and $$(a, -9) $$ then the value of $$a$$ is

A
1

B
2

C
3

D
4

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is:
$$A= \left| \dfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (6,7) $$ ; $$({ x }_{ 2

},{ y }_{ 2 }) = (-4,1) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (a,-9)$$ in the formula for area, we get:

Area of triangle $$ = \left| \dfrac { (6)(1+9)+(-4)(-9-7) + a(7-1) }{ 2 }

\right| = 68 $$
$$ \left| \dfrac { 60 + 64 + 6a }{ 2 } \right| = 68 $$
$$ \dfrac{124 + 6a}{2} = 68 $$
$$ 124 + 6a = 136 $$
$$ 6a = 12 $$
$$ \implies a = 2 $$
Question 7
1 / -0

The co-ordinates of the point which divides the line joining $$(1,-2)$$ and $$(4,7)$$ internally in the ratio $$1:2$$ are

A
$$(1,2)$$

B
$$(-1,-1)$$

C
$$(-1,2)$$

D
$$(2,1)$$

Solution

We know if a point divides the line segment $$AB$$ joining two points $$A\left(x_{1}, y_{1}\right)$$ and $$B\left(x_{2}, y_{2}\right)$$ in given ratio m:n internally, then by section formula, the cordinates of a point are given as
$$=\left (\dfrac {mx_{2}+nx_{1}}{m+n},\dfrac {my_{2}+ny_{1}}{m+n}\right) $$

Since the points $$(1,-2)$$ and $$(4,7)$$ divides internally in the ratio

$$1:2$$, then the coordinates of the point can be determined as follows:

$$(x,y)=\left( \dfrac { 1\times 4+2\times 1 }{ 1+2 } ,\dfrac { 1\times 7+2\times -2 }{ 1+2 } \right) =\left( \dfrac { 4+2 }{ 3 } ,\dfrac { 7-4 }{ 3 } \right) =\left( \dfrac { 6 }{ 3 } ,\dfrac { 3 }{ 3 } \right) =\left( 2,1 \right)$$
Question 8
1 / -0

The point which divides the line segment joining the points (3, 5) and (8, 10) internally in the ratio 2:3 is:

A
3, 5

B
5, 3

C
5, 7

D
7, 5

Solution

Using the section formula, if a point $$(x,y)$$ divides the line joining the

points $$({ x }_{ 1 },{ y }_{ 1 })$$ and $$({ x }_{ 2 },{ y }_{ 2 })$$ internally

in the ratio $$ m:n $$, then $$(x,y) = \left( \frac { m{ x }_{ 2 } + n{ x }_{ 1

} }{ m + n } ,\frac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n }

\right) $$

Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (3,5) $$ and $$({x }_{ 2 },{ y }_{

2 }) = (8,10) $$ and $$ m = 2, n = 3 $$ in the section formula, we get

the point $$ \left( \frac { 2(8) + 3(3) }{ 2 +3 } ,\frac { 2(10) + 3(5)

}{ 2 + 3 } \right) =\left( 5,7 \right) $$
Question 9
1 / -0

The point P divides the line joining the points (5, 0) and (0, 4) in the ratio 2 : 3 internally. The x co-ordinate of P is

A
2

B
1

C
3

D
4

Solution
Question 10
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.

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Coordinate Geometry Test - 30

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Question 1

1, 2

2, 3

-1, 2

None of these

Solution

Question 2

$$2, -2$$

$$2,-3$$

$$1,-3$$

$$1,-2$$

Solution

Question 3

$$\displaystyle x=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}},y=\frac{m_{1}y_{1}+m_{2}y_{2}}{m_{1}+m_{2}}$$

$$\displaystyle x=\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},y=\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$$

x=0,y=0

$$\displaystyle x=\frac{m_{1}x_{1}-m_{2}x_{2}}{m_{1}-m_{2}},y=\frac{m_{1}y_{1}-m_{2}y_{2}}{m_{1}-m_{2}}$$

Solution

Question 4

$$(2,-8)$$

$$(-2,8)$$

$$(-2,-8)$$

$$(2,8)$$

Solution

Question 5

$$\left (0,\dfrac{5}{2}\right )$$

$$(0,2)$$

$$(2,0)$$

$$(-2,0)$$

Solution

Question 6

1

2

3

4

Solution

Question 7

$$(1,2)$$

$$(-1,-1)$$

$$(-1,2)$$

$$(2,1)$$

Solution

Question 8

3, 5

5, 3

5, 7

7, 5

Solution

Question 9

2

1

3

4

Solution

Question 10

2.25

2.5

2.75

3

Solution

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