Coordinate Geometry Test

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Coordinate Geometry Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Three vertices of rhombus taken in order are $$(2, -1), (3, 4)$$ and $$(-2, 3)$$. Find the fourth vertex.

A
$$(1, 2)$$

B
$$(-3, -2)$$

C
$$(3, 2)$$

D
None of these

Solution

Let M be the mid point of one diagonal formed by (2, -1) and (-2, 3) then using mid-point method;
Coordinate of $$M \left (\dfrac {2+( -2)}{2}, \dfrac {-1 + 3}{2}\right ) = (0, 1)$$
$$\Rightarrow \left (\dfrac {x + 3}{2}, \dfrac {y + 4}{2}\right ) = (0, 1)$$
$$\Rightarrow x = -3$$ and $$y = -2$$
$$\therefore (-3, -2)$$ is coordinate of $$D$$
Question 2
1 / -0

Find the midpoint between the coordinates $$(9, 3)$$ and $$(1, 1)$$.

A
$$\left(5, 2\right)$$

B
$$\left(3, 2\right)$$

C
$$\left(5, 1\right)$$

D
$$\left(3, 1\right)$$

Solution

We know the midpoint formula between two points.
$$\left(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\right)$$
Substitute the values, we get
$$=$$ $$\left(\dfrac{9+1}{2},\dfrac{3+1}{2}\right)$$
$$=$$ $$\left(\dfrac{10}{2},\dfrac{4}{2}\right)$$
Therefore, midpoint is $$\left(5, 2\right)$$
Question 3
1 / -0

Find the midpoints between the coordinates $$(2, 3)$$ and $$(1, 0)$$

A
$$\left(\dfrac{1}{2},\dfrac{3}{2}\right)$$

B
$$\left(\dfrac{3}{2},\dfrac{1}{2}\right)$$

C
$$\left(\dfrac{3}{2},\dfrac{4}{2}\right)$$

D
$$\left(\dfrac{3}{2},\dfrac{3}{2}\right)$$

Solution

Midpoint between the coordinates $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is:
$$\left( \dfrac { x_{ 1 }+x_{ 2 } }{ 2 } ,\dfrac { y_{ 1 }+y_{ 2 } }{ 2 } \right)$$
Therefore, the midpoint between the coordinates $$(2,3)$$ and $$(1,0)$$ is:
$$\left( \dfrac { 2+1 }{ 2 } ,\dfrac { 3+0 }{ 2 } \right) =\left( \dfrac { 3 }{ 2 } ,\dfrac { 3 }{ 2 } \right)$$
Therefore, midpoint is $$\left (\dfrac {3}{2}, \dfrac {3}{2}\right)$$.
Question 4
1 / -0

Find the value of $$k$$, so that $$(2, 1)$$ is the midpoint between $$(1, k)$$ and $$(3, 1)$$.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

We know the midpoint of the line joining the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$is given as
$$\left(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\right)$$
Substitute the values, we get
$$\left(\dfrac{1+3}{2},\dfrac{k+1}{2}\right)$$ $$=$$ $$\left(2, 1\right)$$
On equating, we get
$$\left(\dfrac{k+1}{2}\right)$$ $$= 1$$
$$\Rightarrow k + 1 = 2$$
$$\Rightarrow k = 2 - 1$$
$$\Rightarrow k = 1$$
Question 5
1 / -0

Find the centre of circle, if the coordinates of two ends of diameter are $$(-1, 7)$$ and $$(11, 5)$$

A
$$(5, 6)$$

B
$$(-3, 2)$$

C
$$(10, 12)$$

D
$$(12, 2)$$

Solution

Since centre of circle is the mid-point of the diameter of the circle.
$$x = \dfrac {11+( -1)}{2} = 5$$ and $$y = \left (\dfrac {5 + 7}{2}\right )$$ $$= 6$$ (using mid-point formula for centre).
$$\therefore (5, 6)$$ is the centre
Question 6
1 / -0

$$(9, 2), (5, -1) $$ and $$ (7, -5)$$ are the vertices of the triangle. Find its area.

A
10 square units

B
11 square units

C
12 square units

D
13 square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 9$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = -1$$, $$x_{3} = 7$$ and $$y_{3} = -5$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[9(-1 + 5) + 5(-5 - 2) + 7(2 + 1)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 35 + 21]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 22\right|$$
$$=$$ $$\left|11 \right|$$
Area always in absolute value.
So, area of the triangle $$= 11$$ square units.
Question 7
1 / -0

What is the midpoints between the coordinates $$(0, -6)$$ and $$(4, -4)$$?

A
$$\left(-2, -5\right)$$

B
$$\left(2, 5\right)$$

C
$$\left(2, -4\right)$$

D
$$\left(2, -5\right)$$

Solution

We know the midpoint formula between two points.
$$\left(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\right)$$
Substitute the values, we get
$$=$$ $$\left(\dfrac{0+4}{2},\dfrac{-6-4}{2}\right)$$
$$=$$ $$\left(\dfrac{4}{2},\dfrac{-10}{2}\right)$$
Therefore, midpoint is $$\left(2, -5\right)$$.
Question 8
1 / -0

If $$A(3, 5), B (-5, -4), C (7, 10)$$ are the vertices of a parallelogram, taken in the order, then the coordinates of the fourth vertex are

A
$$(10, 19)$$

B
$$(15, 19)$$

C
$$(19, 10)$$

D
$$(19, 15)$$

Solution

The given points are $$A(3,5),B(-5,-4),C(7,10)$$
Let coordinates of fourth vertex are $$D(x, y)$$.

Since, $$ABCD$$ is a parallelogram,
Midpoint of $$AC$$=Midpoint of $$BD$$
By mid point formula,
$$\left(\dfrac{3+7}{2},\dfrac{5+10}{2}\right)=\left(\dfrac{-5+x}{2},\dfrac{-4+y}{2}\right)$$

$$\therefore \dfrac {3+7}{2} = \dfrac {-5+x}{2}$$
$$\Rightarrow 10=-5+x$$
$$\Rightarrow x=15$$

And, $$\dfrac{5+10}{2} = \dfrac {-4+y}{2}$$
$$\Rightarrow 15=-4+y$$
$$\Rightarrow y=19$$

$$\therefore$$ Coordinates of fourth vertex are $$(15, 19)$$.
Question 9
1 / -0

If $$C$$ is a point on the line segment joining $$A\left( -3,4 \right) $$ and $$B\left( 2,1 \right) $$ such that $$AC=2BC$$, then the coordinate of $$C$$ is

A
$$\left( \dfrac { 1 }{ 3 } ,2 \right) $$

B
$$\left( 2,\dfrac { 1 }{ 3 } \right) $$

C
$$\left( 2,7 \right) $$

D
$$\left( 7,2 \right) $$

Solution

Since, $$AC=2BC$$

$$\dfrac {AC}{BC}=\dfrac 21$$

Point $$C$$ divides $$AB$$ in the $$2:1$$ ratio

Giving $$A(-3,4)$$ and $$B(2,1)$$

Here we apply the section formula, i. e. if a point $$C(x,y)$$ divides the line segment $$AB$$ joining the points $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) \& B\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ in the ratio $$m:n$$ then we have

$$\\ x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } \& y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } .\\$$
$$\therefore$$ Coordinate of $$C$$ are

$$\left( \dfrac { 4-3 }{ 2+1 } ,\dfrac { 2+4 }{ 2+1 } \right) $$ ie, $$\left( \dfrac { 1 }{ 3 } ,2 \right) $$
Question 10
1 / -0

In the following figure, F is the midpoint of the line segment GH. Which of the following are the coordinates of F?

A
$$(1, 4)$$

B
$$(2, 5)$$

C
$$\left (\dfrac {3}{2}, 4\right )$$

D
$$\left (\dfrac {3}{2}, \dfrac {9}{2}\right )$$

E
$$\left (2, \dfrac {9}{2}\right )$$

Solution

Given coordinates are $$G(-6,1)$$ and $$H(9,8)$$.
To find out,
The coordinates of $$F$$

Let the coordinates of $$F$$ be $$(x,y)$$.
We know that the coordinates of the midpoint of the line segment joining $$(x_1,y_1) \ and\ (x_2,y_2)$$ are:

$$P(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$$

Here, $$\left( { x }_{ 1, }{ ,y }_{ 1 } \right) =(9,8)$$
and $$\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -6,1 \right) $$

Hence, $$\left( x , y \right) =\left( \dfrac { 9+\left( -6 \right) }{ 2 } ,\dfrac { 8+1 }{ 2 } \right) $$
$$=\left( \dfrac { 3 }{ 2 } ,\dfrac { 9 }{ 2 } \right) $$

Hence, the coordinates of $$F$$ are $$\left( \dfrac { 3 }{ 2 } ,\dfrac { 9 }{ 2 } \right) $$.

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Coordinate Geometry Test - 34

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Question 1

$$(1, 2)$$

$$(-3, -2)$$

$$(3, 2)$$

None of these

Solution

Question 2

$$\left(5, 2\right)$$

$$\left(3, 2\right)$$

$$\left(5, 1\right)$$

$$\left(3, 1\right)$$

Solution

Question 3

$$\left(\dfrac{1}{2},\dfrac{3}{2}\right)$$

$$\left(\dfrac{3}{2},\dfrac{1}{2}\right)$$

$$\left(\dfrac{3}{2},\dfrac{4}{2}\right)$$

$$\left(\dfrac{3}{2},\dfrac{3}{2}\right)$$

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 5

$$(5, 6)$$

$$(-3, 2)$$

$$(10, 12)$$

$$(12, 2)$$

Solution

Question 6

10 square units

11 square units

12 square units

13 square units

Solution

Question 7

$$\left(-2, -5\right)$$

$$\left(2, 5\right)$$

$$\left(2, -4\right)$$

$$\left(2, -5\right)$$

Solution

Question 8

$$(10, 19)$$

$$(15, 19)$$

$$(19, 10)$$

$$(19, 15)$$

Solution

Question 9

$$\left( \dfrac { 1 }{ 3 } ,2 \right) $$

$$\left( 2,\dfrac { 1 }{ 3 } \right) $$

$$\left( 2,7 \right) $$

$$\left( 7,2 \right) $$

Solution

Question 10

$$(1, 4)$$

$$(2, 5)$$

$$\left (\dfrac {3}{2}, 4\right )$$

$$\left (\dfrac {3}{2}, \dfrac {9}{2}\right )$$

$$\left (2, \dfrac {9}{2}\right )$$

Solution

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