Coordinate Geometry Test

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Coordinate Geometry Test - 36

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Weekly Quiz Competition

Question 1
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Find the coordinates of the point $$X$$ that divides the join of $$A(7, 4)$$ and $$B(-3,6)$$ in the ratio $$1:3$$

A
$$\left (\dfrac {9}{2}, \dfrac {11}{2}\right )$$

B
$$\left (\dfrac {9}{2}, \dfrac {5}{2}\right )$$

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

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

Solution

Given the two points $$A(7,4)$$ and $$B(-3,6)$$ and ratio $$m:n=1:3$$
Here we apply the section formula : If a point $$X(x,y)$$ 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

$$\\ x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } , y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } .\\$$
$$\therefore$$ By section formula
$$(x,y)\ =\left( \dfrac { 1\times (-3)+3\times 7 }{ 1+3 } ,\dfrac { 1\times 6+3\times 4 }{ 1+3 } \right) \\ =\left( \dfrac { -3+21 }{ 4 } ,\dfrac { 6+12 }{ 4 } \right)\\ =\left( \dfrac { 18 }{ 4 } ,\dfrac { 18 }{ 4 } \right)\\ =\left( \dfrac { 9 }{ 2 } ,\dfrac { 9 }{ 2 } \right) $$
Question 2
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In the $$xy$$-coordinate plane, the coordinates of three vertices of a rectangle are $$\left(1, 5\right)$$, $$\left(5, 2\right)$$ and $$\left(5, 5\right)$$. What are the coordinates of the fourth vertex of the rectangle?

A
$$\left(1, 2\right)$$

B
$$\left(1, 7\right)$$

C
$$\left(2, 1\right)$$

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

E
$$\left(5, 7\right)$$

Solution

The intersection point of diagonals of rectangle is midpoint of each diagonal.
Let the fourth coordinate be $$(x,y)$$. By the above property we get
$$ \dfrac { x+5 }{ 2 } =\dfrac { 1+5 }{ 2 } $$. which implies $$x=1$$.
$$\dfrac { y+5 }{ 2 } =\dfrac { 2+5 }{ 2 } $$. which implies $$y=2$$).
So, the coordinate $$(x,y)$$ is $$(1,2)$$.
Question 3
1 / -0

In the diagram, $$E$$ is the midpoint of line $$DG$$ and $$F$$ is the midpoint of line $$EG$$.
The coordinates of $$F$$ are:

A
$$(6,3)$$

B
$$(6,2)$$

C
$$(7,3)$$

D
$$(8,3)$$

Solution

Let the coordinate of 'G' be (a,b) & 'F' be (x,y)
As 'E' is be (x,y)
As 'E' is mid lot of OG
$$\Rightarrow \left(\dfrac{a+1}{2},\dfrac{b+6}{2}\right)=(5,4)$$
$$\Rightarrow (a,b)=(9,2)$$
As 'F' is midpoint of EG
$$\Rightarrow \boxed{(x,y)=\left(\dfrac{5+9}{2},\dfrac{4+2}{2}\right)=(7,3)}$$
Question 4
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The coordinates of points $$P(-2, 2), Q(3, 2) $$ and $$R(3, -2)$$ are the vertices of a rectangle $$PQRS$$`. What are the coordinates of S?

A
$$(-3, -2)$$

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

C
$$(3, 2)$$

D
$$(2, 2)$$

Solution

Mid-point of $$PR=\left(\cfrac{-2+3}2,\cfrac{2-2}2\right)=\left(\cfrac12,0\right)$$
Let coordinate of $$S=(x,y)$$
Mid-point of $$QS=\left(\cfrac{3+x}2,\cfrac{2+y}2\right)$$
Mid-point of $$PR=$$Mid-point of $$QS$$
$$\Rightarrow\left(\cfrac12,0\right)=$$$$\left(\cfrac{3+x}2,\cfrac{2+y}2\right)$$
We have $$\cfrac12=\cfrac{3+x}2$$
$$\Rightarrow x=-2$$ and $$0=\cfrac{2+y}2$$
$$\Rightarrow y=-2$$
Coordinate of $$S=(-2,-2)$$
Hence, B is the correct option.
Question 5
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On the Cartesian plane, $$Q$$ is the midpoint of the straight line $$PR$$
Find the values of $$x$$ and $$y$$.

A
$$x=3,y=2$$

B
$$x=4,y=2$$

C
$$x=4,y=3\cfrac{1}{2}$$

D
$$x=8,y=3$$

Solution

By mid point formulae
$$x=\dfrac{7+1}{2}=4$$
$$3=\dfrac{y+4}{2}$$
$$\Rightarrow y=2$$
Question 6
1 / -0

The diagram is on a Cartesian plane.
If the midpoints of $$PQ$$ and $$RS$$ are the same, the coordinates of $$S$$ are:

A
$$(0,2)$$

B
$$(3,4)$$

C
$$(0,6)$$

D
$$(2,0)$$

Solution

Let $$'x'$$ is mid point of $$PQ$$ and $$RS$$
and $$s(x,y)$$ abd $$'x'$$ has $$(a,b)$$ coordinates and
$$P(-2,6)$$ ; $$Q(4,2)$$ ; $$R(2,6)$$
$$\Rightarrow \left(\dfrac{-2+4}{2},\dfrac{6+2}{2}\right)=(a,b)\left(\dfrac{x+2}{2},\dfrac{y+6}{2}\right)\Rightarrow \boxed{x=0\,and\,y=2}$$
Question 7
1 / -0

$$M$$ is the midpoint of the straight line $$PQ$$. If $$P(-2,9)$$ and $$M$$ is $$(4,3)$$, find the coordinates of $$Q$$.

A
$$(1,6)$$

B
$$(10,-3)$$

C
$$(10,6)$$

D
$$(8,-3)$$

Solution

Let the coordinates of point $$Q$$ be $$(h,k).$$

Now, by midpoint formulae,
$$\begin{aligned}{}\left( {\frac{{h - 2}}{2},\frac{{k + 9}}{2}} \right) &= \left( {4,3} \right)\\\frac{{h - 2}}{2} = 4\text{ and }\frac{{k + 9}}{2} &= 3\\h - 2 = 8\text{ and }k + 9 &= 6\\h = 10\text{ and }k &= - 3\end{aligned}$$

Hence, coordinates of point $$Q$$ is $$(10,-3).$$
Question 8
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Identify the true statement.

A
The $$X$$-axis is a vertical line

B
The $$Y$$-axis is a horizontal line

C
The scale on both the axes must be the same in a Cartesian plane

D
The point of intersection between the $$X$$-axis and $$Y$$-axis is called the origin

Solution

The point of intersection between the X-axis and Y-axis is called the origin $$(0,0)$$.
The X-axis is the horizontal line and Y-axis is the vertical line.
Question 9
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In the diagram, $$PQRS$$ is a straight line. $$R$$ is the midpoint of $$QS$$ and $$Q$$ is the midpoint of $$PS$$. $$S$$ is $$(6,5)$$ and $$R$$ is $$(3,5)$$. Find the coordinates of $$P$$.

A
$$(0,5)$$

B
$$(-6,5)$$

C
$$(-3,4)$$

D
$$(-5,6)$$

Solution

According to the midpoint formula, the coordinates of the midpoint of the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is given by $$\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right).$$

Let, the coordinates of the point $$P$$ and $$Q$$ are $$(x,y)$$ and $$(a,b)$$ respectively.

It is given that $$R$$ is the midpoint of $$QS$$ so, by using midpoint formula,
$$\left(\dfrac{a+6}{2},\dfrac{b+5}{2}\right)=(3,5)$$
$$\Rightarrow \dfrac{a+6}{2}=3$$ and $$\dfrac{b+5}{2}=5$$
$$\Rightarrow a+6=6$$ and $$b+5=10$$
$$\Rightarrow a=0$$ and $$b=5$$

Hence, coordinate of the point $$Q$$ is $$(0,5).$$

Similarly, $$Q$$ is the midpoint of $$PS$$ so,
$$\left(\dfrac{x+6}{2},\dfrac{y+5}{2}\right)=(0,5)$$
$$\Rightarrow \dfrac{x+6}{2}=0$$ and $$\dfrac{y+5}{2}=5$$
$$\Rightarrow x+6=0$$ and $$y+5=10$$
$$\Rightarrow x=-6$$ and $$y=5$$

Hence, coordinate of the point $$P$$ is $$(-6,5).$$
Question 10
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On the Cartesian plane, $$Q$$ is the midpoint of the straight line $$PR$$.
Find the value of $$h$$.

A
$$4$$

B
$$5$$

C
$$7$$

D
$$8$$