Coordinate Geometry Test

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Coordinate Geometry Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

$$\triangle OPQ$$ is formed by the coordinates $$P(0,\,5),\, Q(8,\,0)$$ and origin $$O$$. The area of $$\triangle OPQ$$ is ____ square units.

A
$$20$$ sq. units

B
$$10$$ sq. units

C
$$30$$ sq. units

D
None of the above

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Area of $$\triangle OPQ $$
$$=\dfrac 12 \left[ 1(0) - 1(0) +1(-40)\right]$$
$$=\dfrac 12 \times 40 = 20$$ sq. units
Question 2
1 / -0

Find coordinates of point which divides the line segment internally, joining the points $$(5,\,7)$$ and $$(9,\,12)$$ in ratio $$2:4$$.

A
$$\dfrac{26}{3},\,\dfrac{19}{3}$$

B
$$\begin{pmatrix}\dfrac{19}{3},\,\dfrac{26}{3}\end{pmatrix}$$

C
$$19,\,26$$

D
None

Solution

We know that if a point $$P(x,y)$$ 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, by the section formula, we have

$$\\ x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } \& y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } .\\$$

Hence $$m=2,\,n=4,\,x_1=5,\,x_2=9,\,y_1=7,\,y_2=12$$

$$\therefore$$ $$x=\dfrac{(5\times4)+(9\times2)}{6}$$

and

$$y=\dfrac{(7\times4)+(12\times2)}{6}$$

$$x=\dfrac{20+18}{6}$$ and $$y=\dfrac{28+24}{6}$$

$$x=\dfrac{38}{6} = \dfrac {19}{3}$$ and $$y=\dfrac{52}{6}=\dfrac{26}{3}$$

$$(x,\,y)\Rightarrow\,\begin{pmatrix}\dfrac{19}{3},\,\dfrac{26}{3}\end{pmatrix}$$
Question 3
1 / -0

Find coordinates of point which divides the line segment internally, joining the points $$(9,\,18)$$ and $$(1,\,2)$$ in ratio $$3:6$$.

A
$$\dfrac{36}{3},\,\dfrac{20}{3}$$

B
$$\dfrac{20}{3},\,\dfrac{38}{3}$$

C
$$20,\,38$$

D
None

Solution

As we know if a point $$P(x,y)$$ 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, by the section formula,
$$\\ x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } $$
$$ y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } .\\$$
$$m=3,\,n=6,\,x_1=9,\,x_2=1,\,y_1=18,\,y_2=2$$
By section formula for internal division,
$$x=\dfrac{9\times6+3\times1}{9}$$ and $$y=\dfrac{18\times6+2\times3}{9}$$

$$x=\dfrac{54+3}{9}$$ and $$y=\dfrac{108+6}{9}$$

$$x=\dfrac{57}{9}$$ and $$y=\dfrac{114}{9}$$

$$x=\dfrac{19}{3}$$ and $$y=\dfrac{38}{3}$$
Question 4
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The area of triangle with vertices $$A(0,\,9),\,B(0,\,4)$$ and $$C(-5,\,-9)$$ is

A
$$\dfrac{25}{2}$$ sq. units

B
$$\dfrac{23}{2}$$ sq. units

C
$$\dfrac{19}{2}$$ sq. units

D
None of the above

Solution

Area of triangle $$=\dfrac{1}{2}\begin{bmatrix}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\end{bmatrix}$$
$$=\dfrac{1}{2}\begin{bmatrix}0(4+9)+0(-9-9)+(-5)(9-4)\end{bmatrix}$$
$$=\dfrac{1}{2}\begin{bmatrix}-5\times5\end{bmatrix}$$
$$=-\dfrac{25}{2}$$
The area cannot be negative.
$$\therefore$$ It must be $$\dfrac{25}{2}$$ sq. units
So, option A is correct.
Question 5
1 / -0

What is the area of the triangle for the following points $$(6, 2), (5, 4)$$ and $$(3, -1)$$?

A
2.3 square units

B
4.5 square units

C
4.1 square units

D
3.6 square units

Solution

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 $$x_{1} = 6$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = 4$$, $$x_{3} = 3$$ and $$y_{3} = -1$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[6(4 + 1) + 5(-1 - 2) + 3(2 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[30 - 15 - 6]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 9\right|$$
$$=$$ $$\left|4.5 \right|$$
Area always in absolute value.
So, area of the triangle $$= 4.5$$ square units.
Question 6
1 / -0

Calculate mid point of $$A(5,\,3)$$ and $$B(9,\,8)$$

A
$$\dfrac{11}{2},\,7$$

B
$$7,\,\dfrac{11}{2}$$

C
$$7,\,11$$

D
$$14,\,11$$

Solution

The mid point of line segment joining $$(x_1,\,y_1)$$ and $$(x_2,\,y_2)$$ is $$\begin{pmatrix}\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2}\end{pmatrix}$$
Mid point of AB $$\Rightarrow\begin{pmatrix}\dfrac{5+9}{2},\,\dfrac{3+8}{2}\end{pmatrix}$$
$$\Rightarrow\begin{pmatrix}7,\,\dfrac{11}{2}\end{pmatrix}$$
Question 7
1 / -0

The area of the triangle whose vertices are $$(0, 1), (1, 4)$$ and $$(1, 2)$$ is ___ square units.

A
1

B
2

C
3

D
4

Solution

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 $$x_{1} = 0$$, $$y_{1} = 1$$, $$x_{2} = 1$$, $$y_{2} = 4$$, $$x_{3} = 1$$ and $$y_{3} = 2$$
Substitute the values, we get
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[0(4 - 2) + 1(2 - 1) + 1(1 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[0 + 1 - 3]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times -2\right|$$
$$=$$ $$\left|- 1 \right|$$
Area always in absolute value.
So, area of the triangle is $$ 1$$ square units.
Question 8
1 / -0

What is the midpoints between the coordinates $$(-1, 2)$$ and $$(-1, -6)$$?

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

B
$$\left(-1, 2\right)$$

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

D
$$\left(1, -2\right)$$

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
$$=$$ $$\left(\dfrac{-1-1}{2},\dfrac{2-6}{2}\right)$$
$$=$$ $$\left(\dfrac{-2}{2},\dfrac{-4}{2}\right)$$
Therefore, midpoint is $$\left(-1, -2\right)$$
Question 9
1 / -0

Find the vertices of the triangle whose mid point of sides are $$(3, 1), (5, 6)$$ and $$(-3, 2)$$

A
$$(-1, 7) (-5, -3) (6, 5)$$

B
$$(7, 1) (2, 3) (4, 1)$$

C
$$(-1, 7) (-5, -3) (11, 5)$$

D
$$(1, 7) (5, 3) (-11, 5)$$

Solution

Let coordinates of the vertices of the triangle $$A, B$$ and $$C$$ be
$$A (x_{1}, y_{1}) B(x_{2}, y_{2})$$ and $$C (x_{3} , y_{3})$$, respectively.

Now, $$\dfrac {x_{2} + x_{3}}{2} = 3\Rightarrow x_{2} + x_{3} = 6 ..... (I)$$

$$\dfrac {y_{2} + y_{3}}{2} = 1\Rightarrow y_{2} + y_{3} = 2 ...... (II)$$

$$x_{3} + x_{1} = 10 ..... (III)$$

$$y_{1} + y_{3} = 12 ......(IV)$$

$$x_{1} + x_{2} = -6 ......(V)$$

$$y_{1} + y_{2} = 4...... (VI)$$

Adding equations (I), (III) and (V), we get
$$2x_1+2x_2+2x_3 = 10$$
$$x_1+x_2+x_3 = 5$$
$$x_3 = 11$$ ....$$[\because x_1+x_2 = -6]$$
$$x_1 = -1$$ ....$$[\because x_2+x_3 = 6]$$
$$\therefore x_2 = -5$$

Adding equations (II), (IV) and (VI), we get
$$2y_1+2y_2+2y_3 = 18$$

$$y_1+y_2+y_3 = 9$$

$$y_3 = 5$$ ....$$[\because y_1+y_2 = 4]$$
$$y_1 = 7$$ ....$$[\because y_2+y_3 = 2]$$
$$\therefore y_2 = -3$$

Therefore, the coordinates of the vertices are $$A(-1, 7) B(-5, -3)$$ and $$C (11, 5)$$.
Question 10
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What is the area of the triangle whose vertices are: $$(-3, 15), (6, -7) $$ and $$(10, 5)$$?

A
$$94$$ square units

B
$$96$$ square units

C
$$97$$ square units

D
$$98$$ square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is $$A=\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
Here $$x_{1} = -3$$, $$y_{1} = 15$$, $$x_{2} = 6$$, $$y_{2} = -7$$, $$x_{3} = 10$$ and $$y_{3} = 5$$

Substituting the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[-3(-7 - 5) + 6(5 - 15) + 10(15 + 7)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 60 + 220]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 196\right|$$
$$=$$ $$\left|98 \right|$$

Area is always in absolute value.
So, area of the triangle $$= 98$$ square units.

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Coordinate Geometry Test - 33

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Coordinate Geometry Test - 33

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Question 1

$$20$$ sq. units

$$10$$ sq. units

$$30$$ sq. units

None of the above

Solution

Question 2

$$\dfrac{26}{3},\,\dfrac{19}{3}$$

$$\begin{pmatrix}\dfrac{19}{3},\,\dfrac{26}{3}\end{pmatrix}$$

$$19,\,26$$

None

Solution

Question 3

$$\dfrac{36}{3},\,\dfrac{20}{3}$$

$$\dfrac{20}{3},\,\dfrac{38}{3}$$

$$20,\,38$$

None

Solution

Question 4

$$\dfrac{25}{2}$$ sq. units

$$\dfrac{23}{2}$$ sq. units

$$\dfrac{19}{2}$$ sq. units

None of the above

Solution

Question 5

2.3 square units

4.5 square units

4.1 square units

3.6 square units

Solution

Question 6

$$\dfrac{11}{2},\,7$$

$$7,\,\dfrac{11}{2}$$

$$7,\,11$$

$$14,\,11$$

Solution

Question 7

1

2

3

4

Solution

Question 8

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

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

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

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

Solution

Question 9

$$(-1, 7) (-5, -3) (6, 5)$$

$$(7, 1) (2, 3) (4, 1)$$

$$(-1, 7) (-5, -3) (11, 5)$$

$$(1, 7) (5, 3) (-11, 5)$$

Solution

Question 10

$$94$$ square units

$$96$$ square units

$$97$$ square units

$$98$$ square units

Solution

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