Coordinate Geometry Test

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Coordinate Geometry Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Which statement is true?

A
The x-axis is a vertical line

B
The point $$(-2 , 3)$$ lies in the III quadrant

C
Origin is the point of intersection of the x-axis and y-axis

D
The point $$(-3, -4)$$ lies in the II quadrant

Solution

i) The x-axis ,a line parallel to it is called horizontal line
ii) The point (-2,3) lies in II Quadrant
iii) Origin is the point of intersecting of x-axis and y-axis
iv) The point (-3,-4) lies in III quadrant .
Question 2
1 / -0

Given three vertices of a triangle whose coordinates are A (1, 1), B (3, -3) and (5, -3). Find the area of the triangle.

A
3 square units

B
4 square units

C
5 square units

D
6 square units

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
The three vertices of the triangle are $$A(1,1), B(3,-3), C(5,-3)$$
Area of triangle $$=\dfrac{|1(-3-(-3)+3(-3-1)+5(1-(-3))|}{2}$$
$$=\dfrac{|1(0)+3(-4)+4(4)|}{2}$$
$$=\dfrac{|-12+20|}{2}$$
$$=\dfrac{8}{2}$$
$$=4$$ square units.
Question 3
1 / -0

The area of triangle whose vertices are $$A (-3, -1), B(5, 3)$$ and $$C(2, -8)$$ is ____ $$\text{ sq. units}$$.

A
$$34\text{ sq. units}$$

B
$$36\text{ sq. units}$$

C
$$38\text{ sq. units}$$

D
$$32\text{ sq. units}$$

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

The given vertices of the triangle are $$A(-3,-1), B(5,3)$$ and $$C(2,-8)$$.
So, by using the above formula,
$$\begin{aligned}{}\text{Area of the triangle} &= \frac{1}{2} {[ - 3(3 - ( - 8)) + 5( - 8 - ( - 1)) + 2( - 1 - 3)]}\\ &= \frac{1}{2}{[ - 33 - 35 - 8]}\\ &= \frac{{[ - 76]}}{2}\\& = \frac{{76}}{2}\quad\quad\quad\quad\dots[\text{Area can never be negative so, we ignore negative sign}]\\& = 38\text{ sq. units}\end{aligned}$$

So, the area of the triangle is equal to $$38\text{ sq. units}$$.
Question 4
1 / -0

Refer to the following graph, and find the co-ordinates of $$N$$, which is the midpoint of $$\displaystyle \overline { AB } $$.

A
$$(3, k)$$

B
$$\displaystyle \left( \frac { 3 }{ 2 } ,3 \right) $$

C
$$\displaystyle \left( \frac { 3 }{ 2 } ,k \right) $$

D
$$\displaystyle \left( \frac { 3 }{ 2 } ,\frac { k }{ 2 } \right) $$

E
$$\displaystyle \left( \frac { k }{ 2 } ,\frac { k }{ 3 } \right) $$

Solution

Mid-point $$=\left (\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$$
Here $$x_1=-3,x_2=6$$ and $$y_1=0,y_2=k$$
Mid-point $$=\left (\dfrac{-3+6}{2},\dfrac{0+k}{2}\right)$$
$$\Rightarrow\left (\dfrac{3}{2},\dfrac{k}{2}\right)$$
Question 5
1 / -0

Given the points $$A(-1,3)$$ and $$B(4,9)$$.Find the co-ordinates of the mid-point of $$AB$$

A
$$\left(\cfrac{5}{2},-3\right)$$

B
$$(3,-3)$$

C
$$\left(3,\cfrac{-3}{2}\right)$$

D
$$\left(\cfrac{3}{2},6\right)$$

Solution

$$A=(-1,3)$$ and $$B=(4,9)$$
Here, $$x_1=-1,\,y_1=3,\,x_2=4$$ and $$y_2=9$$
Co-ordinates of the mid-point of $$AB=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$$

$$=\left(\dfrac{-1+4}{2},\dfrac{3+9}{2}\right)$$

$$=\left(\dfrac{3}{2},\dfrac{12}{2}\right)$$

$$=\left(\dfrac{3}{2},6\right)$$
Question 6
1 / -0

$$M(2, 6)$$ is the midpoint of $$\overline {AB}$$. If $$A$$ has coordinates $$(10, 12)$$, the coordinates of $$B$$ are

A
$$(6, 10)$$

B
$$(-6, 0)$$

C
$$(-8, -4)$$

D
$$(18, 16)$$

E
$$(22, 18)$$

Solution

Mid point $$=$$ $$\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$$
Here mid-point $$=(2,6)$$
$$x_1=10,y_1=12$$
$$\Rightarrow (2,6)=\dfrac{10+x_2}{2},\dfrac{12+y_2}{2}$$
$$\Rightarrow \dfrac{10+x_2}{2}=2; \dfrac{12+y_2}{2}=6$$
$$\Rightarrow x_2=4-10;y_2=12-12$$
$$\Rightarrow x_2=-6,y_2=0$$
$$\therefore $$ co-ordinates of $$B=(-6,0)$$.
Question 7
1 / -0

The area of the triangle formed by three vertices $$O(0, 0), A(1, 0), B(0, 1)$$ is _____ sq. units.

A
$$0$$

B
$$\dfrac{1}{2}$$

C
$$1$$

D
$$2$$

Solution

Points are given as $$O(0,0), A(1,0), B(0,1)$$.
When plotted on the cartesian plane, these points make a right angled triangle $$OAB$$ with
$$OA=1$$, $$OB=1$$
Area of this right angled triangle $$=\dfrac12\times OA\times OB$$
$$=\cfrac12\times1\times1\ sq.unit$$
$$=\cfrac12\ sq.unit$$
Question 8
1 / -0

If the mid-point between the points $$(a+ b, a- b)$$ and $$(-a, b)$$ lies on the line $$ax + by = k$$, what is $$k$$ equal to?

A
$$\dfrac ab$$

B
$$a + b$$

C
$$ab$$

D
$$a - b$$

Solution

Mid-point of points $$(x_1,y_1)$$ and $$(x_2, y_2)$$ is $$\bigg( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \bigg)$$

Hence, mid-point of $$(a+b,a-b),(-a,b)$$ is $$\left( \dfrac { a+b-a }{ 2 } ,\dfrac { a-b+b }{ 2 } \right) $$

$$=\left(\dfrac { b }{ 2 } ,\dfrac { a }{ 2 } \right)$$

Given: Mid-point lies on line $$ax+by=k$$

$$\Rightarrow a\times \dfrac{b}{2}+b\times \dfrac{a}{2}=k$$

$$\Rightarrow ab=k$$
Question 9
1 / -0

If O(0, 0) and P(-8, 0) then co-ordinates of its midpoint are .............

A
(-4, 0)

B
(4, 0)

C
(0, -4)

D
(0, 0)

Solution

Let $$A(x, y)$$ be the midpoint of line segment joining points $$O$$ and $$P$$.
By midpoint formula we can say that,
$$(x, y) = \left( \dfrac{0 + (-8) }{2}, \dfrac{0 - 0}{2} \right ) = (-4, 0)$$
Hence, the coordinates of midpoint is $$(-4, 0)$$
Question 10
1 / -0

The points A(5, 2), B(3, 4) and C(x, y) are collinear points and AB = BC then find the co-ordinates of C.

A
(1, 6)

B
(2, -2)

C
(4, 3)

D
(8, 6)

Solution

If three points $$P(x_1,y_1), Q, R(x_2,y_2)$$ are collinear.

$$\Rightarrow PQ = QR$$ and $$Q$$ is the midpoint of $$PR$$

The mid point $$Q = \left (\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$$

Given, the points $$A(5, 2), B(3, 4)$$ and $$C(x, y)$$ are collinear and $$AB=BC$$
$$\Rightarrow $$ $$B$$ is the mid-point of $$AC$$.
Using midpoint formula,
$$\left( \dfrac{x + 5}{2} , \dfrac{y + 2}{2} \right ) = (3, 4)$$

$$\implies \dfrac{x + 5}{2} = 3$$ and $$\dfrac{y + 2}{2} = 4$$

$$\implies x + 5 = 6$$ and $$y + 2 = 8$$

$$\implies x = 1$$ and $$y = 6$$
Hence, the coordinates of $$C$$ are $$(1,6)$$.

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Coordinate Geometry Test - 19

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Coordinate Geometry Test - 19

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Question 1

The x-axis is a vertical line

The point $$(-2 , 3)$$ lies in the III quadrant

Origin is the point of intersection of the x-axis and y-axis

The point $$(-3, -4)$$ lies in the II quadrant

Solution

Question 2

3 square units

4 square units

5 square units

6 square units

Solution

Question 3

$$34\text{ sq. units}$$

$$36\text{ sq. units}$$

$$38\text{ sq. units}$$

$$32\text{ sq. units}$$

Solution

Question 4

$$(3, k)$$

$$\displaystyle \left( \frac { 3 }{ 2 } ,3 \right) $$

$$\displaystyle \left( \frac { 3 }{ 2 } ,k \right) $$

$$\displaystyle \left( \frac { 3 }{ 2 } ,\frac { k }{ 2 } \right) $$

$$\displaystyle \left( \frac { k }{ 2 } ,\frac { k }{ 3 } \right) $$

Solution

Question 5

$$\left(\cfrac{5}{2},-3\right)$$

$$(3,-3)$$

$$\left(3,\cfrac{-3}{2}\right)$$

$$\left(\cfrac{3}{2},6\right)$$

Solution

Question 6

$$(6, 10)$$

$$(-6, 0)$$

$$(-8, -4)$$

$$(18, 16)$$

$$(22, 18)$$

Solution

Question 7

$$0$$

$$\dfrac{1}{2}$$

$$1$$

$$2$$

Solution

Question 8

$$\dfrac ab$$

$$a + b$$

$$ab$$

$$a - b$$

Solution

Question 9

(-4, 0)

(4, 0)

(0, -4)

(0, 0)

Solution

Question 10

(1, 6)

(2, -2)

(4, 3)

(8, 6)

Solution

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