Coordinate Geometry Test

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Coordinate Geometry Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

The coordinates of a point on ____ are of the form $$(0, y).$$

A
$$Y\;-\;\text{axis}$$

B
$$X\;-\;\text{axis}$$

C
Both $$A$$ and $$B$$

D
None of the Above

Solution

Since the $$x$$ coordinate of the point is $$0$$, this implies that for any value of $$y$$ the point $$(0,y)$$ will lie on the $$Y\;-\;\text{axis}.$$
Question 2
1 / -0

The horizontal axis is called ______ axis.

A
y-axis

B
x-axis

C
Ambiguous

D
Data insufficient

Solution

The horizontal axis is called x-axis.
Question 3
1 / -0

The distance between the points $$A(0, 6)$$ and $$B (0, 2)$$ is

A
$$6$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

$$ Distance\quad between\quad two\quad points\quad \left( { x }_{ 1 },{ y }_{ 1 } \right) \quad \& \quad \quad \left( { x }_{ 2 },{ y }_{ 2 } \right) \quad is\\ d=\sqrt { { { (x }_{ 2 }-{ x }_{ 1 } })^{ 2 }(-{ { y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \\ $$
$$\therefore AB = \sqrt {(0-0)^2+(6-2)^2}$$

$$\therefore AB = 4$$ units
Question 4
1 / -0

Which of the following statements is true?

A
The $$x-$$axis is a vertical line.

B
The $$y-$$axis is a horizontal line.

C
The scale on both 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

Let us investigate the options.
Option A: $$\longrightarrow $$ Any point on the $$x-$$axis is $$(x,0)$$. $$x-$$axis is horizontal.
So, option A is incorrect.
Option B: $$\longrightarrow $$ Any point on the $$y-$$axis is $$(0,y)$$. $$y-$$axis is vertical.
So, option B is incorrect.
Option C: $$\longrightarrow $$ We can choose any convenient scale on $$x-$$axis and $$y-$$axis.
So, option C is incorrect.
Option D:
$$X-$$axis & $$Y-$$axis are perpendicular to each other and they intersect at O which is the origin.
Question 5
1 / -0

Directions For Questions

Based on this information answer the questions given below.
Given the points $$A(-1, 3)$$ and $$B(4, -9).$$

...view full instructions

Find the co-ordinates of the mid-point of AB.

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

B
(3, -3)

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

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

Solution

Midpoint of AB $$=\left ( \dfrac{-1+4}{2},\dfrac{3+(-9)}{2} \right )$$
$$=\left ( 1\dfrac{1}{2},-3 \right ) or \left ( \dfrac{3}{2},-3 \right )$$
Question 6
1 / -0

Area of a triangle formed by the points A(5, 2), B(4, 7) and C(7, -4) is _____.

A
2 sq. units

B
$$2\sqrt2$$ sq. units

C
4 sq. units

D
$$4\sqrt2$$ sq. units

Solution

Area of a triangle ABC = $$ \frac { 1 }{ 2 } \left|[ x_{ 1 }\left( y_{ 2 }-y_{ 3 }

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

\right) \right]|$$
Substituting the given coordinates, we have
=$$ \frac { 1 }{ 2 } \left[ 5 \left(7+4) \right) -\left( 4+2\right) +7\left( 2-7) \right) \right]$$
$$ =\frac { 1 }{ 2 } \left( 5 ( 11) +4\times( -6)
+7\times( -5)\right)$$
$$ =\frac { 1 }{ 2 } \left[(55) - ( 24) - ( 35) \right]$$
$$ =\frac { 1 }{ 2 } \times |-4| = 2$$ square units.
Question 7
1 / -0

The point at which the two coordinate axes meet is called the

A
Abscissa

B
Ordinate

C
Origin

D
Quadrant

Solution

The point at which the coordinate axes meet is called the origin. The point is denoted by $$O$$ in the above image.
Question 8
1 / -0

The point which divides the line segment joining the points $$(3,4)$$ and $$(7,-6) $$ internally into two parts such that one part is twice the other lies in which quadrant?

A
I

B
II

C
III

D
IV

Solution

If a point divides the line segment $$AB$$ joining two points in given ratio $$m:n$$ internally, by section formula, the co-ordinates of a point are given as $$=\left (\dfrac {mx_{2}+nx_{1}}{m+n},\dfrac {my_{2}+ny_{1}}{m+n}\right) $$

The ratio is $$m:n=2:1$$ and $$A\left(x_{1}, y_{1}\right)$$ = $$\left(3,4\right)$$ and $$B\left(x_{2}, y_{2}\right)$$ = $$\left(7,-6\right)$$

$$\therefore $$ Point $$=\left (\dfrac {mx_{2}+nx_{1}}{m+n},\dfrac {my_{2}+ny_{1}}{m+n}\right) $$

$$=\left (\dfrac {2×7+1×3}{2+1},\dfrac {2×(-6)+1×4}{2+1}\right) $$

$$=\left (\dfrac {17}{3},\dfrac {-8}{3}\right) $$

$$\therefore $$ The point lies in IV quadrant.
Question 9
1 / -0

A triangle has vertices A(1,-1) B(2,4) and C(6,0) The length of the median from A is

A
3

B
$$\displaystyle 3\sqrt{2}$$

C
$$\displaystyle 2\sqrt{3}$$

D
$$\displaystyle 2\sqrt{2}$$

Solution

Midpoint of BC = L = (4, 2)
$$\displaystyle \therefore AL=\sqrt{\left ( 1-4 \right )^{2}+\left ( -1-2 \right )^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$$
Question 10
1 / -0

If the points (1,1) (2,3) and (5,-1) form a right-angled triangle, then the hypotenuse is of length

A
3

B
4

C
6

D
5

Solution

We use the distance formula, i.e.
$$ Distance\quad between\quad two\quad points\quad \left( { x }_{ 1 },{ y }_{ 1 } \right) \quad \& \quad \quad \left( { x }_{ 2 },{ y }_{ 2 } \right) \quad is\\ d=\sqrt { { { (x }_{ 2 }-{ x }_{ 1 } })^{ 2 }(-{ { y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \\ $$
Given points $$A = (1, 1), B = (2, 3), C = (5, -1)$$
Therefore,
$$\displaystyle AB=\sqrt{\left ( 1-2 \right )^{2}+\left ( 1-3 \right )^{2}}=5$$
$$\displaystyle BC=\sqrt{\left ( 2-5 \right )^{2}+\left ( 3+1 \right )^{2}}=25$$
$$\displaystyle AC=\sqrt{\left ( 1-5 \right )^{2}+\left ( 1+1 \right )^{2}}=20$$

$$BC$$ is larger side hence hypotenuse.

Hypotenuse is $$BC = 5$$

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Coordinate Geometry Test - 17

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Coordinate Geometry Test - 17

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Question 1

$$Y\;-\;\text{axis}$$

$$X\;-\;\text{axis}$$

Both $$A$$ and $$B$$

None of the Above

Solution

Question 2

y-axis

x-axis

Ambiguous

Data insufficient

Solution

Question 3

$$6$$

$$4$$

$$3$$

$$2$$

Solution

Question 4

The $$x-$$axis is a vertical line.

The $$y-$$axis is a horizontal line.

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

The point of intersection between the $$x-$$axis and $$y-$$axis is called the origin.

Solution

Question 5

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

(3, -3)

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

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

Solution

Question 6

2 sq. units

$$2\sqrt2$$ sq. units

4 sq. units

$$4\sqrt2$$ sq. units

Solution

Question 7

Abscissa

Ordinate

Origin

Quadrant

Solution

Question 8

I

II

III

IV

Solution

Question 9

3

$$\displaystyle 3\sqrt{2}$$

$$\displaystyle 2\sqrt{3}$$

$$\displaystyle 2\sqrt{2}$$

Solution

Question 10

3

4

6

5

Solution

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