Coordinate Geometry Test

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Coordinate Geometry Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The point P divides the line segment joining the points $$\displaystyle A\left ( 2,1 \right )$$ and $$\displaystyle B\left ( 5,-8\right )$$ such that $$ \dfrac{AP}{AB}=\dfrac{1}{3}$$ If P lies on the line $$\displaystyle 2x+y+k=0$$
then the value of k is-

A
$$-4$$

B
$$4$$

C
$$-3$$

D
$$3$$
Question 2
1 / -0

The midpoints of sides of a triangle are ($$3,4), (4,1)$$ and $$(2,0)$$. Which of the following is not the value of the coordinates of it's vertices?

A
$$(1,3)$$

B
$$(5,3)$$

C
$$(5,5)$$

D
$$(3,-3)$$
Question 3
1 / -0

The area of triangle formed by $$(0, 0), (0, a)$$ and $$(b, 0)$$ is .......... .

A
$$ab$$

B
$$\displaystyle \frac{ab}{2}$$

C
$$\left | \displaystyle \frac{ab}{2} \right |$$

D
$$\left | ab \right |$$

Solution

The area of a triangle formed by joining the points $$(x_1, y_1)$$, $$(x_2, y_2)$$ and $$(x_3, y_3)$$ is

$$A=\dfrac { 1 }{ 2 } \bigg| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \bigg|$$

Therefore, the area of a triangle formed by joining the points $$(0, 0)$$, $$(0, a)$$ and $$(b, 0)$$ is:

$$A=\dfrac { 1 }{ 2 } \bigg| 0(0-b)+a(b-0)+0(0-0) \bigg| $$

$$=\dfrac { 1 }{ 2 } \bigg| 0+ab \bigg| $$

$$=\bigg| \dfrac { ab }{ 2 } \bigg|$$

Hence, the area of the triangle is $$\left| \dfrac { ab }{ 2 } \right|$$
Question 4
1 / -0

The end points of a diagonal of a parallelogram are $$(1, 3)$$ and $$(5, 7)$$, then the mid-point of the other diagonal is ..........

A
$$(1, 7)$$

B
$$(3, 5)$$

C
$$(5, 3)$$

D
$$(7, 1)$$

Solution

Given end points of diagonal of parallelogram $$(1,3)$$ and $$(5,7)$$
We know that the mid point of two point $$(x_1,y_1) \ and \ (x_2,y_2)$$ are
$$\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}$$
$$\Rightarrow \dfrac{1+5}{2},\dfrac{3+7}{2}$$
$$\Rightarrow( 3,5)$$
Question 5
1 / -0

The midpoints of the sides of triangle $$ABC$$ are $$ (-1,-2), (6,1)$$ and $$(3,5) $$. The area of $$\displaystyle \triangle ABC$$ is ____ square units.

A
$$72$$

B
$$73$$

C
$$74$$

D
$$75$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle formed with midpoints
$$\displaystyle =\frac{1}{2}\left [ -1\left ( 1-5 \right )+6\left ( 5+2 \right )+3\left ( -2-1 \right ) \right ]$$
$$\displaystyle =\frac{37}{2}$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta ABC=4\times \frac{37}{2}=74$$ square units.
$$[\because$$ Area of triangle formed by joining mid-points of the sides of given triangle is $$\left(\dfrac{1}{4}\right)^{th}$$ of the area of original triangle $$]$$
Question 6
1 / -0

Area of triangle whose vertices are $$(0, 0), (2, 3), (5, 8)$$ is ____ square units.

A
$$\dfrac 12$$

B
$$1$$

C
$$2$$

D
$$\dfrac 32$$

Solution

Area of the triangle with three vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is
$$A=\dfrac { 1 }{ 2 } \left| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \right|$$
Therefore, the area of triangle with vertices $$(1,2)$$, $$(3,4)$$ and $$(5,6)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| 0(2-5)+3(5-0)+8(0-2) \right| =\dfrac { 1 }{ 2 } \left| 0+15-16 \right| =\dfrac { 1 }{ 2 } \left| -1 \right| =\dfrac { 1 }{ 2 }$$
Hence, the area of the triangle is $$\dfrac { 1 }{ 2 }$$ square units.
Question 7
1 / -0

The area of a triangle with the vertices $$(1, 2), (3, 4)$$ and $$(5, 6)$$, is ____ square units.

A
$$0$$

B
$$1$$

C
$$2$$

D
$$5$$

Solution

Area of the triangle with three vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is

$$A=\dfrac { 1 }{ 2 } \left| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \right|$$

Therefore, the area of triangle with vertices $$(1,2)$$, $$(3,4)$$ and $$(5,6)$$ is:

$$A=\dfrac { 1 }{ 2 } \left| 2(3-5)+4(5-1)+6(1-3) \right| =\dfrac { 1 }{ 2 } \left| -4+16-12 \right| =\dfrac { 1 }{ 2 } \left| 16-16 \right| =\dfrac { 1 }{ 2 } \left| 0 \right| =0$$

Hence, the area of the triangle is $$0$$.
Question 8
1 / -0

The coordinates of the points $$P$$ which divides $$(1, 0)$$ and $$(0, 0)$$ in $$1 : 2$$ ratio are .......... .

A
$$(2, 3)$$

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

C
$$(3, 2)$$

D
$$\left ( \displaystyle \frac{2}{3}, 0 \right )$$

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, $$m = 1, n = 2, A = (1, 0)$$ and $$B = (0, 0)$$
Let $$P(x, y)$$ be the required point.

$$\therefore$$ $$x=\dfrac { (1\times 0)+(2\times 1) }{ 1+2 } =\dfrac { 0+2 }{ 3 } =\dfrac { 2 }{ 3 }$$
and
$$y=\dfrac { (1\times 0)+(2\times 0) }{ 1+2 } =\dfrac { 0+0 }{ 3 } =0$$

Therefore, the coordinates of the point is $$\left( \dfrac { 2 }{ 3 } ,0 \right)$$.
Question 9
1 / -0

Find the area of $$\Delta ABC$$, in which $$A = (2, 1), B = (3, 4)$$ and $$C = (-3, -2).$$

A
$$3$$ square units

B
$$4$$ square units

C
$$6$$ square units

D
$$12$$ square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Therefore, the area of triangle with vertices $$(2,1)$$, $$(3,4)$$ and $$(-3,-2)$$ is:
$$A=\displaystyle \frac { 1 }{ 2 } \left| 2(4-(-2))+3(-2-1)-3(1-4) \right| $$
$$=\dfrac { 1 }{ 2 } \left| 2(6)+3(-3)-3(-3) \right| $$
$$=\dfrac { 1 }{ 2 } \left| 12-9+9 \right|$$
$$ =\dfrac { 1 }{ 2 } \left| 12 \right| =6$$ square units.
Question 10
1 / -0

If coordinates of P,Q, and R are (3,6),(-1,3) and (2,-1) respectively. Then area of $$\displaystyle \triangle PQR$$ is ____ square units.

A
12

B
12.5

C
13

D
14

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of $$\displaystyle \Delta PQR$$ is given by
$$\displaystyle =\frac{1}{2}\left [ 3\left ( 3+1 \right )-1\left ( -1-6 \right )+2\left ( 6-3 \right ) \right ]$$
$$\displaystyle =\frac{1}{2}\left ( 12+7+6 \right )=12.5$$ square units.

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Re-Attempt Weekly Quiz Competition

Coordinate Geometry Test - 31

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Coordinate Geometry Test - 31

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SHARING IS CARING

Question 1

$$-4$$

$$4$$

$$-3$$

$$3$$

Question 2

$$(1,3)$$

$$(5,3)$$

$$(5,5)$$

$$(3,-3)$$

Question 3

$$ab$$

$$\displaystyle \frac{ab}{2}$$

$$\left | \displaystyle \frac{ab}{2} \right |$$

$$\left | ab \right |$$

Solution

Question 4

$$(1, 7)$$

$$(3, 5)$$

$$(5, 3)$$

$$(7, 1)$$

Solution

Question 5

$$72$$

$$73$$

$$74$$

$$75$$

Solution

Question 6

$$\dfrac 12$$

$$1$$

$$2$$

$$\dfrac 32$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

$$5$$

Solution

Question 8

$$(2, 3)$$

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

$$(3, 2)$$

$$\left ( \displaystyle \frac{2}{3}, 0 \right )$$

Solution

Question 9

$$3$$ square units

$$4$$ square units

$$6$$ square units

$$12$$ square units

Solution

Question 10

12

12.5

13

14

Solution

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