Coordinate Geometry Test

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Coordinate Geometry Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Area of the triangle with vertices $$(-2,2), (1,5)$$ and $$(6,-1)$$ is

A
$$15$$

B
$$\dfrac{3}{5}$$

C
$$\dfrac{29}{2}$$

D
$$\dfrac{33}{2}$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$$

$$=\dfrac{1}{2} [-2(5+1) + 1(-1-2)+6(2-5)]$$

$$=\dfrac{1}{2}[-12-3-18]$$

$$=-\dfrac{33}{2}$$

$$\therefore$$ Area $$=\dfrac{33}{2}~sq.\, units$$
Question 2
1 / -0

Select the correct option.
The value of $$p$$, for which the points $$A(3,1) , B (5, p)$$ and $$C (7, -5)$$ are collinear, is

A
$$-2$$

B
$$2$$

C
$$-1$$

D
$$1$$

Solution

$$A (3, 1) \,\, B (5, P) \,\, C (7, -5)$$

If points $$A(x_1,y_1),\,B(x_2,y_2),\,C(x_3,y_3)$$ are collinear then
$$ar(\triangle ABC)=\dfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$$

$$\dfrac{1}{2} [3[5 + p) + 5(-5 - 1) + 7 (1- p)] = 0$$

$$15 + 3p - 30 + 7 - 7p = 0$$

$$4p = -15 + 7$$

$$4p = -8$$

$$p = -2$$
Question 3
1 / -0

Three consecutive vertices of a parallelogram $$ABCD$$ are $$A(3,-2), B(4,0)$$ and $$C(6,-3)$$. The 4th vertex $$D$$ is

A
$$D(5,-5)$$

B
$$D(-5,5)$$

C
$$D(-3,8)$$

D
$$D(3,-8)$$

Solution

Let the 4th vertex be $$D(x,y)$$. Then
mid point of $$AC$$= midpoint of $$BD$$
Midpoint formula = $$\left(\dfrac {x_1+x_2}2,\dfrac {y_1+y_2}2\right)$$
$$\cfrac { 4+x }{ 2 } =\cfrac { 3+6 }{ 2 } ;\cfrac { 0+y }{ 2 } ;\cfrac { -2-3 }{ 2 } \quad \Rightarrow x=(9-4)=5 $$ and $$ y=-5$$
point $$D$$ is $$D(5,-5)$$
Question 4
1 / -0

If the points $$A(-2,3)$$, $$B(1,2)$$ and $$C(k,0)$$ are collinear, then $$k=$$?

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

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

Here $$\left( { x }_{ 1 }=-2,{ y }_{ 1 }=3 \right) ;\left( { x }_{ 2 }=1,{ y }_{ 2 }=2 \right) ;\left( { x }_{ 3 }=k,{ y }_{ 3 }=0 \right) $$

If three points are collinear then, area of triangle formed by them as vertices will be equal to zero

$$Ar\left( \triangle ABC \right) =0\\\Rightarrow \dfrac{1}{2} \left [ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] =0\quad \quad $$

$$\Rightarrow \left[ -2(2-0)+1(0-3)+k(3-2) \right] =0\\\Rightarrow -4-3+k=0\\\Rightarrow k=7$$
Question 5
1 / -0

The vertices of a $$\triangle ABC$$ are $$A(3,8),B(-4,2)$$ and $$C(5,-1)$$. The area of $$\triangle ABC$$ is

A
$$57$$ sq units

B
$$75$$ sq units

C
$$28\cfrac{1}{2}$$ sq units

D
$$37\cfrac{1}{2}$$ sq units

Solution

Here $$\left( { x }_{ 1 }=3,{ y }_{ 1 }=8 \right) ;\left( { x }_{ 2 }=-4,{ y }_{ 2 }=2 \right) ;\left( { x }_{ 3 }=5,{ y }_{ 3 }=-1 \right) $$
$$Ar\left( \triangle ABC \right) =\cfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] $$
$$=\cfrac { 1 }{ 2 } \left[ 3(2+1)-4(1-8)+5(8-2) \right] =\cfrac { 1 }{ 2 } (9+36+30)=\cfrac { 75 }{ 2 } =37\cfrac { 1 }{ 2 } $$
Question 6
1 / -0

The area of quadrilateral ABCD with $$A(1,1),B(7,-3),C(12,2)$$ and $$D(7,21)$$ as its vertices is

A
$$35$$ sq. units

B
$$65$$ sq. units

C
$$85$$ sq. units

D
$$115$$ 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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Area of the $$\left( \triangle ABC \right) =\cfrac { 1 }{ 2 } \left[ 1(-3-2)+7(2-1)+12(1+3) \right] =25\quad $$ sq. units
Area of $$\quad \left( \triangle BCD \right) =\cfrac { 1 }{ 2 } \left[ 7(2-21)+12(21+3)+7(-3-2) \right] =60$$ sq. units
Area of the quad $$ABCD=ar\left( \triangle ABC \right) +ar\left( \triangle BCD \right) =85$$ sq. units
Question 7
1 / -0

In $$\triangle ABC$$ having vertices $$A(-1,3), B(1,-1)$$ and $$C(5,1)$$, the length of the median $$AD$$ is

A
$$3$$ units

B
$$4$$ units

C
$$5$$ units

D
$$6$$ units

Solution

Midpoint formula : $$\left(\dfrac {x_1+x_2}2,\dfrac {y_1+y_2}2\right))$$
Mid point of $$BC$$ is $$D\left( \cfrac { 1+5 }{ 2 } ,\cfrac { -1+1 }{ 2 } \right) \quad \Rightarrow D\left( 3,0 \right) $$
median $$AD=\sqrt { { \left( -1-3 \right) }^{ 2 }+{ \left( 3-0 \right) }^{ 2 } } =\sqrt { { \left( -4 \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 } } =\sqrt { 25 } =5\quad $$ unit
Question 8
1 / -0

In $$\triangle ABC$$, its vertices are $$A(7,-3),B(3,-1)$$ and $$C(5,3)$$. If $$BE$$ is one of it's medians, then $$BE=$$?

A
$$2\sqrt{2}$$ units

B
$$3$$ units

C
$$\sqrt{10}$$ units

D
none of these

Solution

Mid point of $$AC$$ is $$E\left( \cfrac { 7+5 }{ 2 } ,\cfrac { -3+3 }{ 2 } \right) \quad \Rightarrow E\left( 6,0 \right) $$
median $$BE=\sqrt { { \left( 6-3 \right) }^{ 2 }+{ \left( 0+1 \right) }^{ 2 } } =\sqrt { { \left( 3 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 9+1 } =\sqrt{10}$$ units
Question 9
1 / -0

One end of the diameter of a circle is $$A(4,1)$$ and its centre is $$C(3,3)$$. The coordinates of the other end of the diameter is

A
$$B(5,2)$$

B
$$B(-4,1)$$

C
$$B(4,-1)$$

D
$$B(2,5)$$

Solution

Let, the other end of the diameter be $$B(x,y)$$.
We know that centre of the circle is the mid point of the diameter. So, the formula for mid-point can be used to find the co-ordinates of $$B$$ as follows:
$$\cfrac { 4+x }{ 2 } =3$$
$$x+4=6$$
$$x=2$$

$$\cfrac { 1+y }{ 2 } =3$$
$$y+1=6$$
$$y=5$$
Hence, the other end of the diameter is $$B(2,5)$$
Question 10
1 / -0

A point$$P$$ divides the join of $$A(5,-2)$$ and $$B(9,6)$$ in the ratio $$3:1$$. The coordinates of $$P$$ are

A
$$(4,-7)$$

B
$$(-4,7)$$

C
$$(\cfrac{7}{2},4)$$

D
$$(8,4)$$

Solution

coordinates of $$P$$ are $$P(x,y)$$ where
$$x=\cfrac { 3\times 9+1\times 5 }{ 3+1 } =\cfrac { 32 }{ 4 } =8;y=\cfrac { 3\times 6+1\times (-2) }{ 3+1 } =\cfrac { 16 }{ 4 } =4\quad $$
the required point is $$P(8,4)$$

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Re-Attempt Weekly Quiz Competition

Coordinate Geometry Test - 44

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Coordinate Geometry Test - 44

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SHARING IS CARING

Question 1

$$15$$

$$\dfrac{3}{5}$$

$$\dfrac{29}{2}$$

$$\dfrac{33}{2}$$

Solution

Question 2

$$-2$$

$$2$$

$$-1$$

$$1$$

Solution

Question 3

$$D(5,-5)$$

$$D(-5,5)$$

$$D(-3,8)$$

$$D(3,-8)$$

Solution

Question 4

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 5

$$57$$ sq units

$$75$$ sq units

$$28\cfrac{1}{2}$$ sq units

$$37\cfrac{1}{2}$$ sq units

Solution

Question 6

$$35$$ sq. units

$$65$$ sq. units

$$85$$ sq. units

$$115$$ sq. units

Solution

Question 7

$$3$$ units

$$4$$ units

$$5$$ units

$$6$$ units

Solution

Question 8

$$2\sqrt{2}$$ units

$$3$$ units

$$\sqrt{10}$$ units

none of these

Solution

Question 9

$$B(5,2)$$

$$B(-4,1)$$

$$B(4,-1)$$

$$B(2,5)$$

Solution

Question 10

$$(4,-7)$$

$$(-4,7)$$

$$(\cfrac{7}{2},4)$$

$$(8,4)$$

Solution

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