Coordinate Geometry Test

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Coordinate Geometry Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

If $$A$$ and $$B$$ are the points $$(-3,4)$$ and $$(2,1)$$, then the co-ordinates of the point $$C$$ on $$AB$$ produced such that $$AC=2BC$$ are

A
$$(2,4)$$

B
$$(3,7)$$

C
$$(7,-2)$$

D
$$\left(\dfrac{1}{2},\dfrac{5}{2}\right)$$

Solution

Solution:
Given $$AC=2BC$$
$$\dfrac{AC}{BC}=\dfrac{2}{1}$$

So, C divides AB externally by $$2:1$$

Coordinates of $$C=\left(\dfrac{mx_2-nx}{m-n},\dfrac{my_2-ny}{m-n}\right)$$

where $$m:n=2:1$$ then $$(x_1,y_1)=(-3,4)$$, $$(x_2, y_2)=(2,1)$$

$$C=\left(\dfrac{2(2)-1(-3)}{2-1},\dfrac{2(4)-1(4)}{2-1}\right)\\$$
$$C=(7,-2)\\$$
Answer C
Question 2
1 / -0

The coordinate of point which divides the line segment joining points $$A(0,0)$$ and $$B(9,12)$$ in the ratio $$1:2$$, are

A
$$(-3,4)$$

B
$$(3,4)$$

C
$$(3,-4)$$

D
$$none\ of\ these$$

Solution

Using the section formula, if a point $$(x,y)$$ divides the

line joining the points $$A({ x }_{ 1 },{ y }_{ 1 })$$ and $$B({ x }_{ 2 },{ y

}_{ 2 })$$ in the ratio $$ m:n $$, then $$(x,y) = \left( \cfrac { m{ x }_{ 2

}+n{ x }_{ 1 } }{ m+n } ,\cfrac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n }

\right) $$

Let the required point be $$C (x,y) $$ which divides the line segment joining points $$A(0,0)$$ and $$B(9,12)$$ in ratio $$1:2$$
$$m_1=1, m_2=2$$

Then we can calculate the coordinates of $$C$$ using section formula
$$(x,y)=\left[\dfrac{m_1\times x_2+m_2\times x_1}{m_1+m_2}, \dfrac{m_1\times y_2+m_2\times y_1}{m_1+m_2}\right]$$

$$(x,y)=\left[\dfrac{1\times 9+2\times 0}{1+2}, \dfrac{1\times 12+2\times 0}{1+2}\right]$$

$$(x,y)=(3,4)$$

Hence, coordinates of $$C$$ will be $$(3,4)$$.
Question 3
1 / -0

Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are $$(0.-1), (2, 1) and (0, 3)$$

A
$$4$$

B
$$8$$

C
$$1$$

D
$$2$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Also, we know $$\\Area\>of\>triangle\>=4\times\>of\>triangle\>formed\>using\>mid-point\>\\=4\times(\frac{1}{2})[x_1(y_2-y_3)+x_2(y_3+y_1)+x_3(y_1-y_2)]\\=2[0+2(3-1)+0]=8sq\>unit$$
Question 4
1 / -0

If $$(1, 2), (4, y), (x, 6) $$ and $$(3, 5)$$ are the vertices of a parallelogram taken in order find $$x\ and\ y$$

A
$$x=3, y=6$$

B
$$x= -6, y=3$$

C
$$x=6, y=3$$

D
$$x=3, y=-6$$

Solution

Coordinates of M are mid points of BD
$$\therefore M=\left( \cfrac { 4+3 }{ 2 } ,\cfrac { 5+y }{ 2 } \right) \\ $$
Coordinates of M are mid points of AC
$$\therefore M=\left( \cfrac { 1+x }{ 2 } ,\cfrac { 2+6 }{ 2 } \right) \\ $$
$$\therefore \cfrac { 1+x }{ 2 } =\cfrac { y+3 }{ 2 } \\ so\quad x=6\\ \cfrac { 5+y }{ 2 } =\cfrac {2+6 }{ 2 }\\
\therefore y=3$$
Question 5
1 / -0

If $$A(-2, 1), B(a, 0), C(4, b)$$ and $$D(1, 2)$$ are the vertices of a parallelogram ABCD taken in order, find the values of $$a$$ and $$b.$$

A
$$a=1,b=1 $$

B
$$a=2,b=2$$

C
$$a=1,b=4$$

D
$$a=1,b=2$$

Solution

Midpoint formula $$X=\left ( \dfrac{x_1+x_2}{2}\right ) and \ Y=\left ( \dfrac{y_1+y_2}{2}\right )$$

Coordinates of $$M$$ are mid points of $$AC$$
$$ M=\left( \cfrac { 4-2 }{ 2 } ,\cfrac { b+1 }{ 2 } \right) $$
As the diagonals of parallelogram bisect each other,
Coordinates of $$M$$ are mid points of $$BD$$ aswell

$$ M=\left( \cfrac { a+1 }{ 2 } ,\cfrac {0+2}{ 2 } \right) $$

Equating $$X$$- coordinates:
$$ \cfrac { 4-2 }{ 2 } =\cfrac { a+1 }{ 2 } $$
$$a+1 =2$$
$$a=1$$

Equating $$Y$$- coordinates:

$$\cfrac { b+1 }{ 2 } =\cfrac { 0+2 }{ 2 }$$
$$b+1 =2$$
$$ b=1$$
Question 6
1 / -0

Find the area of the triangle whose vertices are $$(-5, -1), (3, -5), (5, 2)$$

A
$$31 sq$$ $$units$$

B
$$32sq$$ $$units$$

C
$$33 sq$$ $$units$$

D
$$\text{no triangle can be formed}$$

Solution

Area of a triangle whose vertices are $$A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$$ is given as,
$$Area=(\dfrac{1}{2})[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] $$
Let $$A$$ be the required area
$$A=(\dfrac{1}{2})[(-5)(-5-2)+3(2+1)+5(-1+5)]\\ (\dfrac{1}{2})[35+9+20]\\=32sq\>unit$$
So, the correct option is (B)
Question 7
1 / -0

Find the Co-ordinate of the points $$O$$ bisection of the line segment joining $$(4, -1)$$ and $$(-2, -3)$$.

A
$$(1,-2)$$

B
$$(1,-3)$$

C
$$(2,-3)$$

D
$$(4,-3)$$

Solution

$$\\ mid-point\\=\left (\dfrac{4+(-2)}{2}),\dfrac{-1+(-3)}{2}\right)\\=(1,-2)$$
Question 8
1 / -0

If $$(3, -4)$$ and $$(-6, 5)$$ are the extremities of a diagonal of a parallelogram and $$(2, 1)$$ is its third vertex, then its fourth vertex is?

A
$$(-1, 0)$$

B
$$(-1, 1)$$

C
$$(0, -1)$$

D
$$(-5, 0)$$

Solution
Question 9
1 / -0

If the vertices of a triangle are $$(1,2),(4,-6)$$ and $$(3,5)$$, then its area is

A
$$\cfrac{23}{2}$$ sq. units

B
$$\cfrac{25}{2}$$ sq. units

C
$$12$$ sq.unit

D
none of these

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore, area of required triangle is given by
Area $$ = \dfrac{1}{2} \times [ 1 (-6 - 5) + 4(5 - 2) + 3(2 + 6) ] $$
$$=\dfrac{1}{2}[-11+12+24]$$
$$=\dfrac{1}{2}\times25$$
$$=\dfrac{25}{2}\ sq.unit$$
Question 10
1 / -0

If $$A(-2,4), B(0,0)$$ and $$C(4,2)$$ are the vertices of a $$\triangle{ABC}$$, then find the length of median through the vertex $$A$$.

A
$$2$$

B
$$4$$

C
$$3$$

D
$$5$$

Solution

$$D=$$slid $$ht$$ of $$BC$$
$$D\cong \left (\dfrac {0+4}{2}, \dfrac {0+2}{2}\right)$$
$$=(2, 1)$$
$$\therefore \ $$ Length of median $$=$$ Light of $$AD$$
$$=\sqrt {(-2 -2)^2 +(4-1)^2}$$
$$=\sqrt {4^2 +3^2}$$
$$=5$$

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Coordinate Geometry Test - 40

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Coordinate Geometry Test - 40

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Question 1

$$(2,4)$$

$$(3,7)$$

$$(7,-2)$$

$$\left(\dfrac{1}{2},\dfrac{5}{2}\right)$$

Solution

Question 2

$$(-3,4)$$

$$(3,4)$$

$$(3,-4)$$

$$none\ of\ these$$

Solution

Question 3

$$4$$

$$8$$

$$1$$

$$2$$

Solution

Question 4

$$x=3, y=6$$

$$x= -6, y=3$$

$$x=6, y=3$$

$$x=3, y=-6$$

Solution

Question 5

$$a=1,b=1 $$

$$a=2,b=2$$

$$a=1,b=4$$

$$a=1,b=2$$

Solution

Question 6

$$31 sq$$ $$units$$

$$32sq$$ $$units$$

$$33 sq$$ $$units$$

$$\text{no triangle can be formed}$$

Solution

Question 7

$$(1,-2)$$

$$(1,-3)$$

$$(2,-3)$$

$$(4,-3)$$

Solution

Question 8

$$(-1, 0)$$

$$(-1, 1)$$

$$(0, -1)$$

$$(-5, 0)$$

Solution

Question 9

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

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

$$12$$ sq.unit

none of these

Solution

Question 10

$$2$$

$$4$$

$$3$$

$$5$$

Solution

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