Coordinate Geometry Test

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Coordinate Geometry Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of $$\Delta DEF$$, the area of triangle DEF is sq. units.

A
$$11$$

B
$$22$$

C
$$48$$

D
$$50$$

Solution

Area of triangle = $$\dfrac{1}{2} {x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$$

Area of triangle $$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$$

$$=|\dfrac{1}{2}(-3+16-35)|$$

$$=11$$ sq. Units.
Question 2
1 / -0

The area of triangle formed by the points $$(p, 2 - 2p), (1 - p, 2p)$$ and $$(-4 - p, 6 - 2p)$$ is $$70$$ sq. units. How many integral values of p are possible ?

A
$$2$$

B
$$3$$

C
$$4$$

D
None of these

Solution

Given the points of triangle, $$A(p,2-2p), B(1-p,2p)$$ and $$C(-4-p,6-2p)$$
Area =70sq.unit
$$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 }){ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] =70\\ \left\{ p\left[ 2p-6+2p \right] +(1-p)\left[ 6-2p+2p \right] +(-4-p)\left[ 2-2p-2p \right] \right\} =140\\ p(4p-6)+(1-p)(4)- \left[(4+p) (2-4p) \right] =140\\ 4{ p }^{ 2 }-6p+4-4p-\left[ 8-16p+2p-4{ p }^{ 2 } \right] =140\\ 4{ p }^{ 2 }-6p+4-4p-8+16p-2p+4{ p }^{ 2 }=140\\ 8{ p }^{ 2 }+4p-144=0\\ 2p^2+p-36=0$$
$$2p^2+9p-8p-36=0$$
$$p(2p+9)-4(2p+9)=0$$
$$(2p+9)=0,(p-4)=0$$
$$ p=-9/2,p=4\\ $$
$$\therefore$$ Only one integral value is possible.
Question 3
1 / -0

$$A(-1, 2), B(4, 1)\ and\ C(7, 6)$$ are three vertices of the parallelogram $$ABCD$$. The coordinates of fourth vertex is _______.

A
$$(7, -2)$$

B
$$(-2, 7)$$

C
$$(7, 2)$$

D
$$(2, 7)$$

Solution

Let the three vertices of parallelogram ABCD be $$A(-1,2) , B(4,1) and C(7,6)$$
Let the fourth vertex be $$D(a,b)$$
By joining AC and BD intersect at point O i.e, diagonals of parallelogram bisect each other
Mid point of AC= $$\left( \dfrac {-1+ 7 }{ 2 } ,\dfrac {2+ 6 }{ 2 } \right) =(3,4)$$
Mid point of BD $$=\left( \dfrac { 4+a }{ 2 } ,\dfrac { 1+b }{ 2 } \right) \\ \dfrac { 4+a }{ 2 } =3\quad \dfrac { 1+b }{ 2 } =4\\ \therefore a=2,b=7$$
$$\therefore$$ The coordinate of fourth vertex =$$(2,7)$$
Question 4
1 / -0

If $$A = (-3, 4), B = (-1, -2), C = (5, 6), D = (x, -4)$$ are vertices of a quadrilateral such that area of triangle $$ABD = 2$$ area of triangle $$ACD$$, then $$x =$$

A
$$6$$

B
$$9$$

C
$$69$$

D
$$96$$

Solution
Question 5
1 / -0

If $$A(2, 2)$$, $$B(- 4, -4)$$ and $$C(5, -8)$$ are the vertices of a triangle, then the length of the median through vertex C is ___________.

A
$$\sqrt {65}\,\,\,units$$

B
$$\sqrt {117}\,\,\,units$$

C
$$\sqrt {85}\,\,\,units$$

D
$$\sqrt {113}\,\,\,units$$

Solution

Given the vertices of $$\triangle$$ABC
i.e, $$A(2,2) , B(-4,-4)$$ and $$C(5,-8)$$
The coordinates of mid-point of $$AB$$,
i.e,$$ D =\left( \dfrac { -4+2 }{ 2 } ,\dfrac { -4+2 }{ 2 } \right) =(-1,-1)$$
hence the length of medium through vertex $$C=CD$$
$$=\sqrt { (5+1)^{ 2 }+(-8+1)^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ 7 }^{ 2 } } =\sqrt { 85 } $$
Question 6
1 / -0

Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points $$(-2,-1),(1,0),(4,3)$$ and $$(1,2)$$ meet.

A
$$(1,1)$$

B
$$(1,3)$$

C
$$(5,1)$$

D
None of these

Solution

The vertices of parallelogram in order are $$A(-2,-1), B(1,0), C(4,3), D(1,2)$$.

So the diagonals will be $$AC$$ and $$BD$$.

Since the diagonals of a parallelogram bisects each other, so mid-point of

$$AC$$ or $$BD$$ will be intersection point of diagonals.

Hence by mid-point theorem, mid-point of $$AC$$ is

$$A(-2,-1) \ and \ C(4,3)$$

$$x=\dfrac{-2+4}{2}=1$$ and $$y=\dfrac{-1+3}{2}=1$$.

so $$(1,1)$$ is required point.
Question 7
1 / -0

If two vertices of a parallelogram are $$(3,2),(-1,0)$$ and the diagonals cut at $$(2,-5)$$, find the other vertices of the parallelogram.

A
$$(1,-14)$$, $$(5,-11)$$

B
$$(2,-12)$$, $$(5,-10)$$

C
$$(1,-12)$$, $$(5,-10)$$

D
None of these

Solution

Since diagonals of a parallelogram bisect each other.

Hence by mid-point theorem for $$AC$$:

$$\Rightarrow$$ $$2=\dfrac{x+3}{2}=>x=1$$ and

$$\Rightarrow$$ $$-5=\dfrac{y+2}{2}=>y=-12$$. So, point $$C$$ is $$(1,-12)$$

Similarly for $$BD$$

$$\Rightarrow$$ $$2=\dfrac{u-1}{2}=>u=5$$ and

$$\Rightarrow$$ $$-5=\dfrac{v+0}{2}=>v=-10$$. So, point $$D$$ IS $$(5,-10)$$.
Question 8
1 / -0

The area of the triangle with vertices at $$(-4, 1), (1, 2)(4, -3)$$ is

A
$$14$$

B
$$16$$

C
$$15$$

D
None of these

Solution
Question 9
1 / -0

The area of the triangle whose vertices are (3,8), (-4,2) and (5,-1) is

A
$$75 \ sq. units$$

B
$$37.5 \ sq. units$$

C
$$45 \ sq. units$$

D
$$22.5 \ sq. units$$

Solution

Let $$A(3,8), B(-4,2),C(5,-1)$$ be the vertices of the given $$\triangle ABC$$.

Then,

$$(x_{1}=3,y_{1}=8),(x_{2}=-4,y_{2}=2),(x_{3}=5,y_{3}=-1)$$

Area of $$\triangle ABC$$ = $$\dfrac{1}{2}|[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]|$$

$$=\dfrac{1}{2}|3[2-(-1)]-4(-1-8)+5(8-2)|$$

$$=\dfrac{1}{2}|9+36+30|=\dfrac{75}{2}=37.5 \ sq. units$$
Question 10
1 / -0

The mid point of line $$AB$$ with $$A(2,3)$$ and $$B(5,6)$$

A
$$(3.5,4.5)$$

B
$$(3,4)$$

C
$$(4,5)$$

D
None of these

Solution

Given points $$A(2,3)\equiv(x_1,y_1)$$ and $$B(5,6)\equiv(x_2,y_2)$$

Mid points given as ,

$$\Rightarrow\left(\dfrac{x_1+x_2}2,\dfrac{y_1+y_2}2\right)$$

$$\Rightarrow\left(\dfrac{2+5}2,\dfrac{3+6}2\right)$$

$$\Rightarrow\left(3.5,4.5\right)$$

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Coordinate Geometry Test - 39

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Coordinate Geometry Test - 39

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SHARING IS CARING

Question 1

$$11$$

$$22$$

$$48$$

$$50$$

Solution

Question 2

$$2$$

$$3$$

$$4$$

None of these

Solution

Question 3

$$(7, -2)$$

$$(-2, 7)$$

$$(7, 2)$$

$$(2, 7)$$

Solution

Question 4

$$6$$

$$9$$

$$69$$

$$96$$

Solution

Question 5

$$\sqrt {65}\,\,\,units$$

$$\sqrt {117}\,\,\,units$$

$$\sqrt {85}\,\,\,units$$

$$\sqrt {113}\,\,\,units$$

Solution

Question 6

$$(1,1)$$

$$(1,3)$$

$$(5,1)$$

None of these

Solution

Question 7

$$(1,-14)$$, $$(5,-11)$$

$$(2,-12)$$, $$(5,-10)$$

$$(1,-12)$$, $$(5,-10)$$

None of these

Solution

Question 8

$$14$$

$$16$$

$$15$$

None of these

Solution

Question 9

$$75 \ sq. units$$

$$37.5 \ sq. units$$

$$45 \ sq. units$$

$$22.5 \ sq. units$$

Solution

Question 10

$$(3.5,4.5)$$

$$(3,4)$$

$$(4,5)$$

None of these

Solution

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