Coordinate Geometry Test

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Coordinate Geometry Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The mid-point of the line $$(a, 2)$$ and $$(3, 6)$$ is $$(2, b)$$. Find the numerical values of $$a$$ and $$b$$.

A
$$a=1$$, $$b=6$$

B
$$a=2$$, $$b=4$$

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

D
$$a=2$$, $$b=6$$

Solution

Mid-point of $$(a,2)$$ and $$(3,6)$$ is $$(2,b)$$
=>$$(2,b)=\left( \cfrac { a+3 }{ 2 } ,\cfrac { 2+6 }{ 2 } \right) \\ =>a=4-3,b=4\\ =>a=1,b=4$$
Question 2
1 / -0

The mid-point of the line segment joining the point $$\left( {3m,6} \right)$$ and $$\left( { - 4,3n} \right)$$ is $$(1,2m - 1)$$. Find $$(m,n)$$.

A
$$(2,0)$$

B
$$(2,2)$$

C
$$(0,0)$$

D
$$(0,2)$$

Solution

Mid-point of any two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is given by
$$(m, n)=\left (\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)$$

Since $$(1,2m-1)$$ is midpoint of $$(3m, 6)$$ and $$(-4, 3n)$$, then :

Taking $$x$$ coordinates,
$$1=\dfrac{3m-4}{2},$$
$$2=3m-4,$$
$$3m=4+2$$
$$3m=6$$
$$m=2$$

Similarly,
Taking $$y$$ coordinates,
$$2m-1=\dfrac{6+3n}{2},$$
$$4m-2=6+3n,$$
$$4\times2-2=6+3n,$$ ($$m=2$$)
$$3n+6=6$$
$$n=0$$

Hence, $$(m ,n)=(2,0)$$
Question 3
1 / -0

If the points $$(2, 5), (4, 6)$$ and $$(a, a)$$ are collinear, then find the value of $$a$$.

A
$$4$$

B
$$-4$$

C
$$-8$$

D
$$8$$

Solution

Points: $$A = (2,5), B = (4,6),C = (a, a)$$ collinear
$$\therefore$$ area of $$\triangle ABC = 0$$
$$\Rightarrow \ \dfrac {1}{2} [x_1(y_2-y_3)+x_2 (y_3-y_1)+x_3(y_1-y_2)]=0$$
$$\Rightarrow \ \dfrac {1}{2} [2(6-a)+4(a-5)+a(5-6)]=0$$
or, $$12-2a+4a-20+5a-6a=0$$
$$-8+a=0$$
$$\boxed {a=8}$$
Option $$D$$ is correct
Question 4
1 / -0

Area of the triangle formed by $$(x_{1,}y_{1}),(x_{2},y_{2}), (3y_{2},(-2y_{1}))$$

A
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

B
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

C
$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

D
$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

We have,

$$ A\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},{{y}_{1}} \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{x}_{2}},{{y}_{2}} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3{{y}_{2}},-2{{y}_{1}} \right) $$

We know that the area of triangle is

$$ Area\,of\,\Delta ABC=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

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

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$
Hence, the is the answer
Question 5
1 / -0

The length of the median from the vertex A of a triangle whose vertices are $$A (-1, 3),$$ B $$(1, -1)$$ and C$$(5,1)$$ is

A
$$5$$

B
$$4$$

C
$$1$$

D
$$3$$

Solution

Length of the median from the vertex $$A$$ of a triangle $$\triangle{ABC}$$
Let $$AD$$ be the median.

$$\Rightarrow\,D$$ is the midpoint of $$BC$$

Using midpoint formula,$$D=\left(\dfrac{1+5}{2},\,\dfrac{-1+1}{2}\right)=\left(3,\,0\right)$$

Length of median $$=AD=\sqrt{{\left(-1-3\right)}^2{}+{\left(0-3\right)}^{2}}=\sqrt{16+9}=\sqrt{25}=5 $$ units.
Question 6
1 / -0

Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are $$(2, 1)$$, $$(-2, 3)$$, $$(4, -3)$$.

A
$$1.5$$ sq. units

B
$$3$$ sq. units

C
$$6$$ sq. units

D
$$20$$ sq. units
Question 7
1 / -0

Find the coordinates of the point $$P$$ which divides the join of $$A(-2,5)$$ and $$B(3,-5)$$ in the ratio $$2:3$$

A
$$(0,1)$$

B
$$(1,0)$$

C
$$(1,1)$$

D
$$(0,0)$$

Solution

Given that $$A(-2,5) \quad B(3,-5)$$

Using section formula (m:n) if ratio, then points are

$$x=\dfrac{mx_{2}+nx_{1}}{m+n}$$

$$x=\dfrac{2\times 3+3(-2)}{2+3}$$

$$=\dfrac{6-6}{5}$$

$$\therefore x=0$$

$$y=\dfrac{my_{2}+ny_{1}}{m+n}$$

$$y= \dfrac{2(-5)+3(5)}{2+3}$$

$$= \dfrac{-10+15}{5}$$

$$\therefore y=\dfrac{5}{5}=1$$

$$\therefore P\rightarrow (0,1)$$
Question 8
1 / -0

The midpoints of the sides of a triangle are $$\left(1,1\right),\left(4,3\right)$$ and $$\left(3,5\right)$$. The area of the triangle is ___ square units.

A
$$14$$

B
$$16$$

C
$$18$$

D
$$20$$

Solution
Question 9
1 / -0

If $$P(1,2),Q(4,6),R(5,7)$$ and $$S(a,b)$$ are the vertices of a parallelogram $$PQRS$$, then

A
$$a=2,b=4$$

B
$$a=3,b=4$$

C
$$a=2,b=3$$

D
$$a=3,b=3$$

Solution

R.E.F image
$$ \dfrac{P+R}{2} = \dfrac{Q+5}{2} $$
$$ \dfrac{1+5}{2} = \dfrac{4+a}{2} \Rightarrow a = 2 $$
$$ \dfrac{b+6}{2}=\dfrac{7+2}{2} \Rightarrow b = 3 $$
Question 10
1 / -0

If A (-2, 1) B (1 , 0), C (x , 3 ) and D (1, Y) are the vertices of a Parallelogram ABCD, then find the values of x and y.

A
$$(4,4)$$

B
$$(4,-3)$$

C
$$(4,3)$$

D
$$(4,-4)$$

Solution

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Coordinate Geometry Test - 41

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Coordinate Geometry Test - 41

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Question 1

$$a=1$$, $$b=6$$

$$a=2$$, $$b=4$$

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

$$a=2$$, $$b=6$$

Solution

Question 2

$$(2,0)$$

$$(2,2)$$

$$(0,0)$$

$$(0,2)$$

Solution

Question 3

$$4$$

$$-4$$

$$-8$$

$$8$$

Solution

Question 4

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

Question 5

$$5$$

$$4$$

$$1$$

$$3$$

Solution

Question 6

$$1.5$$ sq. units

$$3$$ sq. units

$$6$$ sq. units

$$20$$ sq. units

Question 7

$$(0,1)$$

$$(1,0)$$

$$(1,1)$$

$$(0,0)$$

Solution

Question 8

$$14$$

$$16$$

$$18$$

$$20$$

Solution

Question 9

$$a=2,b=4$$

$$a=3,b=4$$

$$a=2,b=3$$

$$a=3,b=3$$

Solution

Question 10

$$(4,4)$$

$$(4,-3)$$

$$(4,3)$$

$$(4,-4)$$

Solution

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