Coordinate Geometry Test

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Coordinate Geometry Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$(a, b)$$ is the mid-point of the line segment joining the points $$A (10, 6)$$ and $$B (k, -4)$$ and $$a+ 2b = 18,$$ find the value of $$k.$$

A
$$11 $$

B
$$16 $$

C
$$ 22 $$

D
$$32 $$

Solution

Let the point $$P(x,y)=(a,b)$$ the line segment $$AB$$ joining the points $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 10,6 \right) \& B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( k,-4 \right)$$ in the ratio $$m:n=1:1$$

Then, by the section formula,
$$x=\dfrac { n{ x }_{ 2 }+m{ x }_{ 1 } }{ m+n } =\dfrac { { x }_{ 2 }+{ x }_{ 1 } }{ 2 }$$

$$\Rightarrow a=\dfrac { 10+k }{ 2 } \Rightarrow k=2a-10$$ ...(i)

$$y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } \Rightarrow b=\dfrac { 6-4 }{ 2 } =1$$

But given that $$a+2b=18\Rightarrow a=18-2b=18-2\times 1=16$$

Substituting $$a=16$$ in (i) we get $$k=2\times 16-10=22$$

Hence, option C.
Question 2
1 / -0

The point which divides the line joining the points $$A(1, 2)$$ and $$B(-1, 1)$$ internally in the ratio $$1:2$$ is

A
$$\left(\dfrac {-1}{3}, \dfrac {5}{3}\right)$$

B
$$\left(\dfrac {1}{3}, \dfrac {5}{3}\right)$$

C
$$(-1, 5)$$

D
$$(1, 5)$$

Solution

Let the given point $$C$$ divide $$A(1,2)$$ and $$B(-1, 1)$$ internally in the ratio $$1:2$$.
Therefore, we have $$m=1$$ and $$n=2$$.
By section formula for internal division, the point $$C(x,y)$$ is
$$x=\dfrac{(mx_{2}+nx_{1}) } {(m+n)}$$ and $$y=\dfrac{(my_{2}+ny_{1}) } {(m+n)}$$
$$x=\dfrac{(-1 \times 1+2 \times 1) } {(1+2)}$$ and $$y=\dfrac{(1 \times 1+2\times 2) } {(1+2)}$$
$$ \therefore x=\dfrac{1}{3}$$ and $$y=\dfrac{5}{3}$$
Question 3
1 / -0

If $$P\left(\dfrac{a}{2},4\right)$$ is the mid-point of the line-segment joining the points A(-6,5) and B(-2,3), then the value of a is

A
$$-8$$

B
$$3$$

C
$$-4$$

D
$$4$$

Solution

Given P is the mid point of AB,where $$A(-6,5)$$ and $$B(-2,3)$$
$$\therefore$$ $$\dfrac { a }{ 2 } =\dfrac { -6+(-2) }{ 2 } $$
$$\therefore a=-8$$
Question 4
1 / -0

The area of triangle ABC (in sq. units) is :

A
$$15$$ sq. units

B
$$10$$ sq. units

C
$$7.5$$ sq. units

D
$$2.5$$ sq. units

Solution

The area of a triangle is given as:
Area $$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$
the given points are $$ A(1,3),~B(-1,0) $$and $$C(4,0)$$
by Substituting ,
$$\text{Area} =\dfrac { 1 }{ 2 } \left[ 1(0-0)+(-1)(0-3)+4(3-0) \right] \\ =\dfrac { 1 }{ 2 } \left[ 3+12 \right] =\dfrac { 15 }{ 2 } =7.5$$
$$\therefore$$ Area of the given triangle ABC is $$7.5$$ sq. units
Question 5
1 / -0

Find the mid point of (3,8) and (9,4).

A
$$(5,6)$$

B
$$(6,6)$$

C
$$(4,4)$$

D
None of the above

Solution

Midpoint formula is given by $$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \right)$$
So midpoint of $$(3,8)$$ and $$(9,4)$$ is $$=\left(\dfrac{3+9}{2},\dfrac{8+4}{2} \right)=(6,6)$$
Question 6
1 / -0

C is the mid-point of PQ, if P is $$(4, x)$$, C is $$(y, -1)$$ and Q is $$(-2, 4)$$, then x and y respectively are-

A
$$-6$$ and $$1$$

B
$$-6$$ and $$2$$

C
$$6$$ and $$-1$$

D
$$6$$ and $$-2$$

Solution

Given C is the mid point of PQ
i.e.P(4,x) and Q(-2,4)
For finding y, equate the (x-coordinates)
y = $$\dfrac { 4+(-2) }{ 2 } =1$$
For finding x,equate the y - coordinates
$$\dfrac { x+4 }{ 2 } =-1\\ \therefore x=-6$$
$$\therefore x = -6$$
$$\therefore$$ x and y are $$-6$$ and $$1$$ respectively.
Question 7
1 / -0

Area of the triangle formed by the points P(-1.5, 3), Q(6, -2) and R(-3, 4) is 0.

A
True

B
False

C
Ambiguous

D
Data insufficient

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Given $$P(-1.5, 3), Q(6, -2)$$ and $$R(-3, 4)$$
Therefore, area is given by
$$= \frac{1}{2}\times[(-1.5)(-2-4) + 6(4-3)+(-3)(3+2)]$$
$$= \frac{1}{2}\times(9 +6-15)$$
$$= 0$$
Question 8
1 / -0

Coordinates of the points which divides the join of $$(-1, 7)$$ and $$(4, -3)$$ in the ratio $$2:3$$ are $$(1, 3).$$

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

We know that, the coordinates of the point dividing the line segment joining the points $$({x_1,y_1})$$ and $$({x_2,y_2})$$ in the ratio m:n is given by $$(\frac {m_1x_2+m_2x_1}{m_1+m_21}, \frac {m_1y_2+m_2y_1}{m_1+m_2})$$

Given $${(x_1,y_1)}= (-1,7)$$; $${(x_2,y_2)} = (4,-3)$$
$${(m_1,m_2)}= 2:3$$
Coordinates of point of intersection of line segment is
$$(\frac {(2)(4)+ (3)(-1)}{2+3}, \frac {(2)(-3)+(3)(7)}{2+3})$$
$$= (5/5,15/5)$$
$$= (1,3)$$
Question 9
1 / -0

If A(2, 4), B(6, 10), then find the midpoint of AB.

A
$$(4,5)$$

B
$$(7,4)$$

C
$$(4,7)$$

D
$$(3,5)$$

Solution

Given points are $$A(2, 4)$$ and $$B(6,10)$$.

To find out,
The coordinates of the midpoint of $$AB$$.

Let the coordinates of the midpoint be $$(h,k)$$.

We know that the coordinates of the midpoint of the line segment joining $$(x_1,y_1) \ and\ (x_2,y_2)$$ are:

$$P(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$$

Here, $$x_1=2,\ y_1=4,\ x_2=6,\ y_2=10,\ x=h,\ y=k$$

Hence, $$(h,k)=\left(\dfrac{2+6}{2}, \dfrac{4+10}{2}\right)$$

$$\Rightarrow (h,k)=\left(\dfrac{8}{2}, \dfrac{14}{2}\right)$$

$$\Rightarrow (h,k)=\left(4, 7\right)$$

Hence, the coordinates of the midpoint of $$AB$$ are $$(4,7)$$.
Question 10
1 / -0

Find the mid point of $$(4,6)$$ and $$(2,-6)$$.

A
$$(3,4)$$

B
$$(2,-2)$$

C
$$(3,0)$$

D
None of the above

Solution

We know that end points of a line segment is $$(a,b)$$ and $$(c,d)$$, then the midpoint of the line segment has the coordinates:
$$\dfrac{a+c}{2},\dfrac{b+d}{2}$$
Then mid point of line segment $$(4,6)$$ and $$(2,-6)$$ is
$$\dfrac{4+2}{2},\dfrac{6-6}{2}$$
$$\Rightarrow \dfrac{6}{2},\dfrac{0}{2}$$
$$\Rightarrow (3,0)$$

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Re-Attempt Weekly Quiz Competition

Coordinate Geometry Test - 23

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Coordinate Geometry Test - 23

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SHARING IS CARING

Question 1

$$11 $$

$$16 $$

$$ 22 $$

$$32 $$

Solution

Question 2

$$\left(\dfrac {-1}{3}, \dfrac {5}{3}\right)$$

$$\left(\dfrac {1}{3}, \dfrac {5}{3}\right)$$

$$(-1, 5)$$

$$(1, 5)$$

Solution

Question 3

$$-8$$

$$3$$

$$-4$$

$$4$$

Solution

Question 4

$$15$$ sq. units

$$10$$ sq. units

$$7.5$$ sq. units

$$2.5$$ sq. units

Solution

Question 5

$$(5,6)$$

$$(6,6)$$

$$(4,4)$$

None of the above

Solution

Question 6

$$-6$$ and $$1$$

$$-6$$ and $$2$$

$$6$$ and $$-1$$

$$6$$ and $$-2$$

Solution

Question 7

True

False

Ambiguous

Data insufficient

Solution

Question 8

True

False

Ambiguous

Data insufficient

Solution

Question 9

$$(4,5)$$

$$(7,4)$$

$$(4,7)$$

$$(3,5)$$

Solution

Question 10

$$(3,4)$$

$$(2,-2)$$

$$(3,0)$$

None of the above

Solution

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