Coordinate Geometry Test

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Coordinate Geometry Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The mid point of $$(2,3)$$ and $$(8,9)$$ is

A
$$(5,6)$$

B
$$(2,8)$$

C
$$(5,7)$$

D
$$(4,6)$$

Solution

The points are $$(x_1,y_1)=(2,3)$$ and $$(x_2,y_2)=(8,9)$$

The mid point is given as $$\left(\dfrac{2+8}2,\dfrac {3+9}2\right)$$

$$=\left(\dfrac {10}2,\dfrac {12}2\right)=(5,6)$$
Question 2
1 / -0

The mid point of $$(4,9) $$ and $$(8,3)$$ is

A
$$(6,6)$$

B
$$(5,7)$$

C
$$(-6,-6)$$

D
None.

Solution

The mid point of $$(4,9) $$ and $$(8,3)$$ is given as
$$\left(\dfrac{4+8}2,\dfrac{9+3}2\right)\\=\left(\dfrac {12}2,\dfrac{12}2\right)=(6,6)$$
Question 3
1 / -0

The mid point of $$(3,4)$$ and $$(1,-2)$$ is

A
(2,1)

B
(1,2)

C
(2,-1)

D
(1,-2)

Solution

The points are $$(3,4)$$ and $$(1,-2)$$

The mid point of $$(3,4)$$ and $$(1,-2)$$ is given by
$$\left(\dfrac {x_1+x_2}2,\dfrac {y_1+y_2}2\right)\\\left(\dfrac{3+1}2,\dfrac {4-2}2\right)=(2,1)$$
Question 4
1 / -0

A(1,-1), B(0,4) and C (-5,3) are vertices of $$\Delta ABC.\;The\;length\;of\;median\;$$$$\overline {AD} $$ is......

A
$$\sqrt {130} $$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{{\sqrt {130} }}{2}$$

D
1

Solution
Question 5
1 / -0

Select the correct alternative from the answer of the question given below. How many midpoints does a segment have?

A
only one

B
two

C
three

D
many

Solution

Every segment has one and only one midpoint.
Hence, the correct answer is only one.
Question 6
1 / -0

Find the area of triangle having vertices are $$A (3, 1), B (12, 2)$$ and $$C (0, 2)$$.

A
$$4$$

B
$$6$$

C
$$12$$

D
$$18$$

Solution

The given points are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,1 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 12,2 \right) \& C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 0,2 \right)$$
Let us obtain $$ar\Delta ABC$$ by applying the area formula.
$$ ar\Delta =\dfrac { 1 }{ 2 } \left\{ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right\}$$
$$\Rightarrow ar\Delta =\dfrac { 1 }{ 2 } \left\{ 3\left( 2-2 \right) +12\left( 2-1 \right) +0\left( 1-2 \right) \right\}$$ units $$=6$$ units
$$\therefore ar\Delta =6$$ units
Hence, option B.
Question 7
1 / -0

The distance of the point P (6, 8) from the origin is

A
$$8$$

B
$$2\sqrt{7}$$

C
$$10$$

D
$$6$$

Solution

$$ The\quad distance\quad p\quad of\quad a\quad point\quad P(x,y)\quad from\quad the\quad origine\quad is\\ p=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } .\\ Here\quad P(x,y)=(6,8).\\ \therefore \quad p=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { 6^{ 2 }+8^{ 2 } } units=\sqrt { 100 } units=10\quad units.\\ Ans-\quad 10\quad units.\\ $$
Question 8
1 / -0

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

A
$$-4$$

B
$$- 12$$

C
$$12$$

D
$$-6$$

Solution

$$P(\dfrac{a}{3} .4)$$ is the mid - point of the line segment joining $$Q (6,5) $$and $$R (2,3)$$.
Using section formula,
co-ordinates of $$P =\left(\dfrac{6+2}{2} , \dfrac{5+3}{2}\right)$$

co-ordinates of $$P =\left(\dfrac{8}{2} ,\dfrac{8}{2}\right)$$

co-ordinates of $$P =\left(4,4\right)$$
$$\dfrac{a}{3}=4$$
$$\Rightarrow a=12$$
Question 9
1 / -0

The point which lies on the perpendicular bisector of the line segment joining the points $$A (2, 5)$$ and $$B (-2,- 5) $$ is:

A
$$(0, 0)$$

B
$$(0, 2)$$

C
$$(-2, 0)$$

D
$$(2, 0)$$

Solution

Let the mid point is $$(x_1 , y_1)$$
$$x_1 =\dfrac{2+(-2)}{2}=\dfrac{0}{2} =0$$
$$y_1=\dfrac{5+(-5)}{0}=\dfrac{0}{2}=0$$
$$(0,0)$$ lies on the perpendicular bisector of $$(2,5)$$ and $$(-2,-5)$$.
Question 10
1 / -0

The area of a triangle with vertices $$(a, b + c), (b, c + a)$$ and $$(c, a + b)$$ is

A
$$a$$

B
$$b$$

C
$$c$$

D
$$0$$

Solution

The area of $$\triangle ABC$$ with vertices $$A\equiv(x_1, y_1)$$, $$B\equiv(x_2, y_2)$$ and $$C\equiv(x_3, y_3)$$ is given as,
$$A(\triangle ABC) = \left|\dfrac12[x_1(y_3-y_2) +x_2(y_1 - y_3) + x_3(y_2-y_1) ]\right|$$
In this problem,
$$A(\triangle ABC) = \left|\dfrac12[a(a+b-(c+a)) +b(b+c - (a+b)) + c(c+a-(b+c)) ]\right|$$
$$\therefore A(\triangle ABC) = \left|\dfrac12[ab-ac + bc-ba + ca-cb]\right|$$
$$\therefore A(\triangle ABC) = 0$$
Hence, the correct Option is D.

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Coordinate Geometry Test - 21

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Coordinate Geometry Test - 21

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Question 1

$$(5,6)$$

$$(2,8)$$

$$(5,7)$$

$$(4,6)$$

Solution

Question 2

$$(6,6)$$

$$(5,7)$$

$$(-6,-6)$$

None.

Solution

Question 3

(2,1)

(1,2)

(2,-1)

(1,-2)

Solution

Question 4

$$\sqrt {130} $$

$$\dfrac{1}{2}$$

$$\dfrac{{\sqrt {130} }}{2}$$

1

Solution

Question 5

only one

two

three

many

Solution

Question 6

$$4$$

$$6$$

$$12$$

$$18$$

Solution

Question 7

$$8$$

$$2\sqrt{7}$$

$$10$$

$$6$$

Solution

Question 8

$$-4$$

$$- 12$$

$$12$$

$$-6$$

Solution

Question 9

$$(0, 0)$$

$$(0, 2)$$

$$(-2, 0)$$

$$(2, 0)$$

Solution

Question 10

$$a$$

$$b$$

$$c$$

$$0$$

Solution

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