Coordinate Geometry Test

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Coordinate Geometry Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

Find the points which divide the line segment joining $$A(-4,0)$$ and $$B(0,6)$$ into two equal parts.

A
$$ (-2,3)$$

B
$$\left(-3,\cfrac{3}{2}\right)$$

C
$$\left(-3,\cfrac{-3}{2}\right)$$

D
None of these

Solution

Given $$A(-4,0)$$ and $$B(0,6)$$. Let $$P(x,y)$$ divide $$AB$$ into two equal parts.

By mid-point theorem:

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

$$x=\dfrac{-4+0}{2}=-2$$ and $$y=\dfrac{0+6}{2}=3$$.

so, $$P(x,y)=(-2,3)$$ is require point.
Question 2
1 / -0

Find the point $$P$$ which divides the line segment joining the points $$A(1,-3)$$ and $$B(-3,9)$$ internally in the ratio $$1:3$$.

A
$$(2,1)$$

B
$$(0,0)$$

C
$$(\cfrac{5}{3},2)$$

D
$$(1,-2)$$

Solution

Given the two points $$A(1,-3)$$ and $$B(-3,9)$$ and ratio $$m:n=1:3.$$
Here we apply the section formula, i. e. if a point $$X(x,y)$$ divides the line segment $$AB$$ joining the points $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) \& B\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ in the ratio $$m:n$$ then we have

$$\\ x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } \ , y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } .\\$$
$$\therefore$$ $$(x,y)\ =\left( \dfrac { 3\times 1+1\times (-3) }{ 1+3 } ,\dfrac {3\times (-3)+1\times 9 }{ 1+3 } \right) = (0,0) $$
Question 3
1 / -0

Area of the triangle formed by the points $$(0,0),(2,0)$$ and $$(0,2)$$ is

A
$$1$$ sq.units

B
$$2$$ sq.units

C
$$4$$ sq.units

D
$$8$$ sq.units

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,
$$=\dfrac{1}{2}\left|0(0-2)+2(2-0)+0(0-0)\right|$$

$$=\dfrac{1}{2}\left|0+4+0\right|$$

$$=\dfrac{1}{2}\times 4=2\ sq.\ units$$

Hence, this is the answer.
Question 4
1 / -0

The midpoint of the line joining $$(a,-b)$$ and $$(3a,5b)$$ is

A
$$(-a,2b)$$

B
$$(2a,4b)$$

C
$$(2a,2b)$$

D
$$(-a,-3b)$$

Solution

Consider the given points.
$$(a, -b)$$ and $$(3a, 5b)$$

So, the midpoint according to midpoint formula,
$$\left(\dfrac{a+3a}{2}, \dfrac{-b+5b}{2}\right)$$

$$= \left(\dfrac{4a}{2}, \dfrac{4b}{2}\right)$$

$$= \left(2a, 2b\right)$$

Hence, this is the answer.
Question 5
1 / -0

Find the value of $$a$$ if area of the triangle is $$17$$ square units whose vertices are $$(0,0), (4,a), (6,4)$$.

A
$$a=-2$$

B
$$a=-3$$

C
$$a=3$$

D
none of the above

Solution

vertices of the triangle are $$A(0,0), B(4,a), C(6,4)$$.

Then area of triangle$$ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$

$$\Rightarrow 17=\dfrac{1}{2}[0(a-4)+4(4-0)+6(0-a)$$

$$\Rightarrow 34=16-6a$$

$$\Rightarrow a=-\dfrac{18}{6}=-3$$
Question 6
1 / -0

A(1, 2) and (3, -2) are the end points of $$\overline{AB}$$. Then the midpoints of $$\overline{AB}$$.

A
(-1, 0)

B
(2, 0)

C
(2, 1)

D
(0, 0)

Solution

Let $$P(x, y)$$ be the midpoint of the line segment joining $$A(1, 2)$$ and $$B(3, -2)$$.
$$\therefore x = \dfrac{1 + 3}{2} $$ and $$y = \dfrac{2- 2}{2}$$
$$\therefore x = \dfrac{4}{2}$$ and $$y = \dfrac{0}{2}$$
$$\therefore x = 2$$ and $$y = 0$$
$$\therefore P(x, y) = (2, 0)$$
Question 7
1 / -0

If $$A = (-3, 4), B = (-1, -2), C = (5, 6), D = (x, -4)$$ are vertices of a quadrilateral such that $$\Delta ABD = 2 \Delta ACD$$. Then $$x$$, is equal to:

A
$$6$$

B
$$9$$

C
$$69$$

D
$$96$$

Solution

Using $$ar\left( \Delta \right) =\cfrac { 1 }{ 2 } \left| { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( y_{ 1 }-{ y }_{ 2 } \right) \right| $$
$$ar\left( \Delta ABD \right) =\cfrac { 1 }{ 2 } \left| \left( -3 \right) \left( -2+4 \right) +\left( -1 \right) \left( -4-4 \right) +x\left( 4-(-2) \right) \right| $$
$$=\cfrac { 1 }{ 2 } \left| 2+6x \right| $$
$$=\left| 1+3x \right| \longrightarrow \left( i \right) $$
$$ar\left( \Delta ACD \right) =\cfrac { 1 }{ 2 } \left| \left( -3 \right) \left( 6+4 \right) +5\left( -4-4 \right) +x\left( 4-6 \right) \right| $$
$$=\cfrac { 1 }{ 2 } \left| -70-2x \right| $$
$$=\cfrac { 1 }{ 2 } \left| 70+2x \right| $$
$$=35+x\longrightarrow \left( ii \right) $$
$$\because \left| -1 \right| =1$$
So,
     $$ar\left( \Delta ABD \right) =2\Delta (ACD)$$
     $$1+3x=70+2x$$
     $$69=x$$
$$\therefore \quad x=69$$
Question 8
1 / -0

For $$A(1,2)$$ and $$B(3,-2)$$, the coordinates of the midpoint of $$AB$$ are is ..........

A
$$(2,2)$$

B
$$(0,0)$$

C
$$(2,0)$$

D
$$(0,2)$$

Solution

Given : $$A=(1,2)\equiv(x_1,y_1)$$ and $$B=(3,-2)\equiv(x_2,y_2)$$

Let $$M=(x,y)$$ be the midpoint of $$AB$$

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

Then, $$x=\dfrac{1+3}{2}$$ and $$y=\dfrac{2-2}{2}$$

$$\implies x=2$$ and $$y=0$$

$$\therefore M=(2,0)$$

$$\therefore$$ Coordinates of the midpoint of $$AB$$ is $$\left( 2,0 \right) $$
Question 9
1 / -0

If $$C$$ is a point on the line segment joining $$A(-3,4)$$ and $$B(2,1)$$ such that $$AC=2BC$$, then the coordinates of $$C$$ is

A
$$\left( \cfrac { 1 }{ 3 } ,2 \right) $$

B
$$\left( 2,\cfrac { 1 }{ 3 } \right) $$

C
$$(2,7)$$

D
$$(7,2)$$

Solution

Here we apply the section formula: If a point $$C(x,y)$$ divides the line segment $$AB$$ joining the points $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) \& B\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ in the ratio $$m:n$$ then we have

$$\\ x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } \& y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } .\\$$
Let the point is C( $$x,y$$), Which divides AB in the ratio of $$2:1$$.
Then the coordinates of C are given by
$$x= \dfrac{(2\times2)+(-3\times 1)}{2+1}=\dfrac{1}{3}$$
$$y=\dfrac{(2\times 1)+(1\times 4)}{3}=2$$
$$C \equiv\left (\dfrac{1}{3},2\right)$$
Question 10
1 / -0

Find the area (in square units) of the triangle whose vertices are $$(a, b+c), (a, b-c) $$ and $$(-a, c). $$

A
$$2ac$$

B
$$2bc$$

C
$$b(a+c)$$

D
$$c(a-h)$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times |[ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] |$$
Given the vertices of triangle,
$$A (a,b+c) ,B(a,b-c) ,C(-a,c)$$
Therefore, area is given by
$$=\dfrac { 1 }{ 2 } |\left[ a\left[ b-c-c \right] +a\left[ c-b-c \right] +(-a)\left[ b+c-b+c \right] \right]| \\ =\dfrac { 1 }{ 2 } |\left[ a(b-2c)+a(-b)-a(2c) \right]| \\ =\dfrac { 1 }{ 2 } |\left[ ab-2ac-ab-2ac \right]| =\left| \dfrac { -4ac }{ 2 } \right| =2ac$$

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Coordinate Geometry Test - 38

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Coordinate Geometry Test - 38

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Question 1

$$ (-2,3)$$

$$\left(-3,\cfrac{3}{2}\right)$$

$$\left(-3,\cfrac{-3}{2}\right)$$

None of these

Solution

Question 2

$$(2,1)$$

$$(0,0)$$

$$(\cfrac{5}{3},2)$$

$$(1,-2)$$

Solution

Question 3

$$1$$ sq.units

$$2$$ sq.units

$$4$$ sq.units

$$8$$ sq.units

Solution

Question 4

$$(-a,2b)$$

$$(2a,4b)$$

$$(2a,2b)$$

$$(-a,-3b)$$

Solution

Question 5

$$a=-2$$

$$a=-3$$

$$a=3$$

none of the above

Solution

Question 6

(-1, 0)

(2, 0)

(2, 1)

(0, 0)

Solution

Question 7

$$6$$

$$9$$

$$69$$

$$96$$

Solution

Question 8

$$(2,2)$$

$$(0,0)$$

$$(2,0)$$

$$(0,2)$$

Solution

Question 9

$$\left( \cfrac { 1 }{ 3 } ,2 \right) $$

$$\left( 2,\cfrac { 1 }{ 3 } \right) $$

$$(2,7)$$

$$(7,2)$$

Solution

Question 10

$$2ac$$

$$2bc$$

$$b(a+c)$$

$$c(a-h)$$

Solution

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