Coordinate Geometry Test

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Coordinate Geometry Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

If $$A$$ and $$B$$ are two points having co-ordinates $$(3,4)$$ and $$(5,-2)$$ respectively and $$P$$ is a point such that $$PA=PB$$ and area of triangle $$PAB=10$$ square units, then the co-ordinates of $$P$$ are

A
$$(7,4)$$ or $$(13,2)$$

B
$$(7,2)$$ or $$(1,0)$$

C
$$(2,7)$$ or $$(4,13)$$

D
$$None\ of\ these$$

Solution
Question 2
1 / -0

Length of the median from $$B$$ on $$AC$$ where $$A(-1, 3), B(1, -1), C(5, 1)$$ is

A
$$\sqrt{18}$$

B
$$\sqrt {10}$$

C
$$2\sqrt 3$$

D
$$4$$

Solution
Question 3
1 / -0

A line intersects the y-axis and x-axis at points P and Q respectively . If $$ (2 , -5) $$ is the mid-point of PQ , then co-ordinates of P and Q are respectively.

A
$$ (0 , -5) $$ and $$ (2 , 0) $$

B
$$ (0 , 10) $$ and $$ (-4 , 0) $$

C
$$ (0 , 4) $$ and $$ (-10 , 0) $$

D
$$ (0 , -10) $$ and $$(4 , 0) $$

Solution

$$P$$ lies on y-axis so co-ordinates of $$P$$ are $$(0, y)$$
Similarly, co-ordinates of $$Q$$ lies on x-axis $$= Q(x, 0)$$
Mid-point of $$PQ$$ is
$$M\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) = M(2,-5)$$ which is given

$$\Rightarrow M\left(\dfrac{0+x}{2} , \dfrac{y+0}{2}\right) = M(2,-5)$$

$$\Rightarrow \left(\dfrac{x}{2}, \dfrac{y}{2}\right) = M(2, -5)$$

Comparing both side, we get
$$\dfrac{x}{2} = 2$$ and $$\dfrac{y}{2} = -5$$

$$\Rightarrow x = 4$$ and $$y = -10$$

Hence, the co-ordinates of $$P(0, -10)$$ and $$Q(4, 0)$$ verifies ans (d)
Question 4
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

The perpendicular bisector of $$AB$$ will pass through the mid-point of $$AB$$ .
Mid-point of $$A$$ $$ (x_1 , y_1) $$ and $$B$$$$(x_2 , y_2) $$ is given by
$$ \left ( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \right ) $$ .

$$ = \left ( \dfrac{-2 + 2}{2} , \dfrac{-5 + 5}{2} \right ) = (0 , 0) $$
So , the perpendicular bisector passes through $$ (0 , 0) $$
Hence, $$(0,0)$$ lies on the perpendicular bisector.
Question 5
1 / -0

the area of a triangle with vertices $$ (-3, 0),(3,0) $$ and $$ (0, k) $$ is $$ 9 $$ sq units. then the value of $$ k $$ will be

A
$$ 9 $$

B
$$ 3 $$

C
$$ -9 $$

D
$$ 6 $$

Solution

We know that , area of a triangle with vertices $$ (a_1,y_1),(x_2,y_2) $$ and $$ (x_3,y_3) $$ is given by
$$ \Delta = \frac {1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| $$
$$ \therefore \Delta = \frac {1}{2} \left| \begin{matrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} \right| $$
Expanding along $$ R_1 $$
$$ 9 = \frac {1}{2} [ -3 ( -k) - 0 +1 ( 3k) ] $$
$$ \Rightarrow 18 = 3k + 3k = 6 k $$
$$ \therefore K = \frac {18 }{6} = 3 $$
Question 6
1 / -0

The coordinates of the point which divides the join of $$(x_1 , y_1)$$ and $$(x_{2} , y_{2})$$ in the ratio $$m_{1} : m_{2}$$ internally are

A
$$ \left [ \dfrac{m_{1} x_{2} + m_{2}x_{1}}{m_{1} + m_{2}} , \dfrac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \right ] $$

B
$$ \left [ \dfrac{m_{1} x_{2} + m_{2}x_{1}}{m_{1} - m_{2}} , \dfrac{m_{1} y_{2} - m_{2} y_{1}}{m_{1} - m_{2}} \right ] $$

C
$$ \left [ \dfrac{m_{1} x_{2} + m_{2}x_{1}}{m_{1} - m_{2}} , \dfrac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} - m_{2}} \right ] $$

D
$$ \left [ \dfrac{m_{1} x_{2} - m_{2}x_{1}}{m_{1} + m_{2}} , \dfrac{m_{1} y_{2} - m_{2} y_{1}}{m_{1} + m_{2}} \right ] $$

Solution

Let $$A\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ and $$B\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ are end points of segment AB.

Let point P divides segment AB internally in the ratio $$m:n$$

Thus, $$x$$ coordinate of point P is given by,
$$x=\frac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n } $$

Similarly, $$y$$ coordinate of point P is given by,
$$y=\frac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } $$

Thus, coordinates of point P are $$P\left( \frac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n },\frac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } \right) $$
Question 7
1 / -0

The coordinates of the midpoint of the line segment joining $$(-8, 13)$$ and $$(x, 7)$$ is $$(4, 10)$$ then what is the value of $$x$$

A
$$16$$

B
$$10$$

C
$$4$$

D
$$8$$

Solution

Given: $$A(-8,13),\;B(x,7)$$ are points.

Midpoint:- $$P(4,10)$$

By using mid-point formula to calculate the $$x$$-coordinate of the midpoint,
$$\begin{aligned}{}\frac{{ - 8 + x}}{2} &= 4\\ - 8 + x& = 8\\x& = 16\end{aligned}$$

Hence the value of $$x$$ is equal to $$16.$$
Question 8
1 / -0

If points $$ (1,2),(-1, x) $$ and (2,3) are collinear, then $$ x $$ will be:

A
2

B
0

C
-1

D
1

Solution

Let the points $$ A(1,2), B(1, x) $$ and $$ C(2,3) $$ are collinear then area of triangle made by these points will be zero.
$$ \Rightarrow \dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] $$
$$ \Rightarrow \dfrac{1}{2}[(x-3)+(-1)(3-2)+2(2-x)]=0 $$
$$ \Rightarrow x-3-1+2(2-x)=0 $$
$$ \Rightarrow x-3-1+4-2 x=0 $$
$$ \Rightarrow-x-4+4=0 $$
$$ \Rightarrow x=0 $$
So, correct choice is (B)
Question 9
1 / -0

P divides internally the line segment which joins the points (5, 0) and (0, 4) in the ratio of 2 : 3 internally. Co-ordinates of point P is:

A
$$ \left (3. \frac{8}{5} \right ) $$

B
$$ \left (2. \frac{8}{5} \right ) $$

C
$$ \left (\frac{5}{2} \frac{3}{5} \right ) $$

D
$$(2,4)$$

Solution

Let point P(x, y) divides the line segment joining the points A(5, 0) and B(0, 4) internally in the ratio 2 : 3.
$$ \mathbf{then} $$ $$ \mathbf{ x = \dfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}} $$
  $$ \mathbf{ = \dfrac{(2)(0) + (3)(5)}{2 + 3}} $$
  $$ \mathbf{ = \dfrac{15}{5}} $$
$$ \Rightarrow   \mathbf{ x = 3} $$
$$ \mathbf{and} $$ $$ \mathbf{ y = \dfrac{m_{1}y_{2} + m_{2}x_{1}}{m_{1} + m_{2}}} $$
  $$ \mathbf{ = \dfrac{(2)(4) + (3)(0)}{2 + 3}} $$
  $$ \mathbf{ = \dfrac{8}{5}} $$
Hence, co-ordinate of P is $$ \left (3. \dfrac{8}{5} \right ) $$
Hence, correct choice is (A)
Question 10
1 / -0

What will the co-ordinate of point A when $$ \mathrm{AB} $$ is a diameter of circle and coordinate of $$ \mathrm{B}(1,4) $$ and $$ \mathrm{co} $$ -ordinate of center $$ \mathrm{O} $$ is (2,-3)

A
(3,-5)

B
(3,-6)

C
(4,-10)

D
(3,-10)

Solution

Let co-ordinate of point A is (x, y)
By the mid point formula
$$ 2=\dfrac{x+1}{2} $$
$$ \Rightarrow x+1=4 $$
$$ \Rightarrow x=4-1=3 $$
and $$ -3=\dfrac{y+4}{2} $$
$$ \Rightarrow y+4=-6 $$
$$ y=-6-4=-10 $$
$$ \therefore $$ co-ordinate of point $$ \mathrm{A}(3,-10) $$
Hence correct choice is (D).

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Coordinate Geometry Test - 45

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Coordinate Geometry Test - 45

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Question 1

$$(7,4)$$ or $$(13,2)$$

$$(7,2)$$ or $$(1,0)$$

$$(2,7)$$ or $$(4,13)$$

$$None\ of\ these$$

Solution

Question 2

$$\sqrt{18}$$

$$\sqrt {10}$$

$$2\sqrt 3$$

$$4$$

Solution

Question 3

$$ (0 , -5) $$ and $$ (2 , 0) $$

$$ (0 , 10) $$ and $$ (-4 , 0) $$

$$ (0 , 4) $$ and $$ (-10 , 0) $$

$$ (0 , -10) $$ and $$(4 , 0) $$

Solution

Question 4

$$ (0 , 0) $$

$$ (0 , 2) $$

$$ (2 , 0) $$

$$ ( -2 , 0)$$

Solution

Question 5

$$ 9 $$

$$ 3 $$

$$ -9 $$

$$ 6 $$

Solution

Question 6

$$ \left [ \dfrac{m_{1} x_{2} + m_{2}x_{1}}{m_{1} + m_{2}} , \dfrac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \right ] $$

$$ \left [ \dfrac{m_{1} x_{2} + m_{2}x_{1}}{m_{1} - m_{2}} , \dfrac{m_{1} y_{2} - m_{2} y_{1}}{m_{1} - m_{2}} \right ] $$

$$ \left [ \dfrac{m_{1} x_{2} + m_{2}x_{1}}{m_{1} - m_{2}} , \dfrac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} - m_{2}} \right ] $$

$$ \left [ \dfrac{m_{1} x_{2} - m_{2}x_{1}}{m_{1} + m_{2}} , \dfrac{m_{1} y_{2} - m_{2} y_{1}}{m_{1} + m_{2}} \right ] $$

Solution

Question 7

$$16$$

$$10$$

$$4$$

$$8$$

Solution

Question 8

2

0

-1

1

Solution

Question 9

$$ \left (3. \frac{8}{5} \right ) $$

$$ \left (2. \frac{8}{5} \right ) $$

$$ \left (\frac{5}{2} \frac{3}{5} \right ) $$

$$(2,4)$$

Solution

Question 10

(3,-5)

(3,-6)

(4,-10)

(3,-10)

Solution

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