Coordinate Geometry Test

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Coordinate Geometry Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

In the standard $$(x,y)$$ coordinate plane, what are the coordinates of the midpoint of a line segment whose endpoints are $$(-3,0)$$ amd $$(7,4)$$?

A
$$(2,2)$$

B
$$(2,3)$$

C
$$(5,2)$$

D
$$(5,4)$$

Solution

Given two points $$(x_{1},y_{1})$$ and $$(x_{2},y_{2})$$, then the coordinates of the midpoint are $$\left (\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\right)$$.
In the above case the points are $$(-3,0)$$ and $$(7,4)$$.
Hence, the midpoint is $$\left (\dfrac{-3+7}{2},\dfrac{0+4}{2}\right)=(2,2)$$.
Question 2
1 / -0

In figure, if the midpoints of segments $$\overline{GH}, \overline{JK}$$, and $$\overline{LM}$$ are connected, calculate the area of the resulting triangle.

A
$$20$$

B
$$23$$

C
$$26$$

D
$$33$$

Solution

Midpoint of $$GH$$ is $$\left (0, \dfrac {15}{4}\right)$$
Midpoint of $$JK$$ is $$(-5,-2)$$
Midpoint of $$ML$$ is $$(3,-2)$$
If the coordinates of triangle is $$(a,b) , (c,d) , (e,f)$$ then the area formed by the coordinates of triangle is $$\dfrac{1}{2} \times |a(d-f) + c(f-b) + e(b-d)|$$
Therefore, the area enclosed by those midpoints will be
$$ \dfrac12 \times \left |0(-2+2) -5\left (-2-\dfrac {15}{4}\right) +3\left (\dfrac {15}{4}+2\right)\right| $$
$$= \dfrac {1}{2} \times |0+46| $$
$$= 23$$
Question 3
1 / -0

Calculate the area of a triangle with vertices $$(1, 1), (3, 1)$$ and $$(5, 7)$$.

A
$$6$$

B
$$7$$

C
$$9$$

D
$$10$$

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Area of triangle $$=\dfrac 12 \left|1(21-5)-1(7-5)+1(1-3) \right|$$
$$=\dfrac 12 \left| 16-2-2\right| = 6$$
Hence, area of the triangle with the given coordinates is $$6$$.
Question 4
1 / -0

Points $$A(\sqrt {2}, 4), B(6, -\sqrt {3})$$ and $$C$$ are collinear. If $$B$$ is the midpoint of line segment $$AC$$, approximately calculate the $$(x, y)$$ coordinates of point $$C$$.

A
$$(3.71, 1.13)$$

B
$$(3.71, 5.73)$$

C
$$(7.41, -7.46)$$

D
$$(10.59, -7.46)$$

Solution

Given points $$A$$ $$(\sqrt{2},4)$$, $$B$$ $$(6,-\sqrt{3})$$ and $$C$$ are col-linear if $$B$$ the mid point $$AC$$.
if (A) $$(3.71,1.13)$$ is point $$C$$, then point $$(x,y) =$$ $$\dfrac{\sqrt{2}+3.71}{2},\dfrac{4+1.13}{2}$$
$$\Rightarrow 2.56,2.56$$ this not point $$B$$
if (B) $$(3.71,5.73)$$ is point $$C$$, then point $$(x,y) =$$ $$\dfrac{\sqrt{2}+3.71}{2},\dfrac{4+5.73}{2}$$
$$\Rightarrow 2.56,4.86$$ this not point $$B$$
if (C) $$(7.41,-7.46)$$ is point $$C$$, then point $$(x,y) =$$ $$\dfrac{\sqrt{2}+7.41}{2},\dfrac{4-7.46}{2}$$
$$\Rightarrow 4.41,-1.73$$ this not point $$B$$
if (D) $$(10.59,-7.46)$$ is point $$C$$, then point $$(x,y) =$$ $$\dfrac{\sqrt{2}+10.59}{2},\dfrac{4-7.46}{2}$$
$$\Rightarrow 6,-1.73$$ this the point $$B (6,-1.73)$$

So (D) $$(10.59,-7.46)$$ is mid point $$B$$ of line $$AC$$.
Question 5
1 / -0

Given point $$A(-3, -8)$$, if the midpoint of segment $$AB$$ is $$(1, -5)$$, calculate the coordinates of point $$B$$.

A
$$(5, -2)$$

B
$$(4, -2)$$

C
$$(-1, -6.5)$$

D
$$(-2, -2)$$

Solution

Given, coordinates $$A (-3,-8)$$ and the mid point of $$AB$$ is $$(1,-5)$$
As per Midpoint formula coordinates of mid point $$=$$ $$\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}$$
Let the coordinates of point $$B$$ be $$(x,y)$$
Then $$(1,-5)=\left [ \left (\dfrac{-3+x}{2}\right),\left (\dfrac{-8+y}{2}\right) \right ]$$
Then $$\dfrac{-3+x}{2}=1$$
$$\Rightarrow -3+x=2$$
$$\Rightarrow x=2+3=5$$
Then$$\dfrac{-8+y}{2}=-5$$
$$\Rightarrow -8+y=-10$$
$$\Rightarrow y=-10+8=-2$$
Then coordinates of $$B$$ is $$(5,-2)$$.
Question 6
1 / -0

The area of the triangle with coordinates $$(1, 2), (5, 5)$$ and $$(k, 2)$$ is $$15$$ square units. Calculate a possible value for $$k$$.

A
$$-10$$

B
$$-9$$

C
$$-5$$

D
$$5$$

E
$$6$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
Area of triangle is $$15 = \dfrac {1}{2} \times |1(5-2) + 5(2-2) + k(2-5)| $$
$$ \Rightarrow \dfrac {1}{2} \times |3+0-3k| = \dfrac {1}{2} \times |3-3k| = 15$$
$$\Rightarrow |3-3k|=30$$ => $$|1-k| = 10$$
$$\Rightarrow 1-k = 10$$ and $$1-k=-10$$
$$\Rightarrow k=-9$$ and $$k=11$$
Question 7
1 / -0

Find the area of a triangle whose vertices are $$(0, 6\sqrt {3}), (\sqrt {35}, 7)$$, and $$(0, 3)$$.

A
$$15.37$$

B
$$17.75$$

C
$$21.87$$

D
$$25.61$$

E
$$39.61$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle whose vertices are $$(0,6\sqrt{3}),(\sqrt{35},7) ,(0,3)$$
$$=\dfrac{1}{2}\left [ 0(7-3)+\sqrt{35}(3-6\sqrt{3})+0(6\sqrt{3}-7)) \right ]$$
$$=\dfrac{1}{2}\left [ \sqrt{35}(3-6\sqrt{3}) \right ]$$
$$=\dfrac{1}{2}\left [ 5.91(3-6\times 1.73) \right ]$$
$$=\dfrac{1}{2}\times 5.91\times 7.40$$
$$=\dfrac{1}{2}\times 43.74$$
$$=21.87$$
Question 8
1 / -0

A square is formed by the points $$(4, 5), (12, 5), (12, -3)$$ and $$(4, -3)$$. Find the coordinates of the point at which the diagonals of the square intersect.

A
$$(8, 5)$$

B
$$(9, 6)$$

C
$$(8, 1)$$

D
$$(12, 1)$$

Solution

We know that the diagonal of square is intersect equal at mid point.
Then mid point of diagonal $$(4,5)$$ and $$(12,-3)$$ is $$\left ( \dfrac{12+4}{2} \right ),\left ( \dfrac{5-3}{2} \right )=\left ( \dfrac{16}{2} \right ),\left ( \dfrac{5-3}{2} \right )= (8,1)$$
Then mid point of diagonal $$(12,5)$$ and $$(14,-3)$$ is $$\left ( \dfrac{12+4}{2} \right ),\left ( \dfrac{5-3}{2} \right )=\left ( \dfrac{16}{2} \right ),\left ( \dfrac{5-3}{2} \right )= (8,1)$$
Then diagonal intersect at point $$(8,1)$$.
Question 9
1 / -0

If the coordinates of the one end of a diameter of a circle are $$(2,3)$$ and the coordinates of its centre are $$(-2,5),$$ then the coordinates of the other end of the diameter are:

A
$$(-6,7)$$

B
$$( 6,-7)$$

C
$$(6,7)$$

D
$$(-6,-7)$$

Solution

Let the coordinates of the one end of a diameter $$AB$$ of a circle are $$A(2,3)$$ and $$B(x,y) .$$
Let the coordinates of the center of the circle $$C(-2,5).$$
As center is the midpoint of the diameter $$AB$$ so the coordinates of center is
$$C(\dfrac{2+x}{2},\dfrac{3+y}{2})$$
Now equating the coordinates of center we get
$$ \dfrac{2+x}{2}=-2$$
$$\Rightarrow 2+x=-4$$
$$\Rightarrow x=-4-2$$
$$\Rightarrow x=-6$$
$$\dfrac{3+y}{2}=5$$
$$\Rightarrow 3+y=10$$
$$\Rightarrow y=10-3$$
$$\Rightarrow y=7$$
$$\therefore$$Coordinates of $$B=(-6,7).$$
Question 10
1 / -0

In fig., the area of triangle ABC (in sq. units) is:

A
$$15$$

B
$$10$$

C
$$7.5$$

D
$$2.5$$

Solution

Given: Coordinates of Point $$A (1,3) ,B (-1,0)$$ and $$C (4,0)$$
Construction: Drop a perpendicular from $$A$$ on $$x-$$ axis, which meets x-axis at $$D\equiv(1,0)$$
Now in $$\Delta ADC, AD = 3, DC = 3$$
Area of $$\Delta ADC = \dfrac12\times DC\times AD$$
$$= \dfrac12\times3\times3 = \dfrac92 \ cm^2$$

Now in $$\Delta ADB, AD = 3, DB = 2$$
Area of $$\Delta ADB = \dfrac12\times DB\times AD$$
$$= \dfrac12\times2\times3 = 3 \ cm^2$$

Area of $$\Delta ABC =$$ Ara of $$\Delta ADC + $$ Area of $$\Delta ABD$$
$$ = \dfrac92 + 3 = \dfrac{15}2 = 7.5\ cm^2$$

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Coordinate Geometry Test - 35

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Coordinate Geometry Test - 35

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SHARING IS CARING

Question 1

$$(2,2)$$

$$(2,3)$$

$$(5,2)$$

$$(5,4)$$

Solution

Question 2

$$20$$

$$23$$

$$26$$

$$33$$

Solution

Question 3

$$6$$

$$7$$

$$9$$

$$10$$

Solution

Question 4

$$(3.71, 1.13)$$

$$(3.71, 5.73)$$

$$(7.41, -7.46)$$

$$(10.59, -7.46)$$

Solution

Question 5

$$(5, -2)$$

$$(4, -2)$$

$$(-1, -6.5)$$

$$(-2, -2)$$

Solution

Question 6

$$-10$$

$$-9$$

$$-5$$

$$5$$

$$6$$

Solution

Question 7

$$15.37$$

$$17.75$$

$$21.87$$

$$25.61$$

$$39.61$$

Solution

Question 8

$$(8, 5)$$

$$(9, 6)$$

$$(8, 1)$$

$$(12, 1)$$

Solution

Question 9

$$(-6,7)$$

$$( 6,-7)$$

$$(6,7)$$

$$(-6,-7)$$

Solution

Question 10

$$15$$

$$10$$

$$7.5$$

$$2.5$$

Solution

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