Coordinate Geometry Test

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Coordinate Geometry Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The area of the triangle formed by the points $$(2, 6), (10, 0)$$ and $$(0, k)$$ is zero square units. Find the value of $$k.$$

A
$$\dfrac {15}{2}$$

B
$$\dfrac 32$$

C
$$\dfrac 72$$

D
$$\dfrac {13}{2}$$

Solution

The area of the triangle is:

$$A=\dfrac { 1 }{ 2 } \left| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \right|$$

We are given that the area of the triangle is $$0$$ and the points are $$(2,6)$$, $$(10,0)$$ and $$(0,k)$$, therefore,

$$\Rightarrow 0=\dfrac { 1 }{ 2 } \left| 6(10-0)+0(0-2)+k(2-10) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-0-8k \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-8k \right| $$

$$\Rightarrow 60-8k=0\\ \\ \Rightarrow 8k=60\\ \Rightarrow k=\dfrac { 60 }{ 8 } =\dfrac { 15 }{ 2 }$$

Hence, $$k=\dfrac { 15 }{ 2 }$$.
Question 2
1 / -0

Area of a triangle whose vertices are (0, 0), (2, 3) , (5, 8) is _______ square units.

A
1/2

B
1

C
2

D
3/2

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y }_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (0,0) $$ ; $$({ x }_{ 2 },{ y }_{ 2 }) = (2,3) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (5,8)$$ in the area formula, we get

Area of triangle ABC $$ = \left| \frac { (0)(3-8)+(2)(8-0)+5(0-3) }{ 2 } \right| = \left| \frac { 16 -15 }{ 2 } \right| = \frac {1}{2} units $$
Question 3
1 / -0

Determine the coordinates of the point which is three-fifth of the line joining the points (2, -5) and (-3, 5).

A
(-1, 1)

B
(-2, -1)

C
(-1, -2)

D
(1, -1)

E
None

Solution

Let $$A(2,-5),\ B(-3,5) \text{ and } P$$ be three points
Let the coordinates of the point $$P$$ be $$(x,y)$$
Given that, $$AP=\dfrac{3}{5}\times AB$$
$$\therefore \ \dfrac{AP}{AB}=\dfrac{3}{5}$$

$$\Rightarrow \dfrac{AB}{AP}=\dfrac{5}{3}$$

$$\Rightarrow \dfrac{AP+BP}{AP}-1=\dfrac{5}{3}-1$$

$$\Rightarrow \dfrac{BP}{AP}=\dfrac{2}{3}$$

$$\therefore \ \dfrac{AP}{BP}=\dfrac{3}{2}$$

Hence, point $$P$$ divides the line segment $$AB$$ in the ratio $$3:2$$

We know that, if a point $$P(x,y)$$ 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 by the section formula, we have:

$$\\ x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } \ \& \ y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } .\\$$

$$\therefore$$ $$(x,y)=\left( \dfrac { 2\times 2+3\times (-3) }{ 3+2 } ,\dfrac { 2\times (-5)+3\times 5 }{ 3+2 } \right) $$

$$\Rightarrow \left (\dfrac{4-9}{5},\dfrac{-10+15}{5}\right )$$

$$\Rightarrow (\dfrac{-5}{5},\dfrac{5}{5})$$

$$\therefore \ (x,y)=(-1,1)$$

Hence, the coordinates of the point which is three-fifth of the line joining the points $$(2,-5)$$ and $$(-3,5)$$ are $$(-1,1)$$.
Question 4
1 / -0

M is the midpoint of $$\displaystyle \overline{AB}$$ The coordinates of A are (-2,3) and the coordinates of M are (1,0) Find the coordinates of B

A
$$(4,-3)$$

B
$$(-3,4)$$

C
$$(-3,5)$$

D
None of the above

Solution
Question 5
1 / -0

If (-2, -4) is the midpoint of (6, -7) and (x, y) then the values of x and y are

A
x = 2, y = 1

B
x = -10, y = -1

C
x = 10, y = -1

D
x = -8 , y = -1

E
none of these

Solution

Since, $$(-2, -4)$$ is the midpoint of $$(6, -7)$$ and $$(x,y).$$
$$\Rightarrow \dfrac{x+6}2=-2$$$$\Rightarrow x=-10$$
and $$ \dfrac{y-7}2=-4$$$$\Rightarrow y=-1$$
Option D is correct.
Question 6
1 / -0

Find the midpoint of the segment joining the points $$(4, -2)$$ and $$(-8,6)$$.

A
$$(6, 4)$$

B
$$(-6, -4)$$

C
$$(2, 2)$$

D
$$(-2, 2)$$

Solution

It is known that if $$(x,y)$$ is the mid-point of the segment joining $$(x_1,y_1)$$ and $$(x_2,y_2)$$ then,
$$x=\dfrac{x_1+x_2}{2}$$ and $$y=\dfrac{y_1+y_2}{2}$$.
The mid point of segment joining the points $$(4,-2)$$ and $$(-8,6) $$ is
$$\left(\dfrac{4-8}2,\dfrac{-2+6}2\right)=(-2,2)$$
Question 7
1 / -0

Mark planted two trees on a planning grid at coordinates $$(0,8)$$ and $$(12,4)$$. Determine the midpoint of the line segment connecting the two trees

A
$$(4,5)$$

B
$$(6,6)$$

C
$$(6,4)$$

D
None of these

Solution
Question 8
1 / -0

Find the length of the median through $$(-2, -5)$$ of the triangle whose vertices are $$(-6, 2),\; (2, -2),$$ and $$(-2, -5).$$

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Let $$A (-6, -2) \, B(2, -2) $$ and $$c(-2, -5)$$ be the vertices of $$\triangle ABC$$.

We have to find the median passing through $$(-2,-5)$$ so, it will also pass through the midpoint of $$AB$$ as we are concerned with the median.

$$\text{midpoint}(AB)= \left(\dfrac{-6 + 2}{2} , \dfrac{2 - 2}{2} \right) $$
$$= (-2, 0)$$

So, from the above diagram we can say that length of $$PC$$ is the required median length,
$$\begin{aligned}{}PC &= \sqrt {{{( - 2 - ( - 2))}^2} + {{( - 5 - 0)}^2}} \\ &= \sqrt {0 + 25} \\ &= 5{\text{ units}}\end{aligned}$$
Question 9
1 / -0

The area of triangle with vertices $$A(5,\,0),\,B(8,\,0)$$ and $$C(9,\,5)$$ is

A
$$7\dfrac{1}{4}$$

B
$$7\dfrac{1}{2}$$

C
$$\dfrac{15}{4}$$

D
None of the above

Solution

Area of triangle $$=\dfrac{1}{2}\begin{bmatrix}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\end{bmatrix}$$
$$A=\dfrac{1}{2}\begin{bmatrix}5(0-5)+8(5-0)+9(0-0)\end{bmatrix}$$
$$A=\dfrac{1}{2}\begin{bmatrix}-25+40\end{bmatrix}=\dfrac{15}{2}$$ square units.
So, option B is correct.
Question 10
1 / -0

The area of the triangle formed from points $$(1, 2), (2, 4)$$ and $$(3, 1)$$ is ____ square units.

A
$$\dfrac{5}{2}$$

B
$$\dfrac{2}{5}$$

C
$$\dfrac{3}{2}$$

D
$$2$$

Solution

Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 1$$, $$y_{1} = 2$$, $$x_{2} = 2$$, $$y_{2} = 4$$, $$x_{3} = 3$$ and $$y_{3} = 1$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[1(4 - 1) + 2(1 - 2) + 3(2 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[3 - 2 - 6]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times -5\right|$$
$$=$$ $$\left|-\frac{5}{2} \right|$$
Area always in absolute value.
So, area of the triangle $$= \dfrac{5}{2}$$ square units.

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Re-Attempt Weekly Quiz Competition

Coordinate Geometry Test - 32

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Coordinate Geometry Test - 32

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SHARING IS CARING

Question 1

$$\dfrac {15}{2}$$

$$\dfrac 32$$

$$\dfrac 72$$

$$\dfrac {13}{2}$$

Solution

Question 2

1/2

1

2

3/2

Solution

Question 3

(-1, 1)

(-2, -1)

(-1, -2)

(1, -1)

None

Solution

Question 4

$$(4,-3)$$

$$(-3,4)$$

$$(-3,5)$$

None of the above

Solution

Question 5

x = 2, y = 1

x = -10, y = -1

x = 10, y = -1

x = -8 , y = -1

none of these

Solution

Question 6

$$(6, 4)$$

$$(-6, -4)$$

$$(2, 2)$$

$$(-2, 2)$$

Solution

Question 7

$$(4,5)$$

$$(6,6)$$

$$(6,4)$$

None of these

Solution

Question 8

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 9

$$7\dfrac{1}{4}$$

$$7\dfrac{1}{2}$$

$$\dfrac{15}{4}$$

None of the above

Solution

Question 10

$$\dfrac{5}{2}$$

$$\dfrac{2}{5}$$

$$\dfrac{3}{2}$$

$$2$$

Solution

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