Coordinate Geometry Test

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Coordinate Geometry Test - 37

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Question 1
1 / -0

The vertices of a triangle ABC are $$(\lambda, 2 - \lambda), (- \lambda + 1, 2 \lambda) $$ and $$(-4 - \lambda, 6 - 2 \lambda)$$. If its area is 70 units$$^2$$, find the number of integral values of $$\lambda$$.

A
$$1$$

B
$$2$$

C
$$4$$

D
$$10$$

Solution

Given the vertices of $$\triangle$$ABC
$$A(\lambda ,2-\lambda ),B(-\lambda +1,2\lambda ),C(-4-\lambda ,6-2\lambda )$$
Given total area $$=70\ unit^{ 2 }$$
Area
$$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] \\ \Rightarrow \dfrac { 1 }{ 2 } \left[ \left[ 2\lambda -6+2\lambda \right] +(1-\lambda )\left[ 6-2\lambda -2+\lambda \right] -(4+\lambda )\left[ 2-\lambda -2\lambda \right] \right] =70\\ \lambda (4\lambda -6)+(1-\lambda )(4-\lambda )-(4+\lambda )(2-3\lambda )=2\times 70\\ 4{ \lambda }^{ 2 }-6\lambda +4-x-4\lambda +{ \lambda }^{ 2 }-\left[ 8-12\lambda +2\lambda -3{ \lambda }^{ 2 } \right] =140\\ 5{ \lambda }^{ 2 }-10\lambda +4-8+12\lambda -2\lambda +3{ \lambda }^{ 2 }=140\\ 8{ \lambda }^{ 2 }+\lambda -4-2\lambda =140\\ \Longrightarrow 8{ \lambda }^{ 2 }-\lambda =136\\ \lambda =4.186,-4.061$$
$$\therefore$$ No.of Intergral value of $$\lambda =1\left[ i.e,\lambda =4.186 \right] $$
Question 2
1 / -0

In the diagram, $$KN$$ is a straight line. $$L$$ and $$M$$ are two points on $$KN$$ such that $$KL=LM$$ and $$KM=MN$$. Find the coordinates of $$L$$.

A
$$(3,2)$$

B
$$(3,4)$$

C
$$(4,4)$$

D
$$(4,2)$$

Solution

As $$KM=MN\Rightarrow 'M'$$ is mid point
$$\Rightarrow M=\left(\dfrac{12+0}{2},\dfrac{10+2}{2}\right)=(6,6)$$
$$KL=LM\Rightarrow 'L'$$ is mid point
$$\Rightarrow L=\left(\dfrac{6+0}{2},\dfrac{6+2}{2}\right)=(3,4)$$
Question 3
1 / -0

$$A(-3,2)$$ and $$B(5,4)$$ are the end points of a line segment, find the coordinates of the midpoints of the line segment.

A
$$(1,3)$$

B
$$(3,3)$$

C
$$(1,1)$$

D
$$(3,1)$$

Solution

Since $$A(-3,2)=(x_1,y_1)$$ and $$B(5,4)=(x_2,y_2)$$ are the end points of a line segment.

The coordinates of the midpoints of the line segment is given by:

$$(x,y)=\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right)$$

$$ \\ \Rightarrow (x,y)=\left( \dfrac { -3+5 }{ 2 } ,\dfrac { 2+4 }{ 2 } \right) \quad $$

$$\\ \Rightarrow (x,y)=\left( \dfrac { 2 }{ 2 } ,\dfrac { 6 }{ 2 } \right) =\left( 1,3 \right)$$
Question 4
1 / -0

The area of a triangle is 5 and its two vertices are A(2, 1) and B(3, -2). The third vertex lies on $$\displaystyle y=x+3$$. What is the third vertex?

A
$$\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right) $$

B
$$\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) $$

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

D
$$\displaystyle (0,0)$$

Solution
Question 5
1 / -0

In the $$xy$$-plane, the vertices of a triangle are $$(-1,3), (6,3)$$ and $$(-1,-4)$$. The area of the triangle is ___ square units.

A
$$10$$

B
$$17.5$$

C
$$24.5$$

D
$$35$$

Solution

If $$(x_1,y_1)$$,$$(x_2,y_2)$$ and $$(x_3,y_3)$$ are the vertices of a triangle then its area is given by,
$$A=\left| \dfrac { 1 }{ 2 } (x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 })) \right| $$
Therefore, with the vertices $$(-1,3)$$, $$(6,3)$$ and $$(-1,-4)$$.
Area of triangle is given by,
$$A=\left| \dfrac { 1 }{ 2 } (-1(3-(-4))+6(-4-3)+(-1)(3-3)) \right|$$
$$ =\left| \dfrac { 1 }{ 2 } (-7-42) \right| $$
$$=\left| \dfrac { 1 }{ 2 } (-49) \right| $$
$$=\left| -24.5 \right|$$
$$ =24.5$$ square units.
Question 6
1 / -0

The vertices of triangle $$PQR$$ are $$P(-3, 2), Q(1, -4)$$, and $$R(7, 0)$$. The altitude drawn from Q intersects the line $$PR$$ at the point

A
$$(1, 2)$$

B
$$(2, 1)$$

C
$$(1, -2)$$

D
$$(-3, 2)$$

E
$$(7, 0)$$

Solution
Question 7
1 / -0

Two vertices of a triangle are (2, 1) and (3, -2). Its third vertex is (x, y) such that $$\displaystyle y=x+3$$. If its area is 5 sq. units, what are the co-ordinates of the third vertex?

A
(3.5, -6.5)

B
(3.5, 6.5)

C
(-1.5, -1.5)

D
(1.5, -1.5)

Solution

Given third vertex such that,
$$y=(x+3)$$
since, Area of triangle $$ABC=55q$$ units
$$\pm \dfrac{1}{2}{x(1+2)+2(-2-y)+3(y-1)}=\xi$$
$$\Rightarrow \pm \dfrac{1}{2}{x+2x-4-2y+3y-3}=\xi$$
$$\Rightarrow {3x+y-7}=\pm10$$
$$\Rightarrow 3x+y-17=0$$ ------------(1)
and $$3x+y+3=0$$ ------------(2)
Given that $$A(x,y)$$ lies any $$=(x+3)$$ ------------(3)
from equation (L) and (3),
$$x=\dfrac{7}{2},y=\dfrac{13}{2}$$
$$\Rightarrow \boxed{x=3.\xi\,and \, y=6.\xi}$$
from equation (2) and (3) we get,
$$x=\dfrac{-3}{2}\,and\, y=1.\xi $$
Question 8
1 / -0

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

A
$$1$$ sq. unit

B
$$2$$ sq. units

C
$$4$$ sq. units

D
$$8$$ sq. units

Solution

Given: $$A=(x_1,y_1)=(0,0)$$

$$B=(x_2,y_2)=(2,0)$$ and

  $$C=(x_3,y_3)=(0,2)$$

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

$$=\dfrac {1}{2}[0(0-2)+2(2-0)+0(0-0)]$$

         $$=\dfrac {1}{2}[2(2)]$$

         $$=2$$ sq. units.
Question 9
1 / -0

Find the midpoint of the line segment joining the points $$(1,-1)$$ and $$(-5,-3)$$.

A
$$(-2,1)$$

B
$$(2,1)$$

C
$$(-2,-1)$$

D
None of these

Solution

Given points are $$(1,-1)$$ and $$(-5,-3)$$

To find out,
The coordinates of the midpoint of the line joining the given points.

Let the coordinates of the midpoint be $$(x,y)$$.
We know that the coordinates of the midpoint of the line segment joining $$(x_1,y_1) \ and\ (x_2,y_2)$$ are:

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

Here, $$(x_1,y_1)=(1,-1)$$ and $$(x_2,y_2)=(-5,-3)$$.

Hence, $$x=\dfrac{1+(-5)}{2}$$ and $$y=\dfrac{-1-(-3)}{2}$$.

$$\Rightarrow x=-2$$ and $$y=1$$

So, $$(x,y)=(-2,1)$$.

Hence, $$(-2,1)$$ are the coordinates of the midpoint joining $$(1,-1)$$ and $$(-5,-3)$$.
Question 10
1 / -0

The centre of a circle is at $$(-6,4)$$. If one end of a diameter of the circle is at the origin, then find the other end.

A
$$(-12,8)$$

B
$$(-12,-8)$$

C
$$(12,8)$$

D
None of these

Solution

Take coordinate of other end of diameter as $$P(x,y)$$.

since, origin divides the diameter of a circle in $$2$$ equal parts, hence origin will be mid-point of $$(0,0)$$ and $$(x,y)$$.

By midpoint theorem:

$$x=\dfrac{x_1+x_2}{2}$$ and $$y=\dfrac{y_1+y_2}{2}$$

$$\implies -6=\dfrac{0+x}{2}=\dfrac{x}{2}$$

$$\implies x=-12$$

And,

$$\implies 4=\dfrac{y+0}{2}=\dfrac{y}{2}$$

$$\implies y=8$$.

So, $$P(x,y)=(-12,8)$$

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Coordinate Geometry Test - 37

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Coordinate Geometry Test - 37

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Question 1

$$1$$

$$2$$

$$4$$

$$10$$

Solution

Question 2

$$(3,2)$$

$$(3,4)$$

$$(4,4)$$

$$(4,2)$$

Solution

Question 3

$$(1,3)$$

$$(3,3)$$

$$(1,1)$$

$$(3,1)$$

Solution

Question 4

$$\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right) $$

$$\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) $$

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

$$\displaystyle (0,0)$$

Solution

Question 5

$$10$$

$$17.5$$

$$24.5$$

$$35$$

Solution

Question 6

$$(1, 2)$$

$$(2, 1)$$

$$(1, -2)$$

$$(-3, 2)$$

$$(7, 0)$$

Solution

Question 7

(3.5, -6.5)

(3.5, 6.5)

(-1.5, -1.5)

(1.5, -1.5)

Solution

Question 8

$$1$$ sq. unit

$$2$$ sq. units

$$4$$ sq. units

$$8$$ sq. units

Solution

Question 9

$$(-2,1)$$

$$(2,1)$$

$$(-2,-1)$$

None of these

Solution

Question 10

$$(-12,8)$$

$$(-12,-8)$$

$$(12,8)$$

None of these

Solution

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