Straight Lines Test 18

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Straight Lines Test 18

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Question 1
1 / -0

In the diagram $$MN$$ is a straight line on a Cartesian plane. The coordinates of $$N$$ are $$(12,13)$$ and $${MN}^{2}=9$$ units. The coordinates of $$M$$ are:

A
$$(21,13)$$

B
$$(12,22)$$

C
$$(12,4)$$

D
$$(3,13)$$

Solution

(As MN is parallel to 'x' axis \therefore 'y' coordinate not change)
Let $$M=(x,13)$$
$$\sqrt[6]{(x-12)^{2}+(13-13)^{2}}=9$$
$$\Rightarrow (x-12)^{2}=9^{2}\Rightarrow x+12=\pm 9$$
$$\Rightarrow x=3,21$$ (As 'M' is left of N )
$$\therefore x=3$$
Question 2
1 / -0

In the figure, $$AC = 9, BC = 3$$ and $$D$$ is $$3$$ times as far from $$A$$ as from $$B$$. What is $$BD$$?

A
6

B
9

C
12

D
15

E
18

Solution

Let $$BD=x$$
It is given that $$AC=9$$, $$BC=3$$ and $$D$$ is $$3$$ times as far from $$A$$ as from $$B$$, therefore $$AD=3BD$$ which means:
$$AD=3BD$$
$$\Rightarrow 9+3+x=3x$$
$$\Rightarrow 12+x=3x$$
$$\Rightarrow 3x-x=12$$
$$\Rightarrow 2x=12$$
$$\Rightarrow x=6$$
Hence, $$BD=6$$
Question 3
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 4
1 / -0

If the area of the triangle shown is $$30$$ square units, what is the value of $$y$$?

A
$$5$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution

Area of triangle $$=\dfrac 12 \left| \begin {matrix} 1 & 1 & 1 \\ 0 & 0 & 5 \\ y & 0 & 0 \end {matrix}\right|$$
$$\Rightarrow 30=\dfrac 12 \left| 1(0-0) - 1(5y-0) +1(0-0)\right|$$
$$\Rightarrow 60 = |-5y|$$
$$\Rightarrow 60 = 5y$$
$$\Rightarrow y =12$$
Question 5
1 / -0

If the distance between the two points $$P(a,3)$$ and $$Q(4,6)$$ is $$5$$, then find $$a$$.

A
$$0$$

B
$$-4$$

C
$$8$$

D
$$-4$$ and $$0$$

E
$$0$$ and $$8$$

Solution

The distance between points $$(a,3)$$ and $$(4,6)$$ is $$\sqrt { { (a-4) }^{ 2 }+{ (3-6) }^{ 2 } } =\sqrt { { (a-4) }^{ 2 }+9 } $$
Given that the distance is $$5$$
So, we have $$\sqrt { { (a-4) }^{ 2 }+9 } =5$$ $$\Rightarrow {(a-4)}^{2}+9=25$$
$$\therefore a-4 = \sqrt{16} = |4|$$
So, we get $$a = 4+|4| = 0 , 8$$
Question 6
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 7
1 / -0

If figure $$\Box ABCD$$ is a parallelogram, what is the x-coordinate of point B?

A
$$2$$

B
$$5$$

C
$$6$$

D
$$8$$

Solution

Looking at the figure, we can say that the $$y-$$ coordinate of point $$B$$ is $$5$$.
Hence, the coordinates of $$B$$ are $$(x,5)$$.
Since, $$\Box ABCD$$ is a paralleogram,
$$AB = CD$$ ...Opposite sides of parallelogram are congruent
By distance formula,
$$(6-2)^2+(-3+3)^2 = (x-4)^2+(5-5)^2$$
$$\Rightarrow 4^2 = x^2-8x+4^2$$
$$\Rightarrow x^2-8x=0$$
$$x(x-8) = 0$$
$$x=0$$ and $$x=8$$
Now, $$x \neq 0$$ since point B is in the 1st quadrant.
Hence, the $$x-$$ coordinate of B is $$8$$.
Question 8
1 / -0

In the diagram, $$PQR$$, is an isosceles triangle and $$QR=5$$ units.
The coordinates of $$Q$$ are:

A
$$(4,5)$$

B
$$(3,4)$$

C
$$(2,4)$$

D
$$(1,4)$$

Solution

As $$PQ+QR$$
Let coordinates of $$'Q'$$ be $$(x,y)$$
$$\sqrt{(x+2)^2+y^2}=\sqrt{(x-4)^2+y^2}$$
Squaring both sides
$$(x+2)^2+y^2=(x-4)^2+y^2$$
$$\Rightarrow x^2+4+4x=x^2+16-8x$$
$$\Rightarrow 12x=12\Rightarrow \boxed{x=1}$$
As $$QR=5$$ units
$$\Rightarrow \sqrt{(x-4)^2+y^2=5}$$
$$\Rightarrow (1-4)^2+y^2=25$$
$$\Rightarrow y^2=16\Rightarrow y=\pm 4$$
As $$Q$$ is above x-axis
So $$y=4$$
$$Q=(1,4)$$
Question 9
1 / -0

In the figure, there is a regular hexagon with sides of length $$6$$. If the coordinate of $$A$$ is $$(9,0)$$, what is the y-coordinate of $$B$$?

A
$$0$$

B
$$3$$

C
$$3\sqrt{2}$$

D
$$3\sqrt{3}$$

Solution

Let the vertex between A and B be C (other than 'A' lying on the x-axis).
Since it is a regular hexagon, hence the length of side AC will be $$6$$.
Therefore, $$AC=6$$. Let the coordinate of C be $$(x,0)$$.
Therefore, application of distance formula gives us
$$9-x=6$$ or $$x=3$$.
Now let the coordinate of B be $$(0,y)$$.
Therefore, $$BC=6$$ $$\Rightarrow \sqrt{3^{2}+y^{2}}=6$$
$$\Rightarrow 9+y^{2}=36$$
$$\Rightarrow y^{2}=27$$
$$\Rightarrow y=3\sqrt{3}$$.
Hence $$B=(0,3\sqrt{3})$$.
Question 10
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,
$$x=\dfrac{-3}{2}\,and\, y=1.\xi $$