Straight Lines Test 12

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Straight Lines Test 12

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Weekly Quiz Competition

Question 1
1 / -0

If coordinates of P,Q, and R are (3,6),(-1,3) and (2,-1) respectively. Then area of $$\displaystyle \triangle PQR$$ is ____ square units.

A
12

B
12.5

C
13

D
14

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of $$\displaystyle \Delta PQR$$ is given by
$$\displaystyle =\frac{1}{2}\left [ 3\left ( 3+1 \right )-1\left ( -1-6 \right )+2\left ( 6-3 \right ) \right ]$$
$$\displaystyle =\frac{1}{2}\left ( 12+7+6 \right )=12.5$$ square units.
Question 2
1 / -0

Find the slopes of the line whose inclination is $$135^\circ$$.

A
$$1$$

B
$$-1$$

C
$$\sqrt{3}$$

D
$$-\sqrt{3}$$

Solution

We know that, if inclination of a line with positive x-axis is $$\theta$$, then slope $$, m=\tan \theta$$
Here $$ \theta=135^o$$ is given
So slope, $$m=\tan 135^o= - 1$$
Question 3
1 / -0

The area of triangle formed by $$(0, 0), (0, a)$$ and $$(b, 0)$$ is .......... .

A
$$ab$$

B
$$\displaystyle \frac{ab}{2}$$

C
$$\left | \displaystyle \frac{ab}{2} \right |$$

D
$$\left | ab \right |$$

Solution

The area of a triangle formed by joining the points $$(x_1, y_1)$$, $$(x_2, y_2)$$ and $$(x_3, y_3)$$ is

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

Therefore, the area of a triangle formed by joining the points $$(0, 0)$$, $$(0, a)$$ and $$(b, 0)$$ is:

$$A=\dfrac { 1 }{ 2 } \bigg| 0(0-b)+a(b-0)+0(0-0) \bigg| $$

$$=\dfrac { 1 }{ 2 } \bigg| 0+ab \bigg| $$

$$=\bigg| \dfrac { ab }{ 2 } \bigg|$$

Hence, the area of the triangle is $$\left| \dfrac { ab }{ 2 } \right|$$
Question 4
1 / -0

The inclination of a line is $$30^{\circ}$$, then the slope of the line is

A
$$0$$

B
$$1$$

C
$$\sqrt{3}$$

D
$$\displaystyle \frac{1}{\sqrt{3}}$$

Solution

We know that, if inclination of a line with positive x-axis is $$\theta$$, then slope $$, m=\tan \theta$$
Here $$ \theta=30^o$$ is given
So slope, $$m=\tan 30^o= \dfrac{1}{\sqrt 3}$$
Question 5
1 / -0

The points $$(6, 2), (2, 5)$$ and $$(9, 6)$$ form the vertices of a __triangle.

A
right

B
equilateral

C
right isosceles

D
scalene

Solution

The distance between the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is given by:
$$D=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$$
Now, we find the distances between the points $$(6,2)$$, $$(2,5)$$ and $$(9,6)$$ as shown below:
$$AB=\sqrt { { (2-6) }^{ 2 }+{ (5-2) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =5$$
$$BC=\sqrt { { (9-2) }^{ 2 }+{ (6-5) }^{ 2 } } =\sqrt { { 7 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 49+1 } =\sqrt { 50 } =5\sqrt {2}$$
$$AC=\sqrt { { (9-6) }^{ 2 }+{ (6-2) }^{ 2 } } =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 9+16 } =\sqrt { 25 } =5$$
Since $$AB=AC$$, therefore, the triangle is an isosceles triangle because it has two congruent sides.
Now consider, $${ c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\\ \Rightarrow { (5\sqrt { 2 } ) }^{ 2 }={ 5 }^{ 2 }+{ 5 }^{ 2 }\\ \Rightarrow 25\times 2=25+25\\ \Rightarrow 50=50$$
Thus, it is a right triangle because its sides satisfy the pythagoras theorem.
Hence, the triangle is a right isosceles triangle.
Question 6
1 / -0

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

A
$$\dfrac 12$$

B
$$1$$

C
$$2$$

D
$$\dfrac 32$$

Solution

Area of the triangle with three vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is
$$A=\dfrac { 1 }{ 2 } \left| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \right|$$
Therefore, the area of triangle with vertices $$(1,2)$$, $$(3,4)$$ and $$(5,6)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| 0(2-5)+3(5-0)+8(0-2) \right| =\dfrac { 1 }{ 2 } \left| 0+15-16 \right| =\dfrac { 1 }{ 2 } \left| -1 \right| =\dfrac { 1 }{ 2 }$$
Hence, the area of the triangle is $$\dfrac { 1 }{ 2 }$$ square units.
Question 7
1 / -0

The area of a triangle with the vertices $$(1, 2), (3, 4)$$ and $$(5, 6)$$, is ____ square units.

A
$$0$$

B
$$1$$

C
$$2$$

D
$$5$$

Solution

Area of the triangle with three vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is

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

Therefore, the area of triangle with vertices $$(1,2)$$, $$(3,4)$$ and $$(5,6)$$ is:

$$A=\dfrac { 1 }{ 2 } \left| 2(3-5)+4(5-1)+6(1-3) \right| =\dfrac { 1 }{ 2 } \left| -4+16-12 \right| =\dfrac { 1 }{ 2 } \left| 16-16 \right| =\dfrac { 1 }{ 2 } \left| 0 \right| =0$$

Hence, the area of the triangle is $$0$$.
Question 8
1 / -0

One end of a diagonal of a square is $$(2, 0)$$ and its slope is $$1$$. If the length of the side of a square is $$2$$ units, then write all the possible coordinates of the other end.

A
$$(4, 2)$$

B
$$(0, 2)$$

C
$$(2, 4)$$

D
$$(-2, 0)$$

Solution

Given
One end of diagonal of square is $$(2,0)$$ and slope is 1
Let other end of diagonal is $$(x,y)$$
then $$m=1$$
$$1=\dfrac{y}{x-2}$$
$$x-2=y$$----(1)
The length of side of square is 2 hence
By pythagoras theoram length of diagonal be $$\sqrt{8}$$
$$\sqrt{(x-2)^2+y^2}=\sqrt{8}$$
On squaring both sides
$$(x-2)^2+y^2=8$$
from eq (1)
$$2y^2=8$$
$$y^2=4$$
$$y=\pm 2=2,-2$$
$$x=4,0$$
Possible coordinates are $$(4,2)$$ and $$(0,-2)$$
Question 9
1 / -0

If $$(1, 2), (3, 4)$$ and $$(0, 6)$$ are the three vertices of a parallelogram taken in that order, then the fourth vertex is .......... .

A
$$(2, 4)$$

B
$$(-2, 4)$$

C
$$(4, 2)$$

D
$$(-4, 2)$$

Solution

Let the vertices of the parallelogram be $$A(1,2)$$, $$B(3,4)$$, $$C(0,6)$$ and $$D(x,y)$$.
Thus, the diagonals are AC and BD.
Now diagonals of a parallelogram bisect each other.
Therefore, the midpoint of AC is the midpoint of BD.

Now the midpoint of AC is $$\left(\dfrac{1+0}{2},\dfrac{2+6}{2}\right)=\left(\dfrac{1}{2},4\right)$$.

Now the midpoint of BD will be $$\left(\dfrac{3+x}{2},\dfrac{4+y}{2}\right)=\left(\dfrac{1}{2},4\right)$$

$$\dfrac{3+x}{2}=\dfrac{1}{2}$$ or $$x=-2$$.

Similarly $$\dfrac{4+y}{2}=4$$ or $$y=4$$

Hence the last vertex is $$(-2,4)$$.
Question 10
1 / -0

Find the area of $$\Delta ABC$$, in which $$A = (2, 1), B = (3, 4)$$ and $$C = (-3, -2).$$

A
$$3$$ square units

B
$$4$$ square units

C
$$6$$ square units

D
$$12$$ square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Therefore, the area of triangle with vertices $$(2,1)$$, $$(3,4)$$ and $$(-3,-2)$$ is:
$$A=\displaystyle \frac { 1 }{ 2 } \left| 2(4-(-2))+3(-2-1)-3(1-4) \right| $$
$$=\dfrac { 1 }{ 2 } \left| 2(6)+3(-3)-3(-3) \right| $$
$$=\dfrac { 1 }{ 2 } \left| 12-9+9 \right|$$
$$ =\dfrac { 1 }{ 2 } \left| 12 \right| =6$$ square units.

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Straight Lines Test 12

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Straight Lines Test 12

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SHARING IS CARING

Question 1

12

12.5

13

14

Solution

Question 2

$$1$$

$$-1$$

$$\sqrt{3}$$

$$-\sqrt{3}$$

Solution

Question 3

$$ab$$

$$\displaystyle \frac{ab}{2}$$

$$\left | \displaystyle \frac{ab}{2} \right |$$

$$\left | ab \right |$$

Solution

Question 4

$$0$$

$$1$$

$$\sqrt{3}$$

$$\displaystyle \frac{1}{\sqrt{3}}$$

Solution

Question 5

right

equilateral

right isosceles

scalene

Solution

Question 6

$$\dfrac 12$$

$$1$$

$$2$$

$$\dfrac 32$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

$$5$$

Solution

Question 8

$$(4, 2)$$

$$(0, 2)$$

$$(2, 4)$$

$$(-2, 0)$$

Solution

Question 9

$$(2, 4)$$

$$(-2, 4)$$

$$(4, 2)$$

$$(-4, 2)$$

Solution

Question 10

$$3$$ square units

$$4$$ square units

$$6$$ square units

$$12$$ square units

Solution

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