Straight Lines Test 14

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Straight Lines Test 14

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Weekly Quiz Competition

Question 1
1 / -0

The perimeter of triangle with vertices $$A(0,\,9),\,B(0,\,0)$$ and $$C(9,\,0)$$

A
$$9\sqrt{2}$$

B
$$9(2+\sqrt{2})$$

C
$$18\sqrt{2}$$

D
None of the above

Solution

AB=$$9$$ units, BC=$$9$$ units( hint: by using distance formula)
AC $$=\sqrt{(9-0)^2+(0-a)^2}=\sqrt{81+81}=\sqrt{81(2)}$$
$$=9\sqrt{2}$$ units
Perimeter $$=9+9+9\sqrt{2}$$
$$=18+9\sqrt{2}$$ units
$$=9(2+\sqrt{2})$$ units
So, option B is correct.
Question 2
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.
Question 3
1 / -0

The perimeter of triangle with vertices $$A(0,\,0),\,B(5,\,7)$$ and $$C(9,\,5)$$

A
$$\sqrt{74}+\sqrt{20}$$

B
$$\sqrt{74}+\sqrt{106}$$

C
$$\sqrt{74}+\sqrt{20}+\sqrt{106}$$

D
None of the above

Solution

P $$=AB+BC+CA$$
$$AB=\sqrt{5^2+7^2}$$ and $$BC=\sqrt{(5-9)^2+(7-5)^2}$$
$$=\sqrt{25+49}$$ and $$BC=\sqrt{16+4}=\sqrt{20}$$
$$=\sqrt{74}$$
CA $$=\sqrt{9^2+(5)^2}=\sqrt{81+25}$$
CA $$=\sqrt{106}$$
$$P=\sqrt{74}+\sqrt{20}+\sqrt{106}$$
So, option C is correct.
Question 4
1 / -0

Are the points $$(5,\,5),\,(8,\,2)$$ and $$(3,\,-4)$$ are vertices of right angled triangle.

A
True

B
False

C
Cant say

D
None

Solution

Let the points $$A(5,\,5),\,B(8,\,2)$$ and $$C(3,\,-4)$$ are vertices of triangle.
$$AB^2=(5-8)^2+(5-2)^2=9+9=18$$
$$BC^2=(8-3)^2+(2+4)^2=25+36=61$$
$$CA^2=(3-5)^2+(-4-5)^2=4+81=85$$
$$AB^2+AC^2\neq BC^2$$
The vertices are not points of a right angled triangle.
Question 5
1 / -0

$$(9, 2), (5, -1) $$ and $$ (7, -5)$$ are the vertices of the triangle. Find its area.

A
10 square units

B
11 square units

C
12 square units

D
13 square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
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 $$x_{1} = 9$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = -1$$, $$x_{3} = 7$$ and $$y_{3} = -5$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[9(-1 + 5) + 5(-5 - 2) + 7(2 + 1)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 35 + 21]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 22\right|$$
$$=$$ $$\left|11 \right|$$
Area always in absolute value.
So, area of the triangle $$= 11$$ square units.
Question 6
1 / -0

What is the area of the triangle whose vertices are: $$(-3, 15), (6, -7) $$ and $$(10, 5)$$?

A
$$94$$ square units

B
$$96$$ square units

C
$$97$$ square units

D
$$98$$ square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is $$A=\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
Here $$x_{1} = -3$$, $$y_{1} = 15$$, $$x_{2} = 6$$, $$y_{2} = -7$$, $$x_{3} = 10$$ and $$y_{3} = 5$$

Substituting the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[-3(-7 - 5) + 6(5 - 15) + 10(15 + 7)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 60 + 220]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 196\right|$$
$$=$$ $$\left|98 \right|$$

Area is always in absolute value.
So, area of the triangle $$= 98$$ square units.
Question 7
1 / -0

Are the points $$(5,\,3),\,(2,\,9)$$ and $$(-8,\,4)$$ are vertices of

A
a right angled triangle.

B
an isosceles triangle.

C
a scalene triangle

D
None of the above

Solution

Let points $$A(5,\,3),\,B(2,\,9)$$ and $$C(-8,\,4)$$ are vertices of right angled triangle.
$$AB^2=(5-2)^2+(3-9)^2=9+36=45$$
$$BC^2=(2+8)^2+(9-4)^2=100+25=125$$
$$CA^2=(5+8)^2+(3-4)^2=169+1^2=170$$
$$AB^2+BC^2=CA^2$$
It is a right angled triangle's vertices.
So, option A is correct.
Question 8
1 / -0

What is the area of the triangle for the following points $$(6, 2), (5, 4)$$ and $$(3, -1)$$?

A
2.3 square units

B
4.5 square units

C
4.1 square units

D
3.6 square units

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 $$x_{1} = 6$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = 4$$, $$x_{3} = 3$$ and $$y_{3} = -1$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[6(4 + 1) + 5(-1 - 2) + 3(2 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[30 - 15 - 6]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 9\right|$$
$$=$$ $$\left|4.5 \right|$$
Area always in absolute value.
So, area of the triangle $$= 4.5$$ square units.
Question 9
1 / -0

The area of the triangle whose vertices are $$(0, 1), (1, 4)$$ and $$(1, 2)$$ is ___ square units.

A
1

B
2

C
3

D
4

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 $$x_{1} = 0$$, $$y_{1} = 1$$, $$x_{2} = 1$$, $$y_{2} = 4$$, $$x_{3} = 1$$ and $$y_{3} = 2$$
Substitute the values, we get
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[0(4 - 2) + 1(2 - 1) + 1(1 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[0 + 1 - 3]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times -2\right|$$
$$=$$ $$\left|- 1 \right|$$
Area always in absolute value.
So, area of the triangle is $$ 1$$ square units.
Question 10
1 / -0

The point on the line $$4x - y - 2 = 0$$ which is equidistant from the points $$(-5, 6)$$ and $$(3, 2)$$ is

A
$$(2, 6)$$

B
$$(4, 14)$$

C
$$(1, 2)$$

D
$$(3, 10)$$

Solution

Let the point on line $$4x-y-2=0$$ be $$P(x,y)$$.
Let $$A \equiv (-5,6)$$ and $$B \equiv (3,2)$$
$$4x-y-2=0$$ ....(1)
Point P is equidistant from points A and B ....Given
$$\therefore AP = PB$$
By distance formula,
$$(x+5)^2 + (y-6)^2 = (x-3)^2 + (y-2)^2$$
$$x^2+10x+25 + y^2 - 12y + 36 = x^2 - 6x +9 + y^2 - 4y + 4$$
$$16x - 8y + 48=0$$
$$4x-2y+12=0$$ ....(2)
Subtract eq(2) from eq (1), we get
$$y=14$$
Substitute in eq(1), we get
$$x=4$$
So, the point on the line is $$(4, 14)$$.

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Straight Lines Test 14

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Straight Lines Test 14

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SHARING IS CARING

Question 1

$$9\sqrt{2}$$

$$9(2+\sqrt{2})$$

$$18\sqrt{2}$$

None of the above

Solution

Question 2

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

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

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

$$2$$

Solution

Question 3

$$\sqrt{74}+\sqrt{20}$$

$$\sqrt{74}+\sqrt{106}$$

$$\sqrt{74}+\sqrt{20}+\sqrt{106}$$

None of the above

Solution

Question 4

True

False

Cant say

None

Solution

Question 5

10 square units

11 square units

12 square units

13 square units

Solution

Question 6

$$94$$ square units

$$96$$ square units

$$97$$ square units

$$98$$ square units

Solution

Question 7

a right angled triangle.

an isosceles triangle.

a scalene triangle

None of the above

Solution

Question 8

2.3 square units

4.5 square units

4.1 square units

3.6 square units

Solution

Question 9

1

2

3

4

Solution

Question 10

$$(2, 6)$$

$$(4, 14)$$

$$(1, 2)$$

$$(3, 10)$$

Solution

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