Straight Lines Test 18

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Straight Lines Test 18

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Weekly Quiz Competition

Question 1
1 / -0

Consider the points $$A( a, b + c)$$, $$B(b, c + a)$$, and $$C(c, a +b)$$ be the vertices of $$\bigtriangleup$$ABC. The area of $$\bigtriangleup$$ABC is:

A
$$2(a^2 + b^2 +c^2)$$

B
$$a^2 + b^2 +c^2$$

C
$$2(ab + bc + ca)$$

D
None of these

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Since, vertices are $$A(a, b + c),~ B(b, c + a),$$ and $$ C(c, a + b)$$
$$\therefore$$ Area $$ \bigtriangleup ABC = \dfrac{1}{2}|a(c + a) -b(b + c) + b(a + b) - c(c + a) +c(b + c) -a(a + b)|=0$$ .
Hence option 'D' is correct.
Question 2
1 / -0

$$(3, 1), (-3, 2)$$ and $$\displaystyle (0,2-\sqrt{3})$$ are the vertices of __________ triangle of area ___________.

A
an isosceles, $$81$$ sq. units

B
a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

C
an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

D
a right angled, $$81$$ sq. units

Solution

Distance between two points $$= \sqrt { \left( { x }_{ 2 }-{x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Distance between the points $$A (3,1) $$ and $$B (-3,2) = \sqrt { \left( -3-3 \right) ^{ 2 }+\left( 2 - 1 \right) ^{ 2 } } = \sqrt { 36 + 1 } = \sqrt { 37 }$$
Distance between the points $$B(-3,2) $$ and $$C (0,2-\sqrt {3}) = \sqrt { \left( 0 + 3 \right) ^{ 2 }+\left(2 - \sqrt {3} - 2\right) ^{ 2 } } = \sqrt { 9 + 3 } = \sqrt { 12 } $$
Distance between the points $$A(3,1) $$ and $$ C(0,2-\sqrt {3})$$
$$ = \sqrt { \left( 0-3\right) ^{ 2 }+\left( 2-\sqrt {3} - 1\right) ^{ 2 } } = \sqrt { 9 + 1 + 3 -2\sqrt {3} } = \sqrt { 13 - 2\sqrt {3} } $$
Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.
Area of a triangle $$= \left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Area of $$\triangle ABC= \left| \cfrac { (3)(2-2+\sqrt {3})+(-3)(2-\sqrt {3} - 1)+0(1-2) }{ 2 } \right| $$
$$ = \left| \cfrac { 3\sqrt {3} -3 + 3\sqrt {3} }{ 2 } \right| $$
$$ = \cfrac{-3 + 6 \sqrt {3}}{2}$$ sq. units
Question 3
1 / -0

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $$(2,2)$$, $$(4,4)$$ and $$(2,6)$$.

A

5

B
3

C
1

D
0

Solution

Let $$ A(2,2), B(4,4) $$ and $$ C(2,6) $$ be the vertices of a given triangle ABC.

Let D, E, and F be the midpoints of AB, BC and CA respectively.

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{

2 }) $$ is calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{ x

}_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) $$

Using this formula,the coordinates of D, E, and F are given as $$\displaystyle D \left (\frac{2+4}{2},\frac {2+4}{2} \right ), E\left (\frac {4+2}{2},\frac {4+6}{2}\right ) $$ and $$ F\left (\dfrac {2+2}{2},\dfrac {2+6}{2} \right ) $$

i.e., $$ D(3,3), E(3,5) and F(2,4) $$

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (3,3) $$ ; $$({ x

}_{ 2 },{ y }_{ 2 }) = (3,5) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (2,4)$$

in the area formula, we get

Area of triangle DEF $$ = \left| \dfrac { 3(5-4)+(3)(4-3)+2(3-5) }{ 2

} \right| = \left| \dfrac { 3 +3 -4 }{ 2 } \right| =

\dfrac {2}{2} = 1 \ sq \ units $$
Question 4
1 / -0

The triangle with vertices A(4, 4), B(-2, -6) and C(4, -1) is shown in the diagram. The area of $$\Delta$$ ABC is _______

A
5 sq. units

B
12 sq. units

C
15 sq. units

D
20 sq. units

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$= \cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

$$=\dfrac{1}{2}|4(-6-(-1))-2(-1-4)+4(4-(-6))|$$

$$=\dfrac{1}{2}|4(-5)-2(-5)+4(10)|$$

$$=\dfrac{1}{2}|30|$$

$$=15 $$ sq. units
Question 5
1 / -0

The coordinates of $$A$$ for which area of triangle, whose vertices are $$A(a, 2a),\ B(-2, 6)$$ and $$C(3, 1)$$ is $$10$$ square units, are:

A
$$(0,\:3)$$

B
$$(5,\:8)$$

C
$$(\displaystyle 3, \frac{8}{3})$$

D
None of these

Solution

Given: Vertices of the triangle are $$A(a, 2a),\ B(-2, 6),\ C(3,1)$$

The area of the triangle is $$10$$ square units.

$$\Rightarrow \dfrac{1}{2}\bigg | a(6-1)+(-2)(1-2a)+3(2a-6) \bigg |=10$$

$$\Rightarrow 5a-2+4a+6a-18=20$$

$$\Rightarrow 15a=40$$

$$\Rightarrow a=\dfrac{40}{15}=\dfrac{8}{3}$$

$$\therefore $$ The coordinates of vertex $$A $$ are $$\displaystyle \left ( \dfrac{8}{3}, \frac{16}{3} \right )$$
Question 6
1 / -0

The number of positive fractions $$\dfrac{m}{n}$$ such that $$\dfrac{1}{3}< \dfrac{m}{n}<1 $$ and having the property that the fraction remains the same by adding some positive integer to the numerator and multiplying the denominator by the same positive integer is:

A
$$1$$

B
$$3$$

C
$$6$$

D
infinite

Solution

Given $$m,n$$ are integers $$\dfrac{1}{3}<\dfrac{m}{n}<1$$
$$\Rightarrow \dfrac{m+x}{nx}=\dfrac{m}{n}$$
$$\Rightarrow m+x=mx$$
$$\Rightarrow x=\dfrac{m}{m-1},$$ $$x$$ must be an integer.
So only possible value of $$m$$ satisfying above equation is $$2.$$
$$\Rightarrow m=2$$
$$\Rightarrow \dfrac{m}{n}>\dfrac{1}{3}\Rightarrow 3m>n\Rightarrow 6>n$$

$$\Rightarrow \dfrac{m}{n}<1\Rightarrow m<n\Rightarrow 2<n$$
$$\Rightarrow n=3,4,5$$
$$\therefore 3$$ Possible fractions.
Hence, the answer is $$3.$$
Question 7
1 / -0

Three points A, B and C have coordinates $$(a, b + c), \ (b, c + a)$$ and $$(c, a + b)$$, respectively. The area of the triangle ABC will be:

A
$$\displaystyle a^{2}+b^{2}+c^{2}$$

B
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$$

C
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$$

D
$$0$$

Solution

Area of triangle $$=\dfrac{ 1 }{ 2 }\left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$

Given the points: $$A(a, b+c) , B(b, c+a) , C(c, a+b)$$
Therefore, area:

$$A =\dfrac{ 1 }{ 2 }\left[ a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a) \right]$$

$$A =\dfrac{ 1 }{ 2}{\left[ a(c-b)+b(a-c)+c(b-a) \right] }$$

$$A=\dfrac{ 1 }{ 2}{\left[ ac-ab+ba-bc+cb-ac \right] }=0$$
Question 8
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)
$$0,\,3,\,9,\,18,\,30,\,?$$

A
$$48$$

B
$$42$$

C
$$36$$

D
$$45$$

Solution

The difference between the consecutive numbers increases by $$3$$.
$$3-0 = 3$$
$$9 - 3 = 6$$
$$18-9 = 9$$
$$30-18 = 12$$.
Pattern is adding multiples of 3
Next term is $$30+15=45$$
Question 9
1 / -0

In of the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?) Choose the correct answer for the term marked by the symbol (?)

A
$$16$$

B
$$10$$

C
$$9$$

D
$$12$$

Solution

$$\rightarrow 16 \times4=64=8^2$$
$$\rightarrow 3 \times27=81=9^2$$
$$\rightarrow 18 \times8=144=12^2$$
Question 10
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)
$$6,\,9,\,18,\,21,\,30,\,?$$

A
$$38$$

B
$$33$$

C
$$36$$

D
$$39$$

Solution

$$6,\,9,\,18,\,21,\,30$$
Alternate series difference is common
$$\therefore\;30+3=33$$

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Straight Lines Test 18

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Straight Lines Test 18

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SHARING IS CARING

Question 1

$$2(a^2 + b^2 +c^2)$$

$$a^2 + b^2 +c^2$$

$$2(ab + bc + ca)$$

None of these

Solution

Question 2

an isosceles, $$81$$ sq. units

a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

a right angled, $$81$$ sq. units

Solution

Question 3

5

3

1

0

Solution

Question 4

5 sq. units

12 sq. units

15 sq. units

20 sq. units

Solution

Question 5

$$(0,\:3)$$

$$(5,\:8)$$

$$(\displaystyle 3, \frac{8}{3})$$

None of these

Solution

Question 6

$$1$$

$$3$$

$$6$$

infinite

Solution

Question 7

$$\displaystyle a^{2}+b^{2}+c^{2}$$

$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$$

$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$$

$$0$$

Solution

Question 8

$$48$$

$$42$$

$$36$$

$$45$$

Solution

Question 9

$$16$$

$$10$$

$$9$$

$$12$$

Solution

Question 10

$$38$$

$$33$$

$$36$$

$$39$$

Solution

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