Straight Lines Test 13

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Straight Lines Test 13

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Weekly Quiz Competition

Question 1
1 / -0

The area of the triangle formed by the points $$(2, 6), (10, 0)$$ and $$(0, k)$$ is zero square units. Find the value of $$k.$$

A
$$\dfrac {15}{2}$$

B
$$\dfrac 32$$

C
$$\dfrac 72$$

D
$$\dfrac {13}{2}$$

Solution

The area of the triangle is:

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

We are given that the area of the triangle is $$0$$ and the points are $$(2,6)$$, $$(10,0)$$ and $$(0,k)$$, therefore,

$$\Rightarrow 0=\dfrac { 1 }{ 2 } \left| 6(10-0)+0(0-2)+k(2-10) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-0-8k \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-8k \right| $$

$$\Rightarrow 60-8k=0\\ \\ \Rightarrow 8k=60\\ \Rightarrow k=\dfrac { 60 }{ 8 } =\dfrac { 15 }{ 2 }$$

Hence, $$k=\dfrac { 15 }{ 2 }$$.
Question 2
1 / -0

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

A
1/2

B
1

C
2

D
3/2

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y }_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (0,0) $$ ; $$({ x }_{ 2 },{ y }_{ 2 }) = (2,3) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (5,8)$$ in the area formula, we get

Area of triangle ABC $$ = \left| \frac { (0)(3-8)+(2)(8-0)+5(0-3) }{ 2 } \right| = \left| \frac { 16 -15 }{ 2 } \right| = \frac {1}{2} units $$
Question 3
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

Let $$ A (6,2), B (2,5) $$ and $$ C (9,6) $$

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Distance between the points A$$ (6,2) $$ and B $$ (2,5) = \sqrt { \left( 2-6 \right) ^{ 2 }+\left( 5-2 \right) ^{ 2 } } = \sqrt { 16+9 } = \sqrt { 25 } = 5 $$
Distance between the points B$$ (2,5) $$ and C $$ (9,6) = \sqrt { \left( 9-2 \right) ^{ 2 }+\left( 6-5 \right) ^{ 2 } } = \sqrt {49+1 } = \sqrt {50 } $$
Distance between the points A$$ (6,2) $$ and C $$ (9,6) = \sqrt { \left( 6-9 \right) ^{ 2 }+\left( 2-6\right) ^{ 2 } } = \sqrt { 9 + 16 } = \sqrt { 25} = 5 $$

Since, two sides are equal in length $$ {(\sqrt { 50 }) }^{2} = {(\sqrt { 25 }) }^{2} + {(\sqrt { 25 }) }^{2} $$, the triangle is a right angled isosceles triangle.
Question 4
1 / -0

The line through (1, 5) parallel to x-axis is __________

A
x = 1

B
y = 5

C
y = 1

D
x = 5

Solution

From the graph, the line through $$ (1,5) $$ and parallel to x-axis is $$ y = 5 $$ as all points along the line has $$ y = 5 $$
Question 5
1 / -0

Find the equation of a line passing through the points A(3, -5) and (4, -8)

A
3x + y = 4

B
3x + 2y = 5

C
x + 3y = 4

D
4x = 3y

Solution

Equation of line joining two points $$ ({x}_{1}, {y}_{1}) $$ and $$ ({x}_{2}, {y}_{2}) $$ is $$ y - {y}_{1} = \frac { {y}_{2} - {y}_{1}}{ {x}_{2} - {x}_{1}} (x-{x}_{1}) $$

So, equation of line joining $$ A(3,-5) ; B (4,-8) $$ is $$ y +5 = \frac {-8+5}{4-3} (x-3) $$
$$ => y + 5 = (-3)(x-3) $$
$$ => => y+5 = -3x + 9 $$
$$ => 3x + y = 4 $$
Question 6
1 / -0

The area of triangle with vertices $$A(5,\,0),\,B(8,\,0)$$ and $$C(9,\,5)$$ is

A
$$7\dfrac{1}{4}$$

B
$$7\dfrac{1}{2}$$

C
$$\dfrac{15}{4}$$

D
None of the above

Solution

Area of triangle $$=\dfrac{1}{2}\begin{bmatrix}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\end{bmatrix}$$
$$A=\dfrac{1}{2}\begin{bmatrix}5(0-5)+8(5-0)+9(0-0)\end{bmatrix}$$
$$A=\dfrac{1}{2}\begin{bmatrix}-25+40\end{bmatrix}=\dfrac{15}{2}$$ square units.
So, option B is correct.
Question 7
1 / -0

The point $$A(-5,\,8)$$ ______ on circle of radius $$6$$ and centre $$(8,\,9)$$.

A
lies

B
lies inside

C
lies outside

D
cant't say

Solution

Radius of circle$$=6$$
Now, the distance between the centre O and point A is
$$OA=\sqrt{(8-(-5))^2+(9-8)^2}$$
$$=\sqrt{(13)^2+(1)^2}=\sqrt{170}>6$$
The point does not lie on circle.
So, option C is correct.
Question 8
1 / -0

$$\triangle OPQ$$ is formed by the coordinates $$P(0,\,5),\, Q(8,\,0)$$ and origin $$O$$. The area of $$\triangle OPQ$$ is ____ square units.

A
$$20$$ sq. units

B
$$10$$ sq. units

C
$$30$$ sq. units

D
None of the above

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Area of $$\triangle OPQ $$
$$=\dfrac 12 \left[ 1(0) - 1(0) +1(-40)\right]$$
$$=\dfrac 12 \times 40 = 20$$ sq. units
Question 9
1 / -0

The area of triangle with vertices $$A(0,\,9),\,B(0,\,4)$$ and $$C(-5,\,-9)$$ is

A
$$\dfrac{25}{2}$$ sq. units

B
$$\dfrac{23}{2}$$ sq. units

C
$$\dfrac{19}{2}$$ sq. units

D
None of the above

Solution

Area of triangle $$=\dfrac{1}{2}\begin{bmatrix}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\end{bmatrix}$$
$$=\dfrac{1}{2}\begin{bmatrix}0(4+9)+0(-9-9)+(-5)(9-4)\end{bmatrix}$$
$$=\dfrac{1}{2}\begin{bmatrix}-5\times5\end{bmatrix}$$
$$=-\dfrac{25}{2}$$
The area cannot be negative.
$$\therefore$$ It must be $$\dfrac{25}{2}$$ sq. units
So, option A is correct.
Question 10
1 / -0

A circle has its centre at origin and point $$P(8,0)$$ lies on it.
The point $$Q(2,\,8)$$ lies

A
inside the circle.

B
outside the circle.

C
on the circle

D
None of the above

Solution

The radius of the circle is the distance between points O and P i.e. $$8$$ units.
Distance between O and Q
$$=\sqrt{(2-0)^2+(8-0)^2}$$
$$=\sqrt{4+64}=\sqrt{68}>8$$
Therefore point Q lies outside the circle.
So, option B is correct.

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 13

Result

Straight Lines Test 13

Score

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Rank

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SHARING IS CARING

Question 1

$$\dfrac {15}{2}$$

$$\dfrac 32$$

$$\dfrac 72$$

$$\dfrac {13}{2}$$

Solution

Question 2

1/2

1

2

3/2

Solution

Question 3

right

equilateral

right isosceles

scalene

Solution

Question 4

x = 1

y = 5

y = 1

x = 5

Solution

Question 5

3x + y = 4

3x + 2y = 5

x + 3y = 4

4x = 3y

Solution

Question 6

$$7\dfrac{1}{4}$$

$$7\dfrac{1}{2}$$

$$\dfrac{15}{4}$$

None of the above

Solution

Question 7

lies

lies inside

lies outside

cant't say

Solution

Question 8

$$20$$ sq. units

$$10$$ sq. units

$$30$$ sq. units

None of the above

Solution

Question 9

$$\dfrac{25}{2}$$ sq. units

$$\dfrac{23}{2}$$ sq. units

$$\dfrac{19}{2}$$ sq. units

None of the above

Solution

Question 10

inside the circle.

outside the circle.

on the circle

None of the above

Solution

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