Straight Lines Test 19

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Straight Lines Test 19

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Weekly Quiz Competition

Question 1
1 / -0

If the graphs of the equations $$2x + 5y = 7$$ and $$6x + cy = 11$$ are parallel lines, c is equal to

A
3

B
5

C
10

D
12

E
15
Solution
- Given that the lines $$2x+5y=7$$ and $$6x+cy=11$$ are parallel
- So there slopes are equal
- We get $$-2/5 = -6/c$$ , which gives $$c = 15$$
Question 2
1 / -0

Let $$S$$ be the set of points whose abscissas and ordinates are natural numbers. Let $$P \in S$$ such that the sum of the distance of $$P$$ from $$(8,0)$$ and $$(0,12)$$ is minimum among all elements in $$S$$. Then the number of such points $$P$$ in $$S$$ is

A
$$1$$

B
$$3$$

C
$$5$$

D
$$11$$

Solution

The sum of the distances of $$P$$ from $$A(8,0),B(0,12)$$ is minimum if $$P$$ lies on line joining $$AB$$ and between them
$$AP+PB=AB$$
$$AP'+P'B>AB$$ by triangle property
So let $$P(x,y)$$
The line $$AB$$ is $$\dfrac { x }{ 8 } +\dfrac { y }{ 12 } =1$$
ie $$3x+2y=24$$
$$\Rightarrow 3x=2(12-y)$$
So $$y$$ can be $$3,6,9$$
$$y=3,x=6\rightarrow 1$$ point
$$y=6,x=4\rightarrow 1$$ point
$$y=9,x=2\rightarrow 1$$ point
So in total three points.
Question 3
1 / -0

If a straight line $$y=2x+k$$ passes through the point $$(1,2)$$ then the value of $$k$$ is equal to:

A
$$0$$

B
$$4$$

C
$$5$$

D
$$-3$$

Solution

$$\text{if straight line y=2x+k passes through the point (1,2), then (1,2) satisfies this line.}$$
$$\Rightarrow 2=2(1)+k$$
$$\Rightarrow k=0$$
Question 4
1 / -0

Which of the following sets of points is collinear

A
$$(1, -1), (-1, 1), (0, 0)$$

B
$$(1, -1), (-1, 1), (0, 1)$$

C
$$(1, -1), (-1, 1), (1, 0)$$

D
$$(1, -1), (-1, 1), (1, 1)$$

Solution

General equation of a line:

$$y = mx + c $$

Now, equation of line passing through $$( 1,-1)$$ and $$(-1,1)$$

Here, both points $$(1,-1)$$ and $$(-1,1)$$ should satisfies the line.

$$(1,-1)$$ gives $$-1 = m+c$$ ... (i)

$$(-1,1)$$ gives $$1 = -m +c $$ ...(ii)

Solving (i) and (ii), we get

$$m = -1$$ and $$c=0$$

Therefore equation of line is $$y = -x$$

Now, $$(0,0)$$ satisfies the line but $$(0,1), (1,0)$$ and $$(1,1)$$ does not satisfies it.

Hence, $$(1,-1), (-1,1), (0,0)$$ is a set of collinear points.
Question 5
1 / -0

In the figure, PB is perpendicular segment from the point A$$(4, 3)$$. If PA = PB then find the coordinate of B.

A
$$(4, -3)$$

B
$$(-4, -3)$$

C
$$(-4, 3)$$

D
None of these

Solution

$$(4,-3)$$ since x co-ordinate will be same since line 11 toy axis the distance above $$=3$$ so distance down will be 3
Question 6
1 / -0

Area of the triangle formed by the points $$\left( 0,0 \right) ,\left( 2,0 \right) $$ and $$\left( 0,2 \right) $$ is

A
$$1$$ sq. unit

B
$$2$$ sq. units

C
$$4$$ sq. units

D
$$8$$ sq. units

Solution

Given: $$A=(x_1,y_1)=(0,0)$$

$$B=(x_2,y_2)=(2,0)$$ and

  $$C=(x_3,y_3)=(0,2)$$

Area of triangle $$=\dfrac {1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)++x_3(y_1-y_2)]$$

$$=\dfrac {1}{2}[0(0-2)+2(2-0)+0(0-0)]$$

         $$=\dfrac {1}{2}[2(2)]$$

         $$=2$$ sq. units.
Question 7
1 / -0

What is the perimeter of the triangle with vertices $$A (-4, 2), B (0, -1)$$ and $$C (3, 3)$$?

A
$$ 7 + 3 \sqrt{2} $$

B
$$ 10 + 5 \sqrt{2} $$

C
$$ 11 + 6 \sqrt{2} $$

D
$$ 5 + 10 \sqrt{2} $$

Solution

Given : $$A (-4, 2), B (0, -1)$$ and $$C (3, 3)$$
Now distance between A and B $$=\sqrt { (4)^{ 2 }+(-3)^{ 2 } } = 5$$
Distance between A and C $$=\sqrt { (7)^{ 2 }+(1)^{ 2 } } = 5\sqrt { 2 }$$
Distance between B and C $$=\sqrt { (3)^{ 2 }+(4)^{ 2 } } =5$$
So perimeter $$=AB+BC+AC=10+5\sqrt2$$
Hence, option B is correct.
Question 8
1 / -0

Area of the triangle formed by the points $$(0,0),(2,0)$$ and $$(0,2)$$ is

A
$$1$$ sq.units

B
$$2$$ sq.units

C
$$4$$ sq.units

D
$$8$$ sq.units

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
$$=\dfrac{1}{2}\left|0(0-2)+2(2-0)+0(0-0)\right|$$

$$=\dfrac{1}{2}\left|0+4+0\right|$$

$$=\dfrac{1}{2}\times 4=2\ sq.\ units$$

Hence, this is the answer.
Question 9
1 / -0

Find the value of $$a$$ if area of the triangle is $$17$$ square units whose vertices are $$(0,0), (4,a), (6,4)$$.

A
$$a=-2$$

B
$$a=-3$$

C
$$a=3$$

D
none of the above

Solution

vertices of the triangle are $$A(0,0), B(4,a), C(6,4)$$.

Then area of triangle$$ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$

$$\Rightarrow 17=\dfrac{1}{2}[0(a-4)+4(4-0)+6(0-a)$$

$$\Rightarrow 34=16-6a$$

$$\Rightarrow a=-\dfrac{18}{6}=-3$$
Question 10
1 / -0

The vertices of a triangle ABC are A(2, 3, 1), B(-2, 2, 0) and C(0, 1, -1). Find the cosine of angle ABC.

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

B
$$\frac{1}{\sqrt{2}}$$

C
$$\frac{2}{\sqrt{6}}$$

D
None of the above

Solution

Given $$A(2, 3, 1), B(-2, 2, 0)$$ and $$C(0, 1, -1)$$

For $$\triangle ABC,$$

$$a=BC=\sqrt { { (-2-0) }^{ 2 }+{ (2-1) }^{ 2 }+{ (0+1) }^{ 2 } } \\ \therefore a=\sqrt { 6 } \\ b=AC=\sqrt { { (2-0) }^{ 2 }+{ (3-1) }^{ 2 }+{ (1+1) }^{ 2 } } \\ \therefore b=\sqrt { 12 } \\ c=AB=\sqrt { { (2+2) }^{ 2 }+{ (3-2 })^{ 2 }+{ (1) }^{ 2 } } \\ \therefore c=\sqrt { 18 } \\ \cos { \angle ABC } =\cfrac { { a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 } }{ 2ac } \\ =\cfrac { 18+6-12 }{ 2\times 3\sqrt { 2 } \times \sqrt { 6 } } \\ =\cfrac { 1 }{ \sqrt { 3 } } $$

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Straight Lines Test 19

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Straight Lines Test 19

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SHARING IS CARING

Question 1

3

5

10

12

15

Solution

Question 2

$$1$$

$$3$$

$$5$$

$$11$$

Solution

Question 3

$$0$$

$$4$$

$$5$$

$$-3$$

Solution

Question 4

$$(1, -1), (-1, 1), (0, 0)$$

$$(1, -1), (-1, 1), (0, 1)$$

$$(1, -1), (-1, 1), (1, 0)$$

$$(1, -1), (-1, 1), (1, 1)$$

Solution

Question 5

$$(4, -3)$$

$$(-4, -3)$$

$$(-4, 3)$$

None of these

Solution

Question 6

$$1$$ sq. unit

$$2$$ sq. units

$$4$$ sq. units

$$8$$ sq. units

Solution

Question 7

$$ 7 + 3 \sqrt{2} $$

$$ 10 + 5 \sqrt{2} $$

$$ 11 + 6 \sqrt{2} $$

$$ 5 + 10 \sqrt{2} $$

Solution

Question 8

$$1$$ sq.units

$$2$$ sq.units

$$4$$ sq.units

$$8$$ sq.units

Solution

Question 9

$$a=-2$$

$$a=-3$$

$$a=3$$

none of the above

Solution

Question 10

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

$$\frac{1}{\sqrt{2}}$$

$$\frac{2}{\sqrt{6}}$$

None of the above

Solution

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