Straight Lines Test -5

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Straight Lines Test -5

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Weekly Quiz Competition

Question 1
1 / -0

The lines x + 2y – 3 = 0, 2x + y – 3 = 0 and the line l are concurrent . If the line I passes through the origin, then its equation is

A
x + 2y = 0

B
x – y = 0

C
x + y + 0

D
none of these

Solution

Equation of a line passing through the intersection of two lies is given by ax₁ +by₁ +c₁ + k(ax₂ +by₂ +c₂) = 0

Hence x+2y-3 + k(2x+y-3) = 0

Since it passes through (0,0)

-3 -3k = 0

This implies k = -1

Sustituting for k we get,

x+2y-3 +(-1)(2x+y-3) = 0

-x +y =0 or x - y = 0
Question 2
1 / -0

The triangle formed by the lines x + y = 1, 2x + 3y – 6 = 0 and 4x – y + 4 = 0 lies in

A
1^th quadrant

B
3^rd quadrant

C
2^nd quadrant

D
4^th quadrant

Solution

On solving line 1 and line 2 we get x = -3 andy =4. Hence the point of intersection is (-3,4)

On solving line 2 and line 3 we get x = (-3/7) and y = 16/7. Hence the point of intersection is (-3/7, 16/7)

On solving line 3 and line 1 we get x = -3/5 and y - 8/5.Hence the point of intersection is (-3/5,8/5)

All the above points lie in the second quadrant. Hence the triangle formed by these lines also lie in the second quadrant.
Question 3
1 / -0

The area of triangle formed by the lines y = x, y = 2x and y = 3x + 4 is

A
9

B
4

C
8

D
7

Solution

Area of the triangle formed by the coordiates( x₁,y₁), (x₂,y₂) and (x₃,y₃) is given by

1/2[x₁(y₂- y₃) + x₂(y₃ - y₁) + x₃(y₁- y₂)]

On solving the lines 1 and 2, the point of intersection is (0,0)

On solving the lines 2 and 3, the point of intersection is (4,8)

\On solving the line 3 and 1 , the point of intersection is (-2,-2)

Now substituting the values to find the area of the triangle,

Area = 1/2| [ 0 + 4(-2-0) +(-2)(0 - 8)] |

= 4 sq units
Question 4
1 / -0

If the points representing the complex numbers - 4 +3i , 2 – 3i and 0 + pi are collinear , then the value of p is

A
- 1

B
1

C
2

D
none of these

Solution

Let us take the coordinates as (-4,3), (2,-3) and (0,p).

If the points are collienear the 1/2|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁-y₂)| = 0

Now substituting the values |-4(-3 - p) +2(p -3) +0(3+3)|= 0

12 +4p +2p -6 +0 = 0

6p +6 = 0

6p = -6

Therefore p = -1
Question 5
1 / -0

The lines ix + my + n = 0 , mx + ny + l = 0 and nx + ly +m = 0 are concurrent if

A
l – m – n = 0

B
l + m + n = 0

C
l + m – n = 0

D
I – m + n = 0

Solution

The required condition for concurrency is a₃(b₁c₂ - b₂c₁) + b₃(c₁a₂ - c₂a₁) + c₃(a₁b₂ - a₂b₁) = 0

Here a₁ = l, a₂=m, a₃ = n and b₁ = m, b₂ = n, b₃ = l and c₁ = n, c₂ = l and c₃ = m

Substituting the values we get

n(ml - n²) + l(nm - l²) +m(ln - m²) = 0

This implies l³ + m³ + n³ - 3lmn = 0

That is (l + m+ n)(l² + m² + n² -lm -mn - nl) = 0

This implies l + m + n = 0
Question 6
1 / -0

The equations of the lines through ( - 1 , - 1 ) and making angles of 45⁰ with the line x + y = 0 are

A
x + 1 = 0 , y + 1 = 0

B
x – 1 = 0 , y – x = 0

C
x + y = 0 , y + 1 = 0

D
none of these

Solution

The lines x+1=0 and y+1=0 are perpendicular to each other.

The slope of the line x+y =0 is -1

Hence the angle made by this line with respect to X axis is 45⁰

In other words the angle made by this line with x+1=0 is 45⁰

Clearly the other line with which it can make 45⁰ is y+1=0
Question 7
1 / -0

Given the 4 lines with equations x + 2y – 3 = 0, 2x + 3y – 4 = 0, 3x + 4y – 5 = 0, 4x + 5y – 6 = 0 , then these lines are

A
the sides of a quadrilateral

B
concurrent

C
sides of a parallelogram

D
none of these

Solution

The lines are concurrent

On solving the lies 1 and 2 we get the point of intersection as (-1,2)

Similarly on solving lines 2 and 3, the point of intersection is (-1,2)

Similarly solving the lines 3 and 4, the point of intersection is (-1,2)

on solving lines 1 and 4 the point of intersection is (-1,2)

Since the point of intersection is the same for all the lines, the lines are concurrent.
Question 8
1 / -0

The locus of the equation xy = 0 is

A
a pair of perpendicular lines

B
a straight line

C
a hyperbola

D
none of these

Solution

The equation of the pair of straight lines paasing through the origin is fiven by is given by ax² + 2hxy + by² =0

The condition for perpendicularity between the pair of lines is a+b = 0

Hence the locus of the pair of these perpendicular straight lines should be xy = 0
Question 9
1 / -0

The locus of the inequation xy ≥ 0 is

A
a pair of straight lines

B
a straight line

C
the set of all points either in the 1st quadrant or in the 3rd quadrant including the points on coordinate axis

D
none of these

Solution

It is the set of all points either in the 1st quadrant or in the 3rd quadrant including the points on coordiate axis. This is because the inequality ≥ indicates that the points belong either to 1st or 3rd quadrant.
Question 10
1 / -0

The diagram shows two points, $$M$$ and $$N$$ on a Cartesian plane.
Calculate the distance between $$M$$ and $$N$$.

A
$$\sqrt{10}$$ units

B
$$\sqrt{12}$$ units

C
$$\sqrt{20}$$ units

D
$$\sqrt{52}$$ units

Solution

Coordinates of $$M$$ and $$N $$ are $$(1,3)$$ & $$(5,1)$$ respectively.
So $$MN=\sqrt{(1-5)^{2}+(3-1)^{2}}=\sqrt{20}$$ units
Question 11
1 / -0

A student moves $$\sqrt {2x} km$$ east from his residence and then moves x km north. He then goes x km north east and finally he takes a turn of $$90^{\circ}$$ towards right and moves a distance x km and reaches his school. What is the shortest distance of the school from his residence?

A
$$(2\sqrt {2} + 1) x\ km$$

B
$$3x\ km$$

C
$$2\sqrt {2}x\ km$$

D
$$3\sqrt {2}x\ km$$

Solution

In triangle $$BCD$$
$$BD^{2} = (x)^{2} + (x)^{2} = 2x^{2}$$
$$BD = x\sqrt {2}$$
As $$AE = BD$$,
$$AE = x\sqrt {2}$$
$$\therefore OE = 2\sqrt {2}x$$
Also $$DE = x$$
$$\therefore OD = \sqrt {x^{2} + (2\sqrt {2x})^{2}} = \sqrt {9x^{2}} = 3x$$
Question 12
1 / -0

On the Cartesian plane. $$PQR$$ is an isosceles triangle.
The perimeter of $$\triangle{PQR}$$ is

A
$$16$$ units

B
$$18$$ units

C
$$20$$ units

D
$$22$$ units

Solution

$$QR=6$$
$$PQ=\sqrt{(5-2)^{2}+4^{2}}$$
$$=\sqrt{3^{2}+4^{2}}=5$$
$$PR=\sqrt{(5-8)^{2}+4^{2}}=5$$
Perimeter = $$PQ+QR+PR$$
$$\Rightarrow 5+5+6=16$$
Question 13
1 / -0

The area of the triangle formed by three vertices $$O(0, 0), A(1, 0), B(0, 1)$$ is _____ sq. units.

A
$$0$$

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

C
$$1$$

D
$$2$$

Solution

Points are given as $$O(0,0), A(1,0), B(0,1)$$.
When plotted on the cartesian plane, these points make a right angled triangle $$OAB$$ with
$$OA=1$$, $$OB=1$$
Area of this right angled triangle $$=\dfrac12\times OA\times OB$$
$$=\cfrac12\times1\times1\ sq.unit$$
$$=\cfrac12\ sq.unit$$
Question 14
1 / -0

Distance between the points $$(2,-3)$$ and $$(5,a)$$ is $$5$$. Hence the value of $$a=$$............

A
$$-1$$

B
$$6$$

C
$$1$$

D
$$7$$

Solution

Let the two points be $$A(2,-3)$$ and $$B(5,a)$$
$$|A-B|=5$$
$$\implies |A-B|^2=25$$
$$\implies {(2-5)}^{2}+{(-3-a)}^{2}=25$$
$$\implies 9+9+6a+{a}^{2}=25$$
$$\implies {a}^{2}+6a-7=0$$
$$\implies {a}^{2}+7a-a-7=0$$
$$\implies (a+7)(a-1)=0$$
$$\implies a=-7$$ or $$a=1$$
But distance can't be negative
Hence $$a=1$$
Question 15
1 / -0

The points $$(5,1), (1,-1)$$ and $$(11,4)$$ are:

A
Collinear

B
Vertices of right angled triangle

C
Vertices of equilateral triangle

D
Vertices of isosceles triangle

Solution

Let the points be $$A(5,1), B(1,-1),$$ and $$C(11,4)$$.
Then,
$$AB=2\sqrt { 5 } ,BC=5\sqrt { 5 }$$ and $$CA=3\sqrt { 5 } $$
Thus, $$AB+CA=BC$$
Therefore, $$A, B,$$ and $$C$$ are collinear.
Question 16
1 / -0

If a plane has X-intercept $$l$$, Y-intercept $$m$$ and Z-intercept $$n$$, and perpendicular distance of plane from origin is $$k$$, then ?

A
$$\displaystyle l^2+m^2+n^2=\frac{1}{k^2}$$

B
$$\displaystyle\frac{1}{l^2}+\frac{1}{m^2}+\frac{1}{n^2}=\frac{1}{k^2}$$

C
$$l^2+m^2+n^2=k^2$$

D
$$\displaystyle\frac{1}{l^2}+\frac{1}{m^2}+\frac{1}{n^2}=k^2$$

Solution

The equation of plane is
$$\dfrac{x}{l}+\dfrac{y}{m}+\dfrac{z}{n}=1$$

$$\implies \dfrac{x}{l}+\dfrac{y}{m}+\dfrac{z}{n}-1=0$$

Distance from origin $$=\dfrac{|\dfrac{0}{l}+\dfrac{0}{m}+\dfrac{0}{n}-1|}{\sqrt{{(\dfrac{1}{l})^2+(\dfrac{1}{m})^2+(\dfrac{1}{n})^2}}}$$

$$\implies k = \dfrac{1}{\sqrt{{(\dfrac{1}{l})^2+(\dfrac{1}{m})^2+(\dfrac{1}{n})^2}}}$$

$$\implies \dfrac{1}{l^2}+\dfrac{1}{m^2}+\dfrac{1}{n^2}=\dfrac{1}{k^2}$$

Hence, the answer is option (B).
Question 17
1 / -0

If the angle $$\theta$$ gives the inclination of a line, then the slope of that line is given by :

A
$$\sin(\theta)$$

B
$$\cos(\theta)$$

C
$$\tan(\theta)$$

D
None of the above.

Solution

Slope of the line is the tangent of the angle it makes with the positive x axis (inclination).
Question 18
1 / -0

A light ray emerging from the point source placed at $$P(2, 3)$$ is reflected at a point $$Q$$ on the y-axis. It then passes through the point $$R(5, 10)$$. The coordinates of $$Q$$ are

A
$$(0, 3)$$

B
$$(0, 2)$$

C
$$(0, 5)$$

D
None of these

Solution

Let the coordinates of $$Q$$ be $$(0,a)$$
$$\therefore$$ Angle which $$PQ$$ makes with X-axis is same as angle which $$QR$$ makes with X-axis.
$$\cfrac{3-a}{2-0}=-\cfrac{10-a}{5-0} \\ \Rightarrow 5(3-a)=-2(10-a) \\ \Rightarrow 15-5a=-20+2a \\ \Rightarrow 7a=35 \\ \Rightarrow a=5$$
Hence coordinates of $$Q$$ are $$(0,5)$$
Question 19
1 / -0

The graph of $$x = 5$$ is perpendicular to _______.

A
x-axis

B
y-axis

C
Line $$y = x$$

D
Line $$y = -x$$

Solution

General equation of a straight line is $$ax + by + c = 0$$ where slope of line is $$\dfrac{-a}{b}$$
Equation of line $$: x = 5$$
Here slope is $$\dfrac{1}{0}$$ which isn't defined, hence line makes angle of $$90^{\circ}$$ with x-axis.

$$\therefore$$ Line is perpendicular to x-axis.

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Straight Lines Test -5

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Question 1

x + 2y = 0

x – y = 0

x + y + 0

none of these

Solution

Question 2

1th quadrant

3rd quadrant

2nd quadrant

4th quadrant

Solution

Question 3

9

4

8

7

Solution

Question 4

- 1

1

2

none of these

Solution

Question 5

l – m – n = 0

l + m + n = 0

l + m – n = 0

I – m + n = 0

Solution

Question 6

x + 1 = 0 , y + 1 = 0

x – 1 = 0 , y – x = 0

x + y = 0 , y + 1 = 0

none of these

Solution

Question 7

the sides of a quadrilateral

concurrent

sides of a parallelogram

none of these

Solution

Question 8

a pair of perpendicular lines

a straight line

a hyperbola

none of these

Solution

Question 9

a pair of straight lines

a straight line

the set of all points either in the 1st quadrant or in the 3rd quadrant including the points on coordinate axis

none of these

Solution

Question 10

$$\sqrt{10}$$ units

$$\sqrt{12}$$ units

$$\sqrt{20}$$ units

$$\sqrt{52}$$ units

Solution

Question 11

$$(2\sqrt {2} + 1) x\ km$$

$$3x\ km$$

$$2\sqrt {2}x\ km$$

$$3\sqrt {2}x\ km$$

Solution

Question 12

$$16$$ units

$$18$$ units

$$20$$ units

$$22$$ units

Solution

1^th quadrant

3^rd quadrant

2^nd quadrant

4^th quadrant