Straight Lines Test -6

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Straight Lines Test -6

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Question 1
1 / -0

The image of the point ( -1 , 2 ) in the origin is

A
( 1 , -2 )

B
( -1 , -2 )

C
( 2 , - 1 )

D
none of these

Solution

when a point M which is in the 2nd quadrant is reflected in the origin, its image is formed in the 4th quadrant whose coordinates are (x,-y)

Hence the image of the point (1,-2) is (1,-2)
Question 2
1 / -0

The orthogonal projection of the point ( 2 , - 3 ) on the line x + y = 0 is

A
( 5/2 , - 5/2 )

B
( 2 , 3 )

C
( 2 , - 3 )

D
( -2 , -3 )

Solution

Equation of the line which perpendicular to the given line is x - y + k = 0

Since this line passes through (2,-3)

2-(-3) + k = 0

This implies k= -5

Hence the equation og the line is x - y =5

On solving the lines x+y=0 and x-y=5, we get the point of intersection as x = 5/2 and y = -5/2

Hence (5/2,-5/2) is the coordinates of orthogonal projection.
Question 3
1 / -0

Three points A , B and C are collinear if the area of triangle ABC is

A
less than zero

B
greater than zero

C
zero

D
none of these

Solution

Only non collinear points can form a triangle. Hence if the three points are collinear a triangle cannot be formed, hence the area of the triangle is zero
Question 4
1 / -0

The number of points on X axis which are at a distance of c units ( c < 3 ) from ( 2 , 3 ) is

A
0

B
1

C
2

D
3

Solution

Let (x,0) be the point.

Then (by distance formula,

(x-2)² +3² = c²

x² -4x +4 + 9 - c² = 0

x² - 4x +13 - c² = 0

x will be real if the b² - 4ac ≥≥0

i.e; 16 - 4(13 - c²) ≥≥ 0

i.e; c² - 9 ≥≥ 0

i.e; |c| ≥≥ 3

Hence there is no point.
Question 5
1 / -0

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

A
x – 1 = 0 , y – 1 = 0

B
x – 1 = 0 , x – y = 0

C
x – y = 0 , y – 1 = 0

D
none of these

Solution

If the lines make equal angles of 45⁰ with the given line, x+y =0.

Then these lines must be perpendicular with each other.

This is possible only when the two lines are parallel to X axis and Y axis.

That is the equations should be x = a constant and y = a constant.

Since it passes through (1,1)

The equations should be x = 1 or x-1=0 and y=1 or y-1 =0
Question 6
1 / -0

The number of lines that are parallel to 2x + 6y – 7 = 0 and have an intercept 10 units between the coordinate axis is :

A
1

B
2

C
3

D
none of these

Solution

Thslope of the given line 2x+6y = 7 is -1/3

Hence the line which is parallel to the above line is

y = (-1/3)x+c

That is the y intercept is (0,c) and the x intercept is (3c,0)

Using the distance formula

d² = (0-3c)² + (3c-0)²

= 10c²

since the distance is 10 is given,

100 = 10c²

therefore c = ±±10

Since two values are possible, two lines can be drawn.
Question 7
1 / -0

The line ( p + 2q ) x + ( p – 3q ) y = p - q for different values of p and q passes through the fixed point

A
( 2/5 , 3/5 )

B
( 3/2 , 5/2 )

C
(3/5 , 3/5 )

D
(2/5 , 2/5 )

Solution

Expanding the given equation

px+2qx+py-3qy = p-q

px+py+2qx-3qy = p-q

p(x+y) -q(-2x+3y) = p-q

Equating the coeffiecients of like terms

x+y=1 and -2x+3y=1

On solving both the equations we get,

x = 2/5 and y = 3/5

Hence the line passes through the fixed point (2/5.3/5)
Question 8
1 / -0

How far is the line 3x - 4y + 15 = 0 from the origin?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

E
Answer Required

Solution

Distance of $$O(0,0)$$ from the line $$3x-14y+15=0$$ is

$$=>r=\dfrac{|3(0)-4(0)+15|}{\sqrt {3^3+4^2}}$$

$$=>r=\dfrac{15}{\sqrt {25}}$$

$$=>r=\dfrac{15}{5}$$

$$=>r=3$$
Question 9
1 / -0

The point $$(0,\,5)$$ lies on the line $$x+y=5$$ or not

A
True

B
False

C
Cant say

D
None

Solution

Given the line $$x+y=5$$
when$$ x=0 ,y =5 \Rightarrow(0,5)$$
$$\therefore$$ The point (0,5) lies on the line$$ x+y=5$$
Question 10
1 / -0

Which of the following is perpendicular to the line x/3 + y/4 = 1?

A
x - 4y - 8 = 0

B
4x - 3y - 6 = 0

C
3x - 4y - 5 = 0

D
4x + 3y - 11 = 0

Solution

Line is
$$\frac{x}{3}+\frac{y}{4}=1$$
$$=>4x+3y=12$$
$$=>3y=-4x+12$$
$$=>y=\frac{-4}{3}+4$$
Here $$m=\frac{-4}{3}$$
$$\therefore$$ Slope of line perpendicular to given line is $$\frac{3}{4}$$
Option A
$$x-4y-8=0$$
$$=>y=\frac{-x}{4}-\frac{8}{4}$$ (not perpendicular)
Option B
$$4x-3y-6=0$$
$$=>y=\frac{4x}{3}-\frac{6}{3}$$ (not perpendicular)
Option C
$$3x-4y-11=0$$
$$=>y=\frac{3}{4}x-\frac{5}{4}$$ (perpendicular)
Option D
$$4x+3y-11=0$$
$$=>y=\frac{-4}{3}x+\frac{11}{4}$$ (not perpendicular)
Question 11
1 / -0

Find the distance from the point (5, -3) to the line 7x - 4y - 28 = 0

A
$$2.62$$

B
$$2.36$$

C
$$2.48$$

D
$$2.54$$

E
none of these

Solution

Distance of point $$(5,-3)$$ from the line $$7x-4y-28=0$$ is

$$=>r=\dfrac{|7(5)-4(-3)-28|}{\sqrt {7^2+4^2}}$$

$$=>r=\dfrac{|35+12-281|}{\sqrt {49+16}}$$

$$=>r=\dfrac{19}{\sqrt 65}$$

$$=>r=\dfrac{19}{8.06}$$

$$=>r=2.36$$
Question 12
1 / -0

Determine the distance from (5, 10) to the line x - y = 0

A
$$3.86$$

B
$$3.54$$

C
$$3.68$$

D
$$3.72$$

E
none of these

Solution

Distance of point $$(5,10)$$ from the line $$x-y=0$$ is

$$r=\dfrac{|5-10|}{\sqrt {1^2+1^2}}$$

$$=>r=\dfrac{|-5|}{\sqrt 2}$$

$$=>r=\dfrac{5}{\sqrt 2}$$

$$=>r=3.54$$
Question 13
1 / -0

A line is of length $$10$$ and one end is $$(2,-3)$$. If the abscissa of the other end is $$10$$, then find its ordinate.

A
$$3$$ or $$-9$$

B
$$-3$$ or $$-9$$

C
$$3$$ or $$9$$

D
None of these

Solution

Coordinates are $$(10,y)$$
$$\Rightarrow \sqrt{(10-2)^{2}+(y+3)^{2}}=10$$
$$\Rightarrow (y+3)^{2}=6^{2}\Rightarrow y+3=\pm 6$$
$$y=3$$ or $$-9$$
Question 14
1 / -0

The distance between $$M(-1,5)$$ and $$N(x,5)$$ is $$8$$ units. The value of $$x$$ is:

A
$$-9$$ or $$9$$

B
$$-7$$ or $$9$$

C
$$-9$$ or $$7$$

D
$$-7$$ or $$-9$$

Solution

$$\sqrt{[x-(-1)]^{2}+(5-5)^{2}}=8$$
$$\Rightarrow (x+1)^{2}=8^{2}\Rightarrow x+1=\pm 8$$
$$\therefore x=-9,7$$
Question 15
1 / -0

In the diagram $$MN$$, is a straight line.
The distance between $$M$$ and $$N$$ is:

A
$$6$$ units

B
$$8$$ units

C
$$9$$ units

D
$$10$$ units

Solution

$$\sqrt{(3+3)^{2}+(2+6)^{2}}$$
$$\Rightarrow \sqrt{36+64}\Rightarrow 10$$
Question 16
1 / -0

$$P$$ is the point $$(-5,3)$$ and $$Q$$ is the point $$(-5,m)$$. If the length of the straight line $$PQ$$ is $$8$$ units, then the possible value of $$m$$ is:

A
$$-5$$ or $$5$$

B
$$-5$$ or $$11$$

C
$$-5$$ or $$-11$$

D
$$5$$ or $$11$$

Solution

$$\sqrt{(-5-(-5))^{2}+(m-3)^{2}}=8$$
$$\Rightarrow (m-3)^{2}=8^{2}$$
$$\Rightarrow m-3=\pm 8$$
$$\Rightarrow m=-5+11$$
Question 17
1 / -0

In the diagram, $$PQ=PR$$
Find the value of $$h$$.

A
$$1$$

B
$$1\cfrac{1}{2}$$

C
$$2$$

D
$$3$$

Solution

As PQ=PR
$$\sqrt{(h+2)^{2}+(5-1)^{2}}=\sqrt{(h-4)^{2}+(5-1)^{2}}$$
squaring both sides
$$(h+2)^{2}+4^{2}=(h-4)^{2}+4^{2}$$
$$h^{2}+4+4h=h^{2}+16-8h$$
$$\Rightarrow 12h=12\Rightarrow h=1$$

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Straight Lines Test -6

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Straight Lines Test -6

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Question 1

( 1 , -2 )

( -1 , -2 )

( 2 , - 1 )

none of these

Solution

Question 2

( 5/2 , - 5/2 )

( 2 , 3 )

( 2 , - 3 )

( -2 , -3 )

Solution

Question 3

less than zero

greater than zero

zero

none of these

Solution

Question 4

0

1

2

3

Solution

Question 5

x – 1 = 0 , y – 1 = 0

x – 1 = 0 , x – y = 0

x – y = 0 , y – 1 = 0

none of these

Solution

Question 6

1

2

3

none of these

Solution

Question 7

( 2/5 , 3/5 )

( 3/2 , 5/2 )

(3/5 , 3/5 )

(2/5 , 2/5 )

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Answer Required

Solution

Question 9

True

False

Cant say

None

Solution

Question 10

x - 4y - 8 = 0

4x - 3y - 6 = 0

3x - 4y - 5 = 0

4x + 3y - 11 = 0

Solution

Question 11

$$2.62$$

$$2.36$$

$$2.48$$

$$2.54$$

none of these

Solution

Question 12

$$3.86$$

$$3.54$$

$$3.68$$