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RD Chapter 23- The Straight Lines Ex-23.4 Interview Questions Answers

Question 1 : Find the equation of the straight line passing through the point (6, 2) and having slope – 3.

Answer 1 :

Given, A straight linepassing through the point (6, 2) and the slope is – 3

By using the formula,

The equation of lineis [y – y1 = m(x – x1)]

Here, the line ispassing through (6, 2)

It is given that, theslope of line, m = –3

Coordinates of lineare (x1, y1) = (6,2)

The equation of line =y – y1 = m(x – x1)

Now, substitute thevalues, we get

y – 2 = – 3(x – 6)

y – 2 = – 3x + 18

y + 3x – 20 = 0

The equation of lineis 3x + y – 20 = 0

Question 2 : Find the equation of the straight line passing through (–2, 3) and indicated at an angle of 45° with the x – axis.

Answer 2 :

Given:

A line which ispassing through (–2, 3), the angle is 45o.

By using the formula,

The equation of lineis [y – y1 = m(x – x1)]

Here, angle, θ =45o

The slope of the line,m = tan θ

m = tan 45o

= 1

The line passingthrough (x1, y1) = (–2, 3)

The required equationof line is y – y1 = m(x – x1)

Now, substitute thevalues, we get

y – 3 = 1(x – (– 2))

y – 3 = x + 2

x – y + 5 = 0

The equation of lineis x – y + 5 = 0

Question 3 : Find the equation of the line passing through (0, 0) with slope m

Answer 3 :

Given:

A straight linepassing through the point (0, 0) and slope is m.

By using the formula,

The equation of lineis [y – y1 = m(x – x1)]

It is given that, theline is passing through (0, 0) and the slope of line, m = m

Coordinates of lineare (x1, y1) = (0, 0)

The equation of line =y – y1 = m(x – x1)

Now, substitute thevalues, we get

y – 0 = m(x – 0)

y = mx

The equation of lineis y = mx.

Question 4 : Find the equation of the line passing through (2, 2√3) and inclined with x – axis at an angle of 75o.

Answer 4 :

Given:

A line which ispassing through (2, 2√3), the angle is 75o.

By using the formula,

The equation of lineis [y – y1 = m(x – x1)]

Here, angle, θ =75o

The slope of the line,m = tan θ

m = tan 75o

= 3.73 = 2 + √3

The line passingthrough (x1, y1) = (2, 2√3)

The required equationof the line is y – y1 = m(x – x1)

Now, substitute thevalues, we get

y – 2√3 = 2+ √3 (x – 2)

y – 2√3 = (2+ √3)x – 7.46

(2 + √3)x – y – 4= 0

The equation of theline is (2 + √3)x – y – 4 = 0

Question 5 : Find the equation of the straight line which passes through the point (1, 2) and makes such an angle with the positive direction of x – axis whose sine is 3/5.

Answer 5 :

A line which ispassing through (1, 2)

To Find: The equationof a straight line.

By using the formula,

The equation of lineis [y – y1 = m(x – x1)]

Here, sin θ = 3/5

We know, sin θ =perpendicular/hypotenuse

= 3/5

So, according toPythagoras theorem,

(Hypotenuse)2 =(Base)2 + (Perpendicular)2

(5)2 =(Base)2 + (3)2

(Base) = √(25 – 9)

(Base)2 =√16

Base = 4

Hence, tan θ =perpendicular/base

= 3/4

The slope of the line,m = tan θ

= 3/4

The line passingthrough (x1,y1) = (1,2)

The required equationof line is y – y1 = m(x – x1)

Now, substitute thevalues, we get

y – 2= (¾) (x – 1)

4y – 8 = 3x – 3

3x – 4y + 5 = 0

The equation of lineis 3x – 4y + 5 = 0


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RD Chapter 23- The Straight Lines Ex-23.4 Contributors

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