Straight Lines Class 11 Notes

Chapter 10
Straight Lines Class 11 Notes

Q. Radius of the circle is 5 cm. find the equation of the tangent line

x cosω + y sinω = p

x cos600 + y sin600 = 5

x .1/2 + y. √3/2 = 5

x+ √3 y = 10

General equation of a line

The general equation of a line is Ax + By + C = 0

Slope = (-A)/B

x intercept = (-C)/A

y intercept = (-C)/B

Q.   Find the slope, x intercept and y intercept of the line 2x+3y – 4 = 0

Slope = (-A)/B=  (-2)/3

x intercept = (-C )/A =  4/2 = 2

y intercept = (-C)/B  = 4/3

Q. Find the equation of a line perpendicular to x – 7y +5 = 0 and having x intercept 3

Slope of the given line = (-A)/B=  (-1)/(-7) = 1/7

Slope of the required line is -7 and passing through (3,0)

y– 0 = -7(x – 3)

y = -7x+21

7x+y-21 = 0

Distance between parallel lines

The lines Ax+By+C1 = 0 and Ax+By+C2 = 0 are parallel

Distance = |(C_1-C_2)/√(A^2+B^2 )|

Q. The lines 3x-4y+15 = 0 and 3x-4y+5 = 0 are tangents of a cirle. Find the radius of the circle

   Diameter  of the circle is the distance between the given parallel lines

Diameter = |(C_1-C_2)/√(A^2+B^2 )|

                 = |(15-5)/√(3^2+〖(-4)〗^2 )| = 2

Radius = 1

Distance of a point from a line

The distance of the point (x1,y1) from the line Ax+By+C = 0 is

|(Ax_(1+By_1+C))/√(A^2+B^2 )|

Q. The line 3x+4y+3 = 0 is a tangent of a  circle with centre (1,1). Find the radius.

The radius of the circle is the distance fron (1,1) to the tangent 

Radius  = |(Ax_(1+By_1+C))/√(A^2+B^2 )|

                = |(3.1+4.1+3)/√(3^2+4^2 )| = 2

Angle between two lines

The acute angle θ between the lines is given by 

tan θ=|m_(2- m_1 )/(1+m_1 m_2 )| where m1 and m2 are slope of the lines

Q. Find the angle between the lines √(3 ) x + y = 1 and x + √(3 )  y=1

m1 = (-A)/B = -√(3 )

m2 = (-1)/√(3 )

tan θ=|m_(2- m_1 )/(1+m_1 m_2 )|

           =|((-1)/(√(3 )  )+√(3 )  )/(1+(-1)/√(3 ).- √(3 ))|

           = |((-1+3)/√(3 ))/2| = |(2/√(3 ))/2| = 1/√(3 )

θ= 300

Q. If the angle between two line is 450 and slope of one of the lines is 1/2 , find the slope of the other line

θ= 450, m1 = 1/2   we have to find m2

tan θ=|m_(2- m_1 )/(1+m_1 m_2 )|

          tan 450  =|(m- 1/2)/(1+1/2.m)| = |(2m-1)/(2+m)|

1 = |(2m-1)/(2+m)|

(2m-1)/(2+m) = ±1

(2m-1)/(2+m) = 1

2m-1 = 2+m

m = 3

(2m-1)/(2+m) = -1

2m-1 = -(2+m)

2m-1 = -2-m

3m = -1

m = (-1)/3

Q. Reduce the equation x – √3 y + 8 = 0 into normal form

      x – √3 y + 8 = 0

x – √3 y =  – 8

-x + √3 y = 8

 √(a^2+b^2 ) = √(〖(-1)〗^2+〖√3〗^2 ) = 2

(-x)/2 + (√3 y)/2 = 4

x cosω + y sinω = p

cosω = (-1)/2

sinω = √3/2

ω = 1800-600 = 1200

P = 4

x cos120^0 + y sin120^0 = 4

Q. Find the foot of the perpendicular from (3,8) to the line x + 3y -7 = 0

Slope of the given line = (-1)/3

Slope of PQ is 3 and passing through (3,8)

Equation of PQ is

y– y1 = m(x – x1)

y– 8 = 3(x – 3)

y-8 = 3x-9

3x-y-1 = 0

Foot of the perpendicular is the point of intersection of the two lines

9x-3y-3 = 0

10x-10 = 0

10x = 10

X = 1

1+3y -7=0

3y-6 = 0

Y=2

Foot of the perpendicular is (1,2)

R(x,y) is the image

Midpoint of PR is Q

((3+x)/2 , (8+y)/2) = (1,2)

(3+x)/2 = 1

3+x = 2

X = -1

(8+y)/2 = 2

8+y = 4

y = -4

Image is (-1,-4)

Q. Find the equation of a line passing through the point of intersection of the lines 2x+y – 3 = 0and 3x-y-2 = 0 and having slope  4

To find the point of intersection, solve the two equations

On solving the above equations we get x=1 and y =1

The required line passing through (1,1) and having slope 4

y– y1 = m(x – x1)

y– 1 = 4(x – 1)

y -1 = 4x – 4

4x-y-3 = 0

Concurrent lines

Three lines meeting at a point are called concurrent lines.

(m_(2-) m_1)/(1+ m_1 m_2 )

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