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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