RD Chapter 23 The Straight Lines Ex 23.7 Solutions
Question - 1 : - Find the equation of a line for which
(i) p = 5, α = 60°
(ii) p = 4, α = 150°
Answer - 1 : -
(i) p = 5, α = 60°
Given:
p = 5, α = 60°
The equation of theline in normal form is given by
Using the formula,
x cos α + y sin α= p
Now, substitute thevalues, we get
x cos 60° + y sin 60°= 5
x/2 + √3y/2 = 5
x + √3y = 10
∴ The equation of linein normal form is x + √3y = 10.
(ii) p = 4, α = 150°
Given:
p = 4, α = 150°
The equation of theline in normal form is given by
Using the formula,
x cos α + y sin α= p
Now, substitute thevalues, we get
x cos 150° + y sin150° = 4
cos (180° – θ) =– cos θ , sin (180° – θ) = sin θ
x cos(180° – 30°)+ y sin(180° – 30°) = 4
– x cos 30° + ysin 30° = 4
–√3x/2 + y/2 = 4
-√3x + y = 8
∴ The equation of linein normal form is -√3x + y= 8.
Question - 2 : - Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x–axis is 30°.
Answer - 2 : -
Given:
p = 4, α = 30°
The equation of theline in normal form is given by
Using the formula,
x cos α + y sin α= p
Now, substitute thevalues, we get
x cos 30° + y sin 30°= 4
x√3/2 + y1/2 = 4
√3x + y = 8
∴ The equation of linein normal form is √3x + y = 8.
Question - 3 : - Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x–axis is 15°.
Answer - 3 : -
Given:
p = 4, α = 15°
The equation of theline in normal form is given by
We know that, cos15° = cos (45° – 30°) = cos45°cos30° + sin45°sin30°
Cos (A – B) = cos Acos B + sin A sin B
So,
And sin 15 = sin(45° – 30°) = sin 45° cos 30° – cos 45° sin 30°
Sin (A – B) = sin Acos B – cos A sin B
So,
Now, by using theformula,
x cos α + y sin α= p
Now, substitute thevalues, we get
(√3+1)x +(√3-1) y =8√2
∴ The equation of linein normal form is (√3+1)x +(√3-1) y = 8√2.
Question - 4 : - Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle α given by tan α = 5/12 with the positive direction of x–axis.
Answer - 4 : -
Given:
p = 3, α = tan-1 (5/12)
So, tan α = 5/12
sin α = 5/13
cos α = 12/13
The equation of theline in normal form is given by
By using the formula,
x cos α + y sin α= p
Now, substitute thevalues, we get
12x/13 + 5y/13 = 3
12x + 5y = 39
∴ The equation of linein normal form is 12x + 5y = 39.
Question - 5 : - Find the equation of the straight line on which the length of the perpendicular from the origin is 2 and the perpendicular makes an angle α with x–axis such that sin α = 1/3.
Answer - 5 : -
Given:
p = 2, sin α = 1/3
We know that cos α =√(1 – sin2 α)
= √(1 – 1/9)
= 2√2/3
The equation of theline in normal form is given by
By using the formula,
x cos α + y sin α= p
Now, substitute thevalues, we get
x2√2/3 + y/3 = 2
2√2x + y = 6
∴ The equation of linein normal form is 2√2x + y = 6.