RD Chapter 23 The Straight Lines Ex 23.4 Solutions
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