RD Chapter 23 The Straight Lines Ex 23.3 Solutions
Question - 1 : - Find the equation of a line making an angle of 150° with the x–axis and cutting off an intercept 2 from y–axis.
Answer - 1 : -
Given: A line whichmakes an angle of 150o with the x–axis and cutting off anintercept at 2
By using the formula,
The equation of a lineis y = mx + c
We know thatangle, θ = 150o
The slope of the line,m = tan θ
Where, m = tan 150o
= -1/ √3
Coordinate ofy–intercept is (0, 2)
The required equationof the line is y = mx + c
Now substitute thevalues, we get
y = -x/√3 + 2
√3y – 2√3 + x = 0
x + √3y = 2√3
∴ The equation of lineis x + √3y = 2√3
Question - 2 : - Find the equation of a straight line:
(i) with slope 2 and y – intercept 3;
(ii) with slope – 1/ 3 and y – intercept – 4.
(iii) with slope – 2 and intersecting the x–axis at a distance of 3 units to the left of origin.
Answer - 2 : -
(i) With slope 2 and y –intercept 3
The slope is 2 and thecoordinates are (0, 3)
Now, the requiredequation of line is
y = mx + c
Substitute the values,we get
y = 2x + 3
(ii) With slope – 1/ 3 andy – intercept – 4
The slope is – 1/3 andthe coordinates are (0, – 4)
Now, the requiredequation of line is
y = mx + c
Substitute the values,we get
y = -1/3x – 4
3y + x = – 12
(iii) With slope – 2 andintersecting the x–axis at a distance of 3 units to the left of origin
The slope is – 2 andthe coordinates are (– 3, 0)
Now, the requiredequation of line is y – y1 = m (x – x1)
Substitute the values,we get
y – 0 = – 2(x + 3)
y = – 2x – 6
2x + y + 6 = 0
Question - 3 : - Find the equations of the bisectors of the angles between the coordinate axes.
Answer - 3 : -
There are two bisectorsof the coordinate axes.
Their inclinationswith the positive x-axis are 45o and 135o
The slope of thebisector is m = tan 45o or m = tan 135o
i.e., m = 1 or m = -1,c = 0
By using the formula,y = mx + c
Now, substitute thevalues of m and c, we get
y = x + 0
x – y = 0 or y = -x +0
x + y = 0
∴ The equation of thebisector is x ± y = 0
Question - 4 : - Find the equation of a line which makes an angle of tan – 1 (3) with the x–axis and cuts off an intercept of 4 units on the negative direction of y–axis.
Answer - 4 : -
Given:
The equation whichmakes an angle of tan – 1(3) with the x–axis and cuts off anintercept of 4 units on the negative direction of y–axis
By using the formula,
The equation of theline is y = mx + c
Here,angle θ = tan – 1(3)
So, tan θ = 3
The slope of the lineis, m = 3
And, Intercept in thenegative direction of y–axis is (0, -4)
The required equationof the line is y = mx + c
Now, substitute thevalues, we get
y = 3x – 4
∴ The equation of theline is y = 3x – 4.
Question - 5 : - Find the equation of a line that has y – intercept – 4 and is parallel to the line joining (2, –5) and (1, 2).
Answer - 5 : -
Given:
A line segment joining(2, – 5) and (1, 2) if it cuts off an intercept – 4 from y–axis
By using the formula,
The equation of lineis y = mx + C
It is given that, c =– 4
Slope of line joining(x1 – x2) and (y1 – y2),
So, Slope of linejoining (2, – 5) and (1, 2),
m = – 7
The equation of lineis y = mx + c
Now, substitute thevalues, we get
y = –7x – 4
y + 7x + 4 = 0
∴ The equation of lineis y + 7x + 4 = 0.