RD Chapter 23 The Straight Lines Ex 23.18 Solutions
Question - 1 : - Find the equation of the straight lines passing through the origin andmaking an angle of 45o with the straight line √3x + y = 11.
Answer - 1 : -
Given:
Equation passesthrough (0, 0) and make an angle of 45° with the line √3x + y = 11.
We know that, theequations of two lines passing through a point x1,y1 and making anangle α with the given line y = mx + c are
Question - 2 : - Find the equations to the straight lines which pass through the originand are inclined at an angle of 75o to the straight line x + y+ √3(y – x) = a.
Answer - 2 : -
Given:
The equation passesthrough (0,0) and make an angle of 75° with the line x + y + √3(y – x) = a.
We know that theequations of two lines passing through a point x1,y1 and making anangle α with the given line y = mx + c are
Question - 3 : - Find the equations of straight lines passing through (2, -1) and making an angle of 45o with the line 6x + 5y – 8 = 0.
Answer - 3 : -
Given:
The equation passesthrough (2,-1) and make an angle of 45° with the line 6x + 5y – 8 = 0
We know that theequations of two lines passing through a point x1, y1 andmaking an angle α with the given line y = mx + c are
Here, equation of thegiven line is,
6x + 5y – 8 = 0
5y = – 6x + 8
y = -6x/5 + 8/5
Comparing thisequation with y = mx + c
We get, m = -6/5
Where, x1 =2, y1 = – 1, α = 45°, m = -6/5
So, the equations ofthe required lines are
x + 11y + 9 = 0 and11x – y – 23 = 0
∴ The equation of givenline is x + 11y + 9 = 0 and 11x – y – 23 = 0
Question - 4 : - Find the equations to the straight lines which pass through the point (h, k) and are inclined at angle tan-1 m to the straight line y = mx + c.
Answer - 4 : -
Given:
The equation passesthrough (h, k) and make an angle of tan-1 m with theline y = mx + c
We know that theequations of two lines passing through a point x1, y1 andmaking an angle α with the given line y = mx + c are
m′ = m
So,
Here,
x1 =h, y1 = k, α = tan-1 m, m′ = m.
So, the equations ofthe required lines are
Question - 5 : - Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of 450 to the lines 3x + y – 5 = 0.
Answer - 5 : -
Given:
The equation passesthrough (2, 3) and make an angle of 450with the line 3x + y – 5= 0.
We know that theequations of two lines passing through a point x1,y1 and making anangle α with the given line y = mx + c are
Here,
Equation of the givenline is,
3x + y – 5 = 0
y = – 3x + 5
Comparing thisequation with y = mx + c we get, m = – 3
x1 =2, y1 = 3, α = 45∘, m = – 3.
So, the equations ofthe required lines are
x + 2y – 8 = 0 and 2x– y – 1 = 0
∴ The equation of givenline is x + 2y – 8 = 0 and 2x – y – 1 = 0