RD Chapter 23 The Straight Lines Ex 23.16 Solutions
Question - 1 : - Determine the distance between the following pair of parallel lines:
(i) 4x – 3y – 9 = 0 and 4x – 3y – 24 = 0
(ii) 8x + 15y – 34 = 0 and 8x + 15y + 31 = 0
Answer - 1 : -
(i) 4x – 3y – 9 = 0 and 4x– 3y – 24 = 0
Given:
The parallel lines are
4x − 3y − 9= 0 … (1)
4x − 3y − 24= 0 … (2)
Let d be the distancebetween the given lines.
So,
∴ The distance betweengivens parallel line is 3units.
(ii) 8x + 15y – 34 = 0 and8x + 15y + 31 = 0
Given:
The parallel lines are
8x +15y − 34 = 0 … (1)
8x + 15y + 31 = 0 …(2)
Let d be the distancebetween the given lines.
So,
∴ The distance betweengivens parallel line is 65/17 units.
Question - 2 : - Find the equation of two straight lines which are parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1).
Answer - 2 : -
Given:
The equation is parallelto x + 7y + 2 = 0 and at unit distance from the point (1, -1)
The equation of givenline is
x + 7y + 2 = 0 … (1)
The equation of a lineparallel to line x + 7y + 2 = 0 is given below:
x + 7y + λ =0 … (2)
The line x + 7y+ λ = 0 is at a unit distance from the point (1, − 1).
So,
1 =
λ – 6 = ± 5√2
λ = 6 + 5√2, 6 – 5√2
now, substitute thevalue of λ back in equation x + 7y + λ = 0, we get
x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2
∴ The required lines:
x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2
Question - 3 : - Prove that the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6.
Answer - 3 : -
Given:
The lines A, 2x + 3y =19 and B, 2x + 3y + 7 = 0 also a line C, 2x + 3y = 6.
Let d1 bethe distance between lines 2x + 3y = 19 and 2x + 3y = 6,
While d2 isthe distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6
Hence proved, thelines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y =6
Question - 4 : - Find the equation of the line mid-way between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Answer - 4 : -
Given:
9x + 6y – 7 = 0 and 3x+ 2y + 6 = 0 are parallel lines
The given equations ofthe lines can be written as:
3x + 2y – 7/3 =0 … (1)
3x + 2y + 6 = 0 … (2)
Let the equation ofthe line midway between the parallel lines (1) and (2) be
3x + 2y+ λ = 0 … (3)
The distance between(1) and (3) and the distance between (2) and (3) are equal.
Now substitute thevalue of λ back in equation 3x + 2y + λ = 0, we get
3x + 2y+ 11/6 = 0
By taking LCM
18x + 12y + 11 = 0
∴ The required equationof line is 18x + 12y + 11 = 0