RD Chapter 23 The Straight Lines Ex 23.12 Solutions
Question - 1 : - Find the equation of a line passing through the point (2, 3) and parallel to the line 3x– 4y + 5 = 0.
Answer - 1 : -
Given:
The equation isparallel to 3x − 4y + 5 = 0 and pass through (2, 3)
The equation of theline parallel to 3x − 4y + 5 = 0 is
3x – 4y+ λ = 0,
Where, λ isa constant.
It passes through (2,3).
Substitute the valuesin above equation, we get
3 (2) – 4 (3) + λ = 0
6 – 12 + λ =0
λ = 6
Now, substitute thevalue of λ = 6 in 3x – 4y + λ = 0, we get
3x − 4y + 6
∴ The required line is3x − 4y + 6 = 0.
Question - 2 : - Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0.
Answer - 2 : -
Given:
The equation isperpendicular to x – 3y + 5 = 0 and passes through (3,-2)
The equation of theline perpendicular to x − 3y + 5 = 0 is
3x + y + λ =0,
Where, λ isa constant.
It passes through(3, − 2).
Substitute the valuesin above equation, we get
3 (3) + (-2) +λ = 0
9 – 2 + λ =0
λ = – 7
Now, substitute thevalue of λ = − 7 in 3x + y + λ = 0, we get
3x + y – 7 = 0
∴ The required line is3x + y – 7 = 0.
Question - 3 : - Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).
Answer - 3 : -
Given:
A (1, 3) and B (3, 1)be the points joining the perpendicular bisector
Let C be the midpointof AB.
So, coordinates of C =[(1+3)/2, (3+1)/2]
= (2, 2)
Slope of AB = [(1-3) /(3-1)]
= -1
Slope of theperpendicular bisector of AB = 1
Thus, the equation ofthe perpendicular bisector of AB is given as,
y – 2 = 1(x – 2)
y = x
x – y = 0
∴ The required equationis y = x.
Question - 4 : - Find the equations of the altitudes of a ΔABC whose vertices are A (1, 4), B (-3, 2) and C (-5, -3).
Answer - 4 : -
Given:
The verticesof ∆ABC are A (1, 4), B (− 3, 2) and C (− 5, − 3).
Now let us find theslopes of ∆ABC.
Slope of AB = [(2 – 4)/ (-3-1)]
= ½
Slope of BC = [(-3 –2) / (-5+3)]
= 5/2
Slope of CA = [(4 + 3)/ (1 + 5)]
= 7/6
Thus, we have:
Slope of CF = -2
Slope of AD = -2/5
Slope of BE = -6/7
Hence,
Equation of CF is:
y + 3 = -2(x + 5)
y + 3 = -2x – 10
2x + y + 13 = 0
Equation of AD is:
y – 4 = (-2/5) (x – 1)
5y – 20 = -2x + 2
2x + 5y – 22 = 0
Equation of BE is:
y – 2 = (-6/7) (x + 3)
7y – 14 = -6x – 18
6x + 7y + 4 = 0
∴ The requiredequations are 2x + y + 13 = 0, 2x + 5y – 22 = 0, 6x + 7y + 4 = 0.
Question - 5 : - Find the equation of a line which is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.
Answer - 5 : -
Given:
The equation isperpendicular to √3x – y + 5 = 0 equation and cuts off anintercept of 4 units with the negative direction of y-axis.
The line perpendicularto √3x – y + 5 = 0 is x + √3y + λ = 0
It is given that theline x + √3y + λ = 0 cuts off an intercept of 4 units with thenegative direction of the y-axis.
This means that theline passes through (0,-4).
So,
Let us substitute thevalues in the equation x + √3y + λ = 0, we get
0 – √3 (4) + λ = 0
λ = 4√3
Now, substitute thevalue of λ back, we get
x + √3y + 4√3 = 0
∴ The required equationof line is x + √3y + 4√3 = 0.