RD Chapter 23 The Straight Lines Ex 23.9 Solutions
Question - 1 : - Reduce the equation √3x + y + 2 = 0 to:
(i) slope – intercept form and find slope and y – intercept;
(ii) Intercept form and find intercept on the axes
(iii) The normal form and find p and α.
Answer - 1 : -
(i) Given:
√3x + y + 2 = 0
y = – √3x – 2
This is the slopeintercept form of the given line.
∴ The slope = – √3and y – intercept = -2
(ii) Given:
√3x + y + 2 = 0
√3x + y = -2
Divide both sides by-2, we get
√3x/-2 + y/-2 = 1
∴ The intercept form ofthe given line. Here, x – intercept = – 2/√3 and y – intercept = -2
(iii) Given:
√3x + y + 2 = 0
-√3x – y = 2
Question - 2 : - Reduce the following equations to the normal form and find p and α in each case:
(i) x + √3y – 4 = 0
(ii) x + y + √2 = 0
Answer - 2 : -
(i) x + √3y – 4 = 0
x + √3y = 4
The normal form of thegiven line, where p = 2, cos α = 1/2 and sin α = √3/2
∴ p = 2 and α = π/3
(ii) x + y + √2 = 0
-x – y = √2
The normal form of thegiven line, where p = 1, cos α = -1/√2 and sin α = -1/√2
∴ p = 1 and α = 225o
Question - 3 : - Put the equation x/a + y/b = 1 the slope intercept form and find its slope and y – intercept.
Answer - 3 : -
Given: the equationis x/a + y/b = 1
We know that,
General equation ofline y = mx + c.
bx + ay = ab
ay = – bx + ab
y = -bx/a + b
The slope interceptform of the given line.
∴ Slope = – b/aand y – intercept = b
Question - 4 : - Reduce the lines 3x – 4y + 4 = 0 and 2x + 4y – 5 = 0 to the normal form and hence find which line is nearer to the origin.
Answer - 4 : -
The normal forms ofthe lines 3x − 4y + 4 = 0 and 2x + 4y − 5 = 0.
Let us find, in givennormal form of a line, which is nearer to the origin.
-3x + 4y = 4
Now 2x + 4y = – 5
-2x – 4y = 5
From equations (1) and(2):
45 < 525
∴ The line 3x − 4y+ 4 = 0 is nearer to the origin.
Question - 5 : - Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.
Answer - 5 : -
Given:
The lines 4x + 3y + 10= 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.
We need to prove that,the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0and 7x + 24y = 50.
Let us write down thenormal forms of the given lines.
First line: 4x + 3y +10 = 0
-4x – 3y = 10
So, p = 2
Second line: 5x − 12y+ 26 = 0
-5x + 12y = 26
So, p = 2
Third line: 7x + 24y =50
So, p = 2
∴ The origin isequidistant from the given lines.