Find the equations of the straight lines each of which passes through

1.	Find the equations of the straight lines each of which passes through the point (3, 2) and cuts off intercepts a and b respectively on x and y-axes such that a – b = 2
Answer» Concept Used: The equation of the line with intercepts a and b is \(\frac{x}{a}+\frac{y}{b}=1\) Given: Here, a – b = 2 ⇒ a = b + 2 ……(i) Explanation: The line is passing through (3,2). \(∴\frac{3}{a}+\frac{2}{b}\)…(ii) From equation (i) and (ii) \(\frac{3}{b+2}+\frac{2}{b}=1\) ⇒ 3b + 2b + 4 = b² + 2b ⇒ b² – 3b – 4 = 0 ⇒ (b – 4)(b + 1) = 0 ⇒ b = 4, -1 Now, from equation (i) For b = 4, a = 4 + 2 = 6 For b = -1, b = -1 + 2 = 1 Thus the equation of line is \(\frac{x}{1}+\frac{y}{-1}=1\) or \(\frac{x}{6}+\frac{y}{4}=1\) ⇒x – y = 1 or 2x + 3y = 12

Find the equations of the straight lines each of which passes through the point (3, 2) and cuts off intercepts a and b respectively on x and y-axes such that a – b = 2

Answer»

Concept Used:

The equation of the line with intercepts a and b is \(\frac{x}{a}+\frac{y}{b}=1\)

Given:

Here, a – b = 2

⇒ a = b + 2 ……(i)

Explanation:

The line is passing through (3,2).

\(∴\frac{3}{a}+\frac{2}{b}\)…(ii)

From equation (i) and (ii)

\(\frac{3}{b+2}+\frac{2}{b}=1\)

⇒ 3b + 2b + 4 = b² + 2b

⇒ b² – 3b – 4 = 0

⇒ (b – 4)(b + 1) = 0

⇒ b = 4, -1

Now, from equation (i)

For b = 4, a = 4 + 2 = 6

For b = -1, b = -1 + 2 = 1

Thus the equation of line is

\(\frac{x}{1}+\frac{y}{-1}=1\) or \(\frac{x}{6}+\frac{y}{4}=1\)

⇒x – y = 1 or 2x + 3y = 12