Question

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 1

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

Here, a − b = 2 

\[\Rightarrow\] a = b + 2        ... (1)
The line passes through (3, 2).

∴ \[\frac{3}{a} + \frac{2}{b} = 1\]   ... (2)
Substituting a = b + 2 in equation (2)

\[\frac{3}{b + 2} + \frac{2}{b} = 1\]

\[ \Rightarrow 3b + 2b + 4 = b^2 + 2b\]

\[ \Rightarrow b^2 - 3b - 4 = 0\]

\[ \Rightarrow \left( b - 4 \right)\left( b + 1 \right) = 0\]

\[ \Rightarrow b = 4, - 1\]

Now, from equation (1)
For b = 4, a = 4 + 2 = 6
For b = − 1, a = − 1 + 2 = 1
Thus, the equations of the lines are

\[\frac{x}{1} + \frac{y}{- 1} = 1 \text { and } \frac{x}{6} + \frac{y}{4} = 1\]

\[ \Rightarrow x - y = 1 \text { and} 2x + 3y = 12\]