Question

If the straight line $\frac{x}{a}+\frac{y}{b}=1$ passes through the point of intersection of the lines x + y = 3 and 2x − 3y = 1 and is parallel to x − y − 6 = 0, find a and b.

Answer 1

The given lines are x + y = 3 and 2x − 3y = 1.

x + y − 3 = 0 ... (1)

2x − 3y − 1 = 0 ... (2)

Solving (1) and (2) using cross-multiplication method:
$$\begin{gathered} \frac{x}{-1-9}=\frac{y}{-6+1}=\frac{1}{-3-2} \\ \Rightarrow x=2,y=1 \end{gathered}$$

Thus, the point of intersection of the given lines is (2, 1).

It is given that the line $\frac{x}{a}+\frac{y}{b}=1$ passes through (2, 1).

$\therefore \frac{2}{a}+\frac{1}{b}=1$ ... (3)

It is also given that the line $\frac{x}{a}+\frac{y}{b}=1$ is parallel to the line x − y − 6 = 0.

Hence, Slope of $\frac{x}{a}+\frac{y}{b}=1$ $\Rightarrow y=-\frac{b}{a}x+b$ is equal to the slope of x − y − 6 = 0 or, y = x − 6

$\therefore -\frac{b}{a}=1$

$\Rightarrow b=-a$ ... (4)

From (3) and (4):
$\frac{2}{a}-\frac{1}{a}=1\Rightarrow a=1$

From (4):
b = −1

∴ a = 1, b = −1