Question: If the 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 parallel to the line y = x -6, then find the values of a and b.

Solution:

Equation of line 1 is y = x -6.
Comparing equation of line 1 with y = mx +c.
Slope of line 1 ($m_1$) = 1

Given,
Line 2 is $\frac{x}{a} + \frac{y}{b} = 1$
Since a point on line 2 passes through the point of intersection of (x +y =3) and (2x -3y =1)

Solving both the equations simultaneously,

$x + y = 3$
$or, x = 3-y$ - (i)

$2x - 3y = 1$
$or, 2x = 1 +3y$ - (ii)

Put value of x from equation (i) in equation (ii), we get,
$or, 2(3-y) = 1 + 3y$
$or, 6 -2y = 1 +3y$
$or, 6-1 = 3y +2y$
$or, 5 = 5y$
$\therefore y = 1$

Put value of y in equation (i), we get,

$or, x = 3 -1$
$\therefore x = 2$

(x,y) = (2,1)
So, the line 2 passes through the point (2,1).

Also, both line 1 and line 2 are parallel so, their slopes are equal. Let $m_2$ represent the slope of line 2 then,
$m_2 = m_1$
$\therefore m_2 = 1$

Also, when a line is in the form of $\frac{x}{a} + \frac{y}{b} = 1$, the slope of the line is represented by $\frac{-b}{a}$

From above two conditions, $m_2 = \frac{-b}{a}$
$or, 1 = \frac{-b}{a}$
$or, a = -b$

Put value of (x,y) = (2,1) and a = -b in the equation of $\frac{x}{a} + \frac{y}{b} = 1$

$or, \dfrac{x}{a} + \dfrac{y}{b} = 1$

$or, \dfrac{2}{-b} + \dfrac{1}{b} = 1$

$or, \dfrac{1}{b} - \dfrac{2}{b} = 1$

$or, \dfrac{1 -2}{b} = 1$

$or, -1 = b$

$\therefore b = -1$

Put value of b in (a = -b)
$or, a = -(-1)$
$\therefore a = 1$

Therefore, the required values of a and b are 1 and -1 respectively.