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 x - 3y = 1 and is parallel to x - y - 6 = 0, find a and b.

Answer 1

If point of intersection of lines $x+y=3and2x-3y=1is$

$x=3-y$

$2(3-y)-3y=1$

$6-2y-3y=1$

$-5y=-5$

$y=1$

$\Rightarrow x=3-1=2$

$\therefore$ Point is (2, 1)

Any line parallel to $x-y-6=0$

will have the same slope = 1

$\therefore$ Equation of line parring through (2,1) and having slope = 1

$isy-y_1=m(x-x_1)$

$y-1=1(x-2)$

$y-1=x-2$

$y-x=-2+1$

$y-x=-1$

$x-y=1$

$\therefore a=1,b=-1$ $(Comparingwith\frac{x}{a}+\frac{y}{b}=1)$