Question
If the straight line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ 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
If point of intersection of lines $x + y = 3$ and $2x - 3y = 1$ is
$x = 3 - y$
$2(3 - y) - 3y = 1$
$6 - 2y - 3y = 1$
$-5y = - 5$
$y = 1$
$\Rightarrow x = 3 - 1 = 2$
$\therefore$ 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 $= 1$
is $y - y_1= m(x - x_1)$
$y - 1 = 1(x - 2)$
$y - 1 = x - 2$
$y - x = -2 + 1$
$y - x = -1$
$x - y = 1$
$\therefore$ $a = 1, b = -1$ $\Big($ Comparring with $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\Big)$
$x = 3 - y$
$2(3 - y) - 3y = 1$
$6 - 2y - 3y = 1$
$-5y = - 5$
$y = 1$
$\Rightarrow x = 3 - 1 = 2$
$\therefore$ 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 $= 1$
is $y - y_1= m(x - x_1)$
$y - 1 = 1(x - 2)$
$y - 1 = x - 2$
$y - x = -2 + 1$
$y - x = -1$
$x - y = 1$
$\therefore$ $a = 1, b = -1$ $\Big($ Comparring with $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\Big)$
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