Calculus: Early Transcendentals 8th Edition

(a) $\frac{x}{a}+\frac{y}{b} = 1$ (b) $4x-3y -24 = 0$
(a) Let the equation of a straight line be $~~y = mx+b$ Suppose the x-intercept is $a$. $y = mx+b$ $0 = m(a)+b$ $ma = -b$ $m = -\frac{b}{a}$ Then: $y = mx+b$ $y = (-\frac{b}{a})x+b$ $y-b = -\frac{bx}{a}$ $-\frac{y}{b}+1 = \frac{x}{a}$ $\frac{x}{a}+\frac{y}{b} = 1$ (b) $\frac{x}{a}+\frac{y}{b} = 1$ $\frac{x}{6}+\frac{y}{(-8)} = 1$ $4x-3y = 24$ $4x-3y -24 = 0$