, 2013

The slope of the line passing through the points $(2,0)$ and $(0,4)$ is $\dfrac{4-0}{0-2}=\dfrac{4}{-2}=-2$.

Since the line passes through the point $(0,4)$, the $y$-intercept of this line is 4.

Therefore, an equation of the line is $y=-2x+4$.
Rearranging the equation from part (a), $y=-2x+4$ becomes $2x+y=4$.

Dividing both sides of the equation by 4 we get $\dfrac{2x+y}{4}=\dfrac{4}{4}$ or $\dfrac{2x}{4}+\dfrac{y}{4}=1$ and so the required form of the equation is $\dfrac{x}{2}+\dfrac{y}{4}=1$.
To determine the $x$-intercept, we set $y=0$ and solve for $x$.

Thus, $\dfrac{x}{3} + \dfrac{y}{10} = 1$ becomes $\dfrac{x}{3} + \dfrac{0}{10} = 1$ or $\dfrac{x}{3} = 1$, and so $x=3$.

The $x$-intercept is 3.

To determine the $y$-intercept, we let $x=0$ and solve for $y$.

Thus, $\dfrac{x}{3} + \dfrac{y}{10} = 1$, becomes $\dfrac{0}{3} + \dfrac{y}{10} = 1$ or $\dfrac{y}{10} = 1$, and so $y=10$.

The $y$-intercept is 10.

(Note that the intercepts are the denominators of the two fractions.)
Solution 1

The slope of the line passing through the points $(8,0)$ and $(2,3)$ is $\dfrac{3-0}{2-8}=\dfrac{3}{-6}=-\dfrac{1}{2}$.

Thus, an equation of the line is $y=-\dfrac{1}{2}x+b$.

To find the $y$-intercept $b$, we substitute $(8,0)$ into the equation and solve for $b$.

The equation becomes, $0=-\dfrac{1}{2}(8)+b$, or $0=-4+b$ and so $b=4$.

Therefore an equation of the line is $y=-\dfrac{1}{2}x+4$.

Rearranging this equation, $y=-\dfrac{1}{2}x+4$ becomes $\dfrac{1}{2}x+y=4$.

Multiplying both sides of the equation by 2, we get $x+2y=8$.

Dividing both sides of the equation by 8 we get, $\dfrac{x+2y}{8}=\dfrac{8}{8}$ or $\dfrac{x}{8}+\dfrac{2y}{8}=1$ and so the required form of the equation is $\dfrac{x}{8}+\dfrac{y}{4}=1$.

Solution 2

We recognize from the previous parts of the question that a line with equation written

in the form $\dfrac{x}{e} + \dfrac{y}{f} = 1$, has $x$-intercept $e$ and $y$-intercept $f$.

Since the line passes through $(8,0)$, then its $x$-intercept is 8 and so $e=8$.

Substituting the point $(2,3)$ into the equation $\dfrac{x}{8} + \dfrac{y}{f} = 1$ gives $\dfrac{2}{8} + \dfrac{3}{f} = 1$ or $\dfrac{3}{f}=1-\dfrac{1}{4}$ or $\dfrac{3}{f}=\dfrac{3}{4}$, and so $f=4$.

Therefore, the equation of the line is $\dfrac{x}{8} + \dfrac{y}{4} = 1$.