Track / Stage 2 / 67 of 320 #387 of 2604

Problem 387

Algebra Difficulty 2.0 Prove it CEMC Galois · Canada · 2013

Find an equation of the line that passes through the points $(2,0)$ and $(0,4)$ .
Rewrite the equation of the line from part (a) in the form $\dfrac{x}{c} + \dfrac{y}{d} = 1$ , where $c$ and $d$ are integers.
State the $x$ -intercept and the $y$ -intercept of the line $\dfrac{x}{3} + \dfrac{y}{10} = 1$ .
Determine the equation of the line that passes through the points $(8,0)$ and $(2,3)$ written in the form $\dfrac{x}{e} + \dfrac{y}{f} = 1$ , where $e$ and $f$ are integers.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

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 $\dfrac{x}{8} + \dfrac{y}{4} = 1$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.