How can I find the equation of a line given its x-intercept and y-intercept?

[mathdad]

Find an equation of the line whose x-intercept is -6 and y-intercept is sqrt{6}.

1. The x-intercept is the point (-6, 0) and y-intercept can be written as the point (0, sqrt{6}).

Yes?

2. After finding my two points, I must find the slope.

True?

3. I then plug the slope and one of the points into the point-slope formula and solve for y.

Yes?

 

[Greg]

Find an equation of the line whose x-intercept is -6 and y-intercept is sqrt{6}.

1. The x-intercept is the point (-6, 0) and y-intercept can be written as the point (0, sqrt{6}).

Yes?

Yes.

2. After finding my two points, I must find the slope.

True?

Yes.

3. I then plug the slope and one of the points into the point-slope formula and solve for y.

Yes?

Yes.

 

[MarkFL]

It may interest you to know that the two-intercept form of a line may be written:

\(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

Where the $x$-intercept is $(a,0)$ and the $y$-intercept is $(0,b)$. If you're interested, see if you can derive the above formula...:)

 

[mathdad]

It may interest you to know that the two-intercept form of a line may be written:

\(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

Where the $x$-intercept is $(a,0)$ and the $y$-intercept is $(0,b)$. If you're interested, see if you can derive the above formula...:)

To derive the formula, must I use the points (a,0) and (0,b)? Must I also use y = mx + b?

 

[MarkFL]

To derive the formula, must I use the points (a,0) and (0,b)? Must I also use y = mx + b?

You can use the two given intercepts to determine the slope, and then use the slope-intercept form, yes.

 

[mathdad]

The x-intercept is the point (-6, 0) and y-intercept can be written as the point (0, sqrt{6}).

Let m = slope

m = (sqrt{{6} - 0)/(0 -(-6))

m = sqrt{6}/6

I will use (-6,0).

y - 0 = (sqrt{6}/6)(x - (-6))

y = (sqrt{6}/6)(x + 6)

y = (sqrt{6})x + sqrt{6}

Correct?

 

[mathdad]

You can use the two given intercepts to determine the slope, and then use the slope-intercept form, yes.

Let m = slope

m = (b - 0)/(0 - a)

m = b/-a

y = mx + b

y = (b/-a)x + b

Where do I go from here?

 

[MarkFL]

Let m = slope

m = (b - 0)/(0 - a)

m = b/-a

y = mx + b

y = (b/-a)x + b

Where do I go from here?

Divide through by $b$...

 

[mathdad]

Let m = slope

m = (b - 0)/(0 - a)

m = b/-a

y = mx + b

y = (b/-a)x + b

y/b = (b/-a)x(1/b) + (b/b)

y/b = -x/a + 1

(x/a) + (y/b) = 1

- - - Updated - - -

It may interest you to know that the two-intercept form of a line may be written:

\(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

Where the $x$-intercept is $(a,0)$ and the $y$-intercept is $(0,b)$. If you're interested, see if you can derive the above formula...:)

Can you give me an example using (x/a) + (y/b) = 1 to solve a problem?

 

[MarkFL]

...Can you give me an example using (x/a) + (y/b) = 1 to solve a problem?

The problem you originally posted in post #1 would be such an example. :)

 

[mathdad]

Cool. I will post several more questions from the David Cohen Precalculus 3rd Edition later tonight. I believe you also have this textbook.