# Thread: Equation of the Line

1. ## Equation of the Line

Find an equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

What does it mean that cuts off the least area from the first quadrant?

2. ## Re: Equation of the Line

Originally Posted by harpazo
Find an equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

What does it mean that cuts off the least area from the first quadrant?
Okay, a couple of things.

First, you certainly can use the usual y = mx + b for this but it's a bit less transparent at how to use the equation. I'm going to use what is called the intercept-intercept form. This will more easily help us with the area thing.

The intercept-intercept form talks about both the y-intercept (0, b) and the x-intercept (a, 0). (These are simply points at which the line is crossing the relevant axis.) This equation has the form $\displaystyle y = -\left ( \dfrac{b}{a} \right ) x + b$. Why use this form? Look at the graph. (It looks better blown up.)

The two points on the axes are the x and y intercepts. By doing things this way we can easily come up with the area of the triangle: $\displaystyle A = \dfrac{1}{2}bh$ (b as in "base length of the triangle" and not y-intercept b.) In the example I have set a = 1, b = 5. So what's the area of the triangle? It's $\displaystyle A = \dfrac{1}{2}(1)(5) = \dfrac{5}{2}$.

Using different a and b you get different triangular areas. What is the smallest you can get?

See what you can do with this.

-Dan

3. ## Re: Equation of the Line

Any line, other than a vertical line, that passes through (3, 5), crosses both x-axis and y-axis. Let's say that the x-intercept is $\displaystyle (x_0, 0)$ and the y-intercept is $\displaystyle (0, y_0)$. The are "cut off" by that line in the first quadrant is the triangle with vertices $\displaystyle (0, 0)$, $\displaystyle (x_0, 0)$, and $\displaystyle (0, y_0)$. That has area $\displaystyle \frac{1}{2}x_0y_0$. Any non-vertical line through (3, 5) is of the form y= a(x- 3)+ 5. That crosses the x-axis at (3- 5/a, 0) and the y-axis at (0, 5- 3a) so the area of the triangle "cut off" is $\displaystyle \frac{1}{2}(3- 5/a)(5- 3a)= frac{1}{2}(30- 9a- 25/a)$. The derivative of that with respect to a is $\displaystyle \frac{1}{2}(-9+ 25/a^2$. That will be 0 when $\displaystyle 9a^2= 25$ so (a must, or course, be positive) a= 5/3.

4. ## Re: Equation of the Line

Oops! I forgot about the (3, 5) thing. (I'll go get the fifty lashings with the wet textbook now.)

-Dan

5. ## Re: Equation of the Line

Didn't you tell us you've decided to concentrate on basic algebra?

6. ## Re: Equation of the Line

Originally Posted by HallsofIvy
Any line, other than a vertical line, that passes through (3, 5), crosses both x-axis and y-axis.
No, a horizontal line that passes through (3, 5), also does not cross both the x-axis and the y-axis.

7. ## Re: Equation of the Line

Originally Posted by DenisB
Didn't you tell us you've decided to concentrate on basic algebra?
A friend requested math help. I am currently working on algebra 1 material.

8. ## Re: Equation of the Line

Originally Posted by topsquark
Oops! I forgot about the (3, 5) thing. (I'll go get the fifty lashings with the wet textbook now.)

-Dan
We all make mistakes and forget things.

9. ## Re: Equation of the Line

Thank you everyone.

10. ## Re: Equation of the Line

Originally Posted by harpazo
A friend requested math help. I am currently working on algebra 1 material.
I would use Lagrange Multipliers for this problem. Let a and b be positive real numbers representing the x and y intercepts respectively. Suppose the line is constrained to pass though the quadrant I point:

$\displaystyle \left(x_1,y_1\right)$

We have the objective function (the area bounded by the line and the coordinate axes in the first quadrant):

$\displaystyle f(a,b)=\frac{1}{2}ab$

Subject to the constraint:

$\displaystyle g(a,b)=\frac{x_1}{a}+\frac{y_1}{b}-1=0$

This implies the system:

$\displaystyle \frac{1}{2}b=\lambda\left(-\frac{x_1}{a^2}\right)$

$\displaystyle \frac{1}{2}a=\lambda\left(-\frac{y_1}{b^2}\right)$

And this system implies:

$\displaystyle b=\frac{y_1a}{x_1}$

Substituting into the constraint, there results:

$\displaystyle \frac{x_1}{a}+\frac{x_1}{a}-1=0\implies a=2x_1\implies b=2y_1$

And so we find the line we want:

$\displaystyle \frac{x}{2x_1}+\frac{y}{2y_1}=1$

Or:

$\displaystyle y=-\frac{y_1}{x_1}x+2y_1$

In this problem, we are given:

$\displaystyle \left(x_1,y_1\right)=(3,5)$

Hence:

$\displaystyle y=-\frac{5}{3}x+10$

11. ## Re: Equation of the Line

This is over my head. Back to Algebra 1.

12. ## Re: Equation of the Line

Originally Posted by harpazo
This is over my head. Back to Algebra 1.
Hopefully, you pointed your friend to the help posted in this thread, and it is not over their head, as it is a problem likely given in a calculus course.

13. ## Re: Equation of the Line

Originally Posted by MarkFL
Hopefully, you pointed your friend to the help posted in this thread, and it is not over their head, as it is a problem likely given in a calculus course.
I told my friend about the MHF.