another quadratics question

[momomath]

Determine the minimum distance from (0,0) to the line 3x +2y -12 = 0.

This is in the quadratics section of my book, so I have to solve it by getting the info into a quadratic equation in vertex form.

First I labeled the length from (0,0) to 3x + 2y - 12 = 0 "c".

Then I reworked the equation to say y= -(3/2)x + 6.

Then I realized that "c" was the hypotenuse for a right triangle, so I wrote c^2 = x^2 + y^2 .

Which I re-wrote as c^2 = x^2 + (-(3/2)x + 6)^2.

Which equals c^2 = x^2 + (9/4)x^2 - 18x + 36.

which equals c^2 = (13/4)x^2 - 18x + 36.

But now i feel confused. Thanks for any help you can give.

 

[HallsofIvy]

Let (x, y) be the point on the line 3x+ 2y- 12= 0. Then y= 6- (3/2)x so that the distance from (0, 0) to (x, y) is $\displaystyle \sqrt{x^2+ y^2}= \sqrt{x^2+ 36- 18x+ (9/4)x^2}= \sqrt{(13/4)^2- 18x+ 36}$. Of course that square root will be minimized if and only if $\displaystyle (13/4)x^2- 18x+ 36$ is, exactly what you have. Now complete the square. That is, find a and b so that $\displaystyle (13/4)x^2- 18x+ 36= (13/4)(x- a)^2+ b$. Since a square is never negative, that will have its smallest value, b, when x= a. I recommend first factoring that "13/4" out of the first two terms- $\displaystyle (13/4)(x^2- (72/13)x)+ 36$.
Now compare $\displaystyle x^2- (72/13)x$ with $\displaystyle x^2- 2ax+ a^2$. What must a be equal to so that 2a= 72/13? And then what is $\displaystyle a^2$?

 

[Bob Brown MSEE]

(13/4)x^2 - 18x + 36-c*c=0

You want a c that yields only one real answer.
Use the quadratic formula.
Look at the discriminant.

 

[momomath]

One thing that confuses me is this c^2. I am confused about the connection between the minimum value from c^2 = A(x-p) + q and the minimum value of c.

 

[momomath]

I keep getting the answer 11.42 when my book says it's 3.33

 

[soroban]

Hello, momomath!

Determine the minimum distance from $\displaystyle (0,0)$ to the line $\displaystyle 3x + 2y - 12 \:=\:0.$

The minimum distance is the perpendicular distance from $\displaystyle (0,0)$ to the line.

The equation of the line is: .$\displaystyle y \:=\:-\frac{3}{2}x+6$
. . Its slope is $\displaystyle -\frac{3}{2}$
The line through $\displaystyle (0,0)$ with slope $\displaystyle \frac{2}{3}$ is: .$\displaystyle y \:=\:\frac{2}{3}x$

Their intersection: .$\displaystyle \frac{2}{3}x \:=\:-\frac{3}{2}x + 6 \quad\Rightarrow\quad \frac{2}{3}x + \frac{3}{2}x \:=\:6$

. . . . . $\displaystyle \frac{13}{6}x \:=\:6 \quad\Rightarrow\quad x \:=\:\frac{36}{13}$

Then: .$\displaystyle y \:=\:\frac{2}{3}\left(\frac{36}{13}\right) \:=\:\frac{24}{13}$

The minimum distance is: .$\displaystyle d \;=\;\sqrt{\left(\frac{36}{13}\right)^2 + \left(\frac{24}{13}\right)^2}$

. . $\displaystyle d \:=\:\sqrt{\frac{1872}{169}} \:=\:\sqrt{\frac{144}{13}} \:=\:\frac{12}{\sqrt{13}} \:=\:\frac{12\sqrt{13}}{13} \:\approx\:3.328201177$

 

[Bob Brown MSEE]

0 = bb-4ac =discriminant

0 =(18)(18)-4(13/4)(36-cc)
cc = 11.077
c = 3.328