another quadratics question

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momomath

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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.
 
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 \(\displaystyle a^2\)?
 
(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.

 
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.
 
I keep getting the answer 11.42 when my book says it's 3.33
 
Last edited:
Hello, momomath!

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

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

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