1. ## Analytic

Corners of a triangle ABC is A(-2,5),B(4,2) and C(x,0).For the
perimeter of the triangle to be minimum what must be the value of x?

MY WORK:

I found the slope of AB as 3/2. For perimeter to be minimum A-B-C must
be collinear. So -2/x+4=3/2,and x=-16/7

2. ## Re: Analytic

Originally Posted by kastamonu
Corners of a triangle ABC is A(-2,5),B(4,2) and C(x,0).For the
perimeter of the triangle to be minimum what must be the value of x?

MY WORK:

I found the slope of AB as 3/2. For perimeter to be minimum A-B-C must
be collinear. So -2/x+4=3/2,and x=-16/7

Looks like you missed a negative sign. Probably B should be (-4,2)? Otherwise, the slope of AB is $\dfrac{2-5}{4-(-2)} = \dfrac{-3}{6} = -\dfrac{1}{2}$.

Then, $\dfrac{0-2}{x-4} = -\dfrac{1}{2}$ implies $-4=4-x$, so $x=8$.

4. ## Re: Analytic

Yes it is B(-4,2). I made a typo.

5. ## Re: Analytic

If $A(-2,5)$ and $B(-4,2)$ then

$A-D-C$ must be collinear where $D(-4,-2)$

so $x=-24/7$

6. ## Re: Analytic

Why do we have to find symmetric of B to y-axis?

7. ## Re: Analytic

I meant x-axis.

8. ## Re: Analytic

We don't have to do it this way but this is an alternative way to solve the problem.

since $CB=CD$, minimizing $CA+CB$ is equivalent to minimizing $CA+CD$ which occurs when $ACD$ is a straight line

slope of $AD=7/2$

so slope of $CA=-5/(x+2)=7/2$

solve for $x$

Many Thanks.