Find the value of x such that the area of a triangle whose vertices have coordinates (6, 5), (8, 2) and (x, 11) is 15 square units

I just did a whole page of latex and lost the whole lot so now I will do a condensed version. 

Heureka, Your answer looks great but I don't understand it so I will do it my way.

Maybe you can explain your way?

Anyway here goes- again

Find the value of  x such that the area of a triangle whose vertices have coordinates

P(6, 5), Q(8, 2) and R( x, 11) is 15 square units.

distance PQ=sqrt(13)

Equation of line PQ is

3x+2y-28=0

Now I need the perp distance from R to PQ

a=3, b=2, c=-28, x1=x, y1=11

$$\\d=\frac{|3x+22-28|}{\sqrt{13}}\\\\
d=\frac{|3x-6|}{\sqrt{13}}\\\\
d=\frac{|3x-6|}{\sqrt{13}}\\\\$$

$$\\\mbox{perpendicular distance from R to PQ}=\frac{|3x-6|}{\sqrt{13}}\\\\
Area=\frac{1}{2}\times \sqrt{13} \times \frac{|3x-6|}{\sqrt{13}}\\\\
15=\frac{1}{2} \times |3x-6|\\\\
30=|3x-6|\\\\
30=3x-6\qquad or \qquad 30=6-3x\\\\
36=3x\qquad\qquad or \qquad 24=-3x\\\\
x=12\qquad\qquad \;\;or \qquad x=-8\\\\$$

Below are the 2 possible triangles.  :)

Find the value of  x such that the area of a triangle whose vertices have coordinates (6, 5), (8, 2) and ( x, 11) is 15 square units.

$$area = \pm15 =
\dfrac{
\left|
\left[
\left(\begin{array}{c}
x\\
11
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\times\left[
\left(\begin{array}{c}
8\\
2
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\right|
}
{2}
\\\\
\pm30 =
\left|
\left[
\left(\begin{array}{c}
x\\
11
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\times\left[
\left(\begin{array}{c}
8\\
2
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\right|
} \\
\pm30 =
\left|
\left(\begin{array}{c}
x-6\\
11-5
\end{array} \right)
\times
\left(\begin{array}{c}
8-6\\
2-5
\end{array} \right)
\right|
} \\
\pm30 =
\left|
\left(\begin{array}{c}
x-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\\\
x_1 =?\\
30 =
\left|
\left(\begin{array}{c}
x_1-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\
(x_1-6)(-3)-6*2=30\\
(x_1-6)(-3)=42\\
(x_1-6)=-14\\
\textcolor[rgb]{1,0,0}{x_1 =-8}\\\\$$

$$x_2 =?\\
-30 =
\left|
\left(\begin{array}{c}
x_2-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\
(x_2-6)(-3)-6*2=-30\\
(x_2-6)*3+12=30\\
(x_2-6)*3=18\\
x_2-6=6\\
\textcolor[rgb]{1,0,0}{ x_2=12 }$$

a. is okay

Melody....how did you obtain the fraction    l 3x - 6 l / √13     as a perpendicular distance from R to PQ ????.....I'm not seeing how you got that

Sorry Chris that was on the Latex that i lost.

But here is the formula

I know this formula....but I don't see how you're getting this......can you "connect the dots" more fully for me??

Thanks

Thanks Heureka, that will give me plenty to think about :))

I've added some more dots but if you still want more let me know. :)))

I added it in my original explanation.

Well I thought I did but it does not appear to be there.

this is not my night for presentation.  LOL

I'll try again. :))

ok I redid it AGAIN. :)))))

If you still have questions just ask :))

OK, Melody...got it.....

Well that is good - NOW we both have to 'get' Heureka's answer  LOL 

I actually recall (kinda') that method from Linear Algebra........don't remember how to calculate that cross-product......maybe I'll go back and review it....

Hey.....I'm still working out how to write an equation of a line......!!!!

Yes I too have only the vaguest recollection of this stuff :))

Here's a graphical solution to save having to bother with all that messy stuff!

.

Alternatively, if you prefer a numerical solution, continue using Heron's formula for the area and:

.