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!

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

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