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.

a. | -8 or 12 | c. | 16, -8 |

b. | 6, -12 | d. | 4, |

mishforever21
Nov 17, 2014

#2**+10 **

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. :)

Melody
Nov 18, 2014

#1**+5 **

**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**

heureka
Nov 18, 2014

#2**+10 **

Best Answer

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. :)

Melody
Nov 18, 2014

#3**0 **

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

CPhill
Nov 18, 2014

#5**0 **

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

Thanks

CPhill
Nov 18, 2014

#8**0 **

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. :))

Melody
Nov 18, 2014

#12**0 **

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......!!!!

CPhill
Nov 18, 2014