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Triangle $ABC$ has vertices $A(0,8)$, $B(2,0)$, $C(8,0)$. A vertical line intersects $AC$ at $R$ and $\overline{BC}$ at $S$, forming triangle $RSC$. If the area of $\triangle RSC$ is 12.5, determine the positive difference of the $x$ and $y$ coordinates of point $R$.

Guest Jun 24, 2018
 #1
avatar+2117 
+1

Constructing a diagram is a crucial step here; digesting all that information visually is extremely difficult without a visual aid. I have concocted one that I am proud of. I have tried to label everything in this diagram for your convenience. This problem most likely has many avenues toward achieving the correct answer, but I will present my approach for you:
 

 

In the diagram above, I have added point D, which is the intersection of point A and point C. The point lies on the origin. Let's focus on some characteristics regarding \(\triangle ACD\) .

 

Point A and point D have the same x-coordinate, so a vertical segment connects them. Point C and point D share the same y-coordinate, so they must be connected with a horizontal segment. \(\overline{AD}\perp\overline{DC}\) because horizontal and vertical lines are always perpendicular; thus, \(m\angle ADC=90^{\circ}\).

 

\(AD=DC=8\), so \(\triangle ACD\) is also an isosceles triangle.

 

A triangle that is both right and isosceles is also known as a 45-45-90 triangle. It is possible to observe these similarities with \(\triangle RSC\).

 

\(\overline{AD}\parallel\overline{RS}\) because they are both vertical lines. Whenever a parallel line cuts through a triangle, the resulting triangles are similar. In this case, \(\triangle ADC\sim\triangle RSC\)

 

\(DC=8\) and \(DS=a\) , so \(CS=8-a\) . Because \(\triangle RSC\) is isosceles, \(RS=CS=8-a\) . The area of this triangle, according to the original problem, is 12.5. We know the lengths of the side lengths, so we can determine the value of a.

 

\(12.5=\frac{1}{2}(8-a)(8-a)\)Multiply by 2 on both sides to solve for a. We are using the area of a triangle to generate this equation.
\(25=(8-a)^2\)Take the square root of both sides. 
\(5=|8-a|\) 
  

 

Yes, it is possible to solve for a here, and it is relatively simple from here on out. However, we can take a slight shortcut. The end goal is to determine the difference of x- and y-coordinates, but 8-a represents the height of the triangle. Since 8-a=5, we know that the y-coordinate is 5. 

 

 (3,5) is the coordinate of point R. \(|3-5|=2\)  is the positive difference.

 


 

TheXSquaredFactor  Jun 24, 2018
 #2
avatar+87334 
+1

Thanks, X2 !!!....here's another approach.....

 

Let  SC  be the base of triangle RSC.....and let  its length  =  8 - x

 

The slope of the line segment  AC   =    (0 - 8) / (8 - 0)  =  -8/8  = -1

And the equation of the line containing this segment is   y = - x + 8  ⇒  y  = 8 - x

So...let the height of  triangle  RSC = SR   = 8 - x

 

And we have that the area of triangle RSC =

 

(1/2) (8 -x) (8 -x)  =  12.5     multiply both sides by 2

(8 - x) ( 8 - x)  = 25     simplify

x^2 - 16x + 64  = 25

x^2 - 16x + 39 =  0     factor

(x - 3) ( x - 13)  = 0

 

Setting both factors to 0  and solving for x produces

x = 13   (reject)   or

x  = 3   (accept)

 

So.... since R  lies on the line  y  = 8 - x, its coordinates are  ( 3, 5)

And the positive difference of its x and  y coordinates  =  5 - 3 =  2

 

 

cool cool cool

CPhill  Jun 24, 2018
edited by CPhill  Jun 24, 2018
edited by CPhill  Jun 24, 2018

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