# [SOLVED] Right angles? (OG Page311)

• Nov 21st 2008, 07:05 PM
fabxx
[SOLVED] Right angles? (OG Page311)
In the xy-plane, point R (2,3) and point S (5,6) are two vertices of triangle RST. If the sum of the slope of the sides of the triangle is 1, why can't angle T be a right angle?

• Nov 21st 2008, 07:49 PM
o_O
Edit: Hmm ... this seemed shorter in my head. Please bear the long-windedness of this post ...
__________________________________

Let the point $\displaystyle T$ be arbitrarily labeled $\displaystyle (x,y)$.

Now if you form $\displaystyle \triangle RST$, it is essentially composed of 3 line segments: $\displaystyle \overline{RT} , \ \overline{RS}, \ \overline{ST}$

Let $\displaystyle m_{RT}, \ m_{RS}, \ m_{ST}$ be the respective slopes of $\displaystyle \overline{RT} , \ \overline{RS}, \ \overline{ST}$.

We have: $\displaystyle m_{RT} = \frac{y-3}{x-2}$.....$\displaystyle m_{RS} = \frac{6-3}{5-2} = 1$.....$\displaystyle m_{ST} = \frac{y - 6}{x-5}$

Now, we're told that the sums of the slopes equal 1, i.e. $\displaystyle \color{magenta}m_{RT} + m_{RS} + m_{ST} = 1$. Since $\displaystyle m_{RS} = 1$, we have: $\displaystyle {\color{magenta}m_{RT} + m_{ST} = 0} \ \Leftrightarrow \ {\color{red}\frac{y-3}{x-2} = -\frac{y-6}{x-5}}$

Also, note that $\displaystyle \angle STR = 90^{\circ}$ is the same thing as saying that $\displaystyle \overline{RT} \perp \overline{ST}$. If you remember, in order for two line segments to be perpendicular to one another, their slope one of line segment must be the negative reciprocal of the other.

So in order for $\displaystyle \angle STR = 90^{\circ}$, we must have: $\displaystyle m_{RT} = -\frac{1}{m_{ST}} \ \iff \ {\color{blue}\frac{y-3}{x-2} = -\frac{x-5}{y-6}}$

Equate the red and blue to get: $\displaystyle \frac{x-5}{y-6} = \frac{y-6}{x-y}$

Notice that they're reciprocals of each other. The only value that is equal to its own reciprocal is 1 and -1.

So either: $\displaystyle \frac{x-5}{y-6} = \frac{y-6}{x-y} = 1$ .....or.....$\displaystyle \frac{x-5}{y-6} = \frac{y-6}{x-y} = -1$

But we're told that the sum of the slopes is equal 1 (from above in magenta): $\displaystyle \color{magenta}m_{RT} + m_{RS} + m_{ST} = 1 \ \Leftrightarrow \ m_{RT} + m_{ST} = 0$

Do you see how this can never happen?
• Nov 21st 2008, 08:26 PM
Soroban
Hello, fabxx!

Quote:

In the xy-plane, point $\displaystyle R (2,3)$ and point $\displaystyle S (5,6)$ are two vertices of $\displaystyle \Delta RST.$

If the sum of the slopes of the sides of the triangle is 1, why can't $\displaystyle \angle T$ be a right angle?

Code:

        |         |            S(5,6)         |              o         |            *  *  m2         |          *      *         |        *          o T(x,y)         |      *      *         |    *  *    m1         |  o         | R(2,3)         |       - + - - - - - - - - - - - - -         |

$\displaystyle \text{Slope of RS} \:=\:\frac{6-3}{5-2} \:=\:\frac{3}{3}\:=\:1$

Let: .$\displaystyle \begin{array}{ccc}\text{slope of RT} \:=\:m_1 \\ \text{slope of ST}\:=\:m_2 \end{array}$

Sum of the slopes is 1: .$\displaystyle m_1+m+2+1\:=\:1 \quad\Rightarrow\quad m_1+m_2 \:=\:0$ .[1]

If $\displaystyle m_1 \perp m_2$, then: .$\displaystyle m_2 \:=\:-\frac{1}{m_1}$ .[2]

Substitute [2] into [1]: .$\displaystyle m_1 - \frac{1}{m_1} \:=\:0 \quad\Rightarrow\quad m_1^2 - 1 \:=\:0 \quad\Rightarrow\quad m_1 \:=\:\pm1$

If $\displaystyle m_1 = 1,\;RT$ is parallel to (coincides with) $\displaystyle RS,$
. . and we don't have a triangle.
(In other words: .$\displaystyle \angle R = 0^o,\;\angle S = 0^o,\;\angle T = 180^o$)

If $\displaystyle m_1 = -1,\:RT \perp RS \quad\Rightarrow\quad \angle SRT = 90^o$
. . and a triangle cannot have two right angles.