<p>In the xy-plane, point R (2,3) and point S (5,6) are two vertices of triangle RST. If the sum of the slopes...?
In the xy-plane, point R (2,3) and point S (5,6) are two vertices of triangle RST. If the sum of the slopes of the sides of the triangle is 1, which of the following angles could be a right angle?
I. R
II. S
III. T
A. None
B. I only
C. III only
D. I and II only
E. I, II, and III</p>
<p>I think it's C, because the slope of RS is already 1, but if point T was at (5,3), this would form a right triangle with the hypotenuse having the slope of one and the other two legs would have a slope of undefined and 0. Disregarding the undefined, this would add up to 1. However, my book says the answer is none. Am I right?</p>
<p>Undefined doesn’t count. The slope of a vertical line is essentially +/- infinity.</p>
<p>To solve this, note that the sum of the slopes of RT and ST is zero. This means that their slopes are m and -m for some m.</p>
<p>Assume that R is a right angle. Then RT must have a slope of -1, and ST must have a slope of 1 (so that the slopes of RT and ST add to zero). However, this implies that ST and RS are parallel, contradiction.</p>
<p>Similarly, if we assume that S is a right angle, we reach the same contradiction.</p>
<p>If we assume that T is a right angle, then T must lie on the circle with diameter RS. We want to find a T such that the slopes of RT and ST add up to zero. It can easily be shown by inspection that this is impossible (e.g. |slope of ST| > 1 and |slope of RT| < 1, or vice versa).</p>
<p>Therefore the answer is A) none.</p>
<p>You learn in calculus that infinity plus anything can equal everything and anything but it definitely doesn’t equal 1. How can infinity plus 1 equal one? In that case, infinity would have to be equal to zero!</p>
<p>How is ∞ + 1 equal to 1?</p>
<p>In any case, T(5,3) or T(2,6) doesn’t work.</p>
<p>how do they not work</p>
<p>Umm we just answered that like 3 posts ago…</p>
<p>becuz their slopes are infinity? ok cool i said tht</p>
<p>Was this actually on the SAT? If so, shoot me now.</p>
<p>It seems SAT-appropriate, maybe on the more difficult side.</p>
<p>Yeah it wasn’t too hard actually. I was trying to solve it in my head at first, but once I drew a little diagram it was easier. I’m not exactly a math genius compared to most of the people here though (660, trying to get it to a 700). Much prefer my CR (760).</p>
<p>Ah okay…I’m the complete opposite. Math’s my strong point, so I can score 800’s consistently, but I’m pretty weak at CR (~600ish).</p>