<ol>
<li>In the xy-plane, line L passes through the origin and is perpendicular to the line 4x + y = k, where k is a constant. If the two lines intersect at the point (t, t+1), what is the value of t?</li>
</ol>
<p>A. -4/3
B. -5/4
C. 3/4
D. 5/4
E. 4/3</p>
<p>Answer is (A), explain please.</p>
<p>We want to put the given equation into slope intercept form.</p>
<p>4x+y=k
y=-4x+k</p>
<p>It is important to realize k does not matter at all, we are only concerned with slope.</p>
<p>Now the given point of intersection is (t,t+1) and the first line passes through the origin (0,0). We are given the slope of the line perpendicular to the one passing through the origin, and we know that if m is a slope then -1/m is the slope of the line perpendicular to the given slope.</p>
<p>-4 is the slope of the line passing through the origin. So 1/4 is the slope of the line passing through the origin.</p>
<p>From here we can plug the slope and the 2 points into the formula for slope. (y(1)-y(2)/(x(1)-x(2))=m</p>
<p>t+1-0/t-0=1/4</p>
<p>t+1/t=1/4</p>
<p>Cross multiply</p>
<p>4t+4=t</p>
<p>3t=-4</p>
<p>t=-4/3</p>
<p>Hope this clears stuff up.</p>
<p>4x + y = k</p>
<p>Put it into y = mx + b format…</p>
<p>y = k - 4x
y = -4x + k</p>
<p>A line perpendicular to this has the slope 1/4, and since it passes through the origin, has a y intercept of 0.</p>
<p>y = 1/4x + 0
y = 1/4x</p>
<p>Equal the two equations together to find the intersection point.</p>
<p>1/4x = -4x + k
1/4*t = -4t + k
4.25t = k</p>
<p>Plug this k value into the other equation</p>
<p>t+1 = -4t + 4.25t
t+1 = 0.25t
1 = -.75t
-4/3 = t</p>
<p>(A)</p>
<p>=</p>
<p>Yep, that works. Mine was simpler…:-P</p>
<p>There are tons of ways to do this. The shortest one doesn’t even involve the perpendicular line, or k. They’re simply testing your comprehension of functions</p>
<p>Ah waow, now I see. Thanks a bunch guys. I rarely can’t figure out an SAT math problem, but this one just got me.</p>
<p>Oh and lol, I understand Bassir’s better hah. When I was working on it I just assumed k could = 0, so that got me nowhere. Owen, in yours why do you plug in the slope? I know because it gets you the answer but how would you know to do that?</p>
<p>He knows to do that because a slope can get you from one point of a line to another point of a line.</p>
<p>I just knew to do that from past experiences with similar problems. I saw I had the origin, a point with some variables in it, and a slope. So I just plugged it into the slope formula.</p>
<p>Oh, and my feelings are hurt.</p>
<p>;-)</p>