<p>Question #17 PT3 S2 from the Blue book just seems so difficult for me! I've been looking at it without figuring it out for a while now.
here it is:</p>
<p>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?
(a) -4/3
(b) -5/4
(c) 3/4
(d) 5/4
(e) 4/3</p>
<p>can any kind soul help me out here?</p>
<p>We know that line L is perpendicular to line 4x+y=k, therefore, line L has to be y=1/4x+k.</p>
<p>We also know that line L passes through the origin, so: y=1/4x+0 </p>
<p>Because the intersection at (t, t+1), let us substitute these values into our line L equation.</p>
<p>t+1=1/4t
solve for t:
4t+4=t
3t=-4
t=-4/3</p>
<p>Small correction to the solution above: line L has to be y=1/4x + n
(n is not necessarily equal to k from 4x + y = k ).</p>
<p>Not really a big deal for this particular problem, since n works out = 0.</p>
<p>ETS often sets you on a longer path - finding extraneous values, for example.</p>
<p>In this question we should not be concerned with k (or equation for L).</p>
<p>Let's call the second line M.
Slope of M is -4.</p>
<p>Slope of a line perpendicular to M is inverse opposite (important fact to memorize): 1/4.</p>
<p>Slope of L is (t+1)/t.</p>
<p>L is perpendicular to M, thus
(t+1)/t = 1/4
1 + 1/t = 1/4
1/t = -3/4
t = -4/3</p>