<p>hey i found this one in the blue book and need some help. thogh it was marked a Medium question so im prob missing something here...</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?</p>
<p>answer= (-4/3)</p>
<p>Is there a diagram? I don't think there's a unique solution otherwise.</p>
<p>ehhhhhh 10 characteristix</p>
<p>there's probably a graph..........</p>
<p>There isn't a diagram, but you don't need one to solve the problem.</p>
<p>Ok, first of all, write down what you know about the two lines, L and J (the name of the second line doesn't matter). </p>
<p>For Line J you know it's equation is 4x+y=k, which when rearranged into slope-intercept form is y=-4x+k. Therefore the slope of line J is -4.</p>
<p>Line L:
Write it's equation in the form y=mx+b, where m equals the slope and b is a constant. Since it passes, through the origin, plug in (0,0) for x and y, giving you 0=0+b. Therefore b is zero! So the equation for L then becomes simply y=mx. We also know that line L is perpendicular to line J, which has slope -4, therefore, line L's slope is the negative reciprocal of -4 (make it negative and put it under 1). This equals 1/4. Therefore, the equation for L is y=1/4x. </p>
<p>Now it says that J and L intersect at (t, t+1), so plug in those values for x and y, respectively. </p>
<p>y=1/4x becomes t+1=1/4t, which equals 3/4t=-1, t=-1/(3/4) or -4/3.</p>
<p>The line 4x+y=k can also be written as y=k-4x, so has a slope of -4. A line perpendicular to it has a slope of 1/4 (negative reciprocal), so that the line through the origin is y=x/4. Substituting, we get t+1=t/4 or 4t+4=t or 3t= -4 or t=-4/3.</p>
<p>Whoops, I see jlauer beat me to the post, and with a clearer explanation too!</p>
<p>Lol, we did it the same way johnshade. I just wrote out every painstaking step.</p>