<p>In the xy-plane, line l passes trough the origin and is perpendicular to the line 4x+y=k ,
where k is a constant. If the two lines intersect at some point (t,t+1), what is the value of t.</p>
<p>A) -4/3
B) -5/4
C) 3/4
D) 5/4
E) 4/3</p>
<p>I am very grateful if there's somebody to help me out with this question.
Ive already asked a lot of people, but no 1 could give me the solution.</p>
<p>i got the answer to be -1/5
u sure thats not one of the answer choices?</p>
<p>u sure its asking for value of t.. not the value of the something else</p>
<p>perpendicular line: y = x/4 + 17/20
line: y = -4x</p>
<p>with intersection point: (-1/5 , 4/5)</p>
<p>it passes through the origin so k has to equal to 0</p>
<p>where did u get this problem?</p>
<p>4x+y=k - - slope = (-4)</p>
<p>slope of other line is (1/4)</p>
<p>other line is y=(x/4)+C</p>
<p>Since it passes through origin, c=0</p>
<p>other line is y=x/4</p>
<p>put (t, t+1) in it;</p>
<p>t+1=t/4
4t+4=t
3t+4=0
t=-4/3</p>
<p>which would be (A)</p>
<p>alebed01 - knot=0 since given line does not pass through orgin; perp line does</p>
<p>The answer is 4/3, i.e. the answer is E.</p>
<p>I don't know what you're worrying about. Easy question.</p>
<br>
<blockquote> <p>t=-4/3<<</p> </blockquote>
<br>
<p>Shash, it cannot be negative, because the lines intersect in first quater.</p>
<p>Sorry!!</p>
<p>I recalculated everything on paper and got -4/3.</p>
<p>Shash-rao, you was right. When I calculated in my head, I took k as positive.</p>
<p>Dont complicate the problem unnecessarily - ignore k. </p>
<p>Dont try to find k since all you need is the slope of the given line to completely describe the perpedicular one, as a point on it (o,o) is already given in the question.</p>
<p>And yes, the question is rather easy.</p>
<p>No, I ignored k, but I wrongly transfered x from one side to another, and then imagined the picture to check myself. And it was right with k positive, but not negative..</p>
<p>i didnt even have this question.......</p>
<p>Can someone now explain clearly how to slve this question easily?</p>
<p>Original function:
y=-4x+k</p>
<p>Perpendicular, going through origin then: y=1/4x</p>
<p>Now put t instead of x, and t+1 instead of y in perpendicular one:</p>
<p>t+1=t/4
3t/4=-1
t=-4/3</p>
<p>It's the easiest way.</p>
<p>how can you now that's Y=1/4x ???</p>
<p>First of all, understan what is given and what you need to find out. You have a line whose slope is known, but intercept is not - that means that you dont know the line completely. What you do know that the 2nd line passes through the origin, and is perpendicular to this. Since product of slopes of 2 perpendicular lines is -1, you can find the unknow slope - meaning you now have enough info to determine the 2nd line</p>
<p>Also, the point lies on both lines nd is in terms of a parameter t. Actually, this parametric representation is also that of a line, but you dont need to know that to solve the problem. Since the 1st line can be anything (due to the arbitrary nature of k), the point (t,t+1) needs to satisfy only the second line. Substitute the value of x=t and y=t+1 in the 2nd equation and simply solve the linear equation for t!</p>
<p>
[quote]
how can you now that's Y=1/4x ???
[/quote]
</p>
<p>the general form is y=mx+c</p>
<p>we know that m=1/4 and that the line passes through (0,0)</p>
<p>therefore, x=0,y=0 satisfies y=x/4+c
solve for C (you know one ordered pair of x,y which satisfies)</p>
<p>So Is It Positive Or Negative?</p>
<p>It's not so bad. Since we know that the line is perpendicular to 4x+y=k, or in other words y=-4x+k, the slope will be the negative reciprocal of that line. Furthermore, since it passed through the origin, it has no "y" intercept. So the perpendicular line's equation is:
y=(1/4)x
Now plug in "t" for "x" and "t+1" for "y." Since there's only one unknown, we can solve quite easily:
y=(1/4)x
t+1=(t/4)
4t+4=t
4=-3t
t=-(4/3)</p>
<p>It's negative</p>
<p>sorry.. misread.. the perpendicular passes.. my bad.
it was 4 AM</p>