0
$\begingroup$

The question is to find the equations of the lines (plural) that are parallel to the graph of $y=-x+6$ and perpendicular to the graph of $y=1/x$. I understand that to be parallel to the line, the slope would be $-1$. The derivative of the curve is $-1/x^2$. The tangent would be opposite reciprocal...so $x^2$ Now I'm stuck. I can easily calculate the points of intersection of the line and curve as $(5.828, 0.172)$ and $(0.172, 5.828)$, but not clear how that helps.

Could I just choose a random point, say $(0,0)$ and calculate a line parallel to the given line to be $y=-x$. That line is parallel to the given line. So are an infinite number of others.

Then to find a line perpendicular to the curve I could use $(1,1)$ (random choice) and get $y'=1/1$, so $y=x$.

These lines seem to answer the question but seems too arbitrary.

$\endgroup$
  • 2
    $\begingroup$ So a line that satisfies the description requires that you satisfy satisfy $x^2 = -1$. What can you conclude about such lines? $\endgroup$ – Jim H Aug 26 '17 at 22:26
  • $\begingroup$ I would conclude that it doesn't exist. However, it looked like the question allowed for 2 different lines (perhaps I'm misreading) and so I'm not sure I'd need to set the slope of one of them ($-1)$ to the slope of the other $x^2.$ $\endgroup$ – user163862 Aug 26 '17 at 22:36
1
$\begingroup$

Your reasoning in the first paragraph is quite correct.

Also, consider carefully the sign of the derivative of $\frac{1}{x}$. Since the derivative is always negative, every perpendicular must have positive slope.

The given line does not have positive slope.

Try applying your reasoning to the similar problem of finding all lines parallel to $y=4x+3$ and perpendicular to $y=\frac{1}{x}$.

$\endgroup$
0
$\begingroup$

Let the set of line be L y= mx+c

here $mm_1 = -1$ where $m_1$ is the slope of the curve. $m_1 = -\frac{1}{x^2}$

and $m$ and $m_2$ must be the same where $m_2$ is the slope of the line $y = -x+6$.

$ = -1$

$m(-\frac{1}{x^2}) = -1 $

$m = x^2$

and hence as @Jim H has said, you would have all lines that satisfy $x^2=-1$ which do not exist.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.