Why is the equations for a perpendicular line $-\frac{1}{m}$? Why is it just $-m$? Lets say $y=x$ and the $y$ intercept is at $0$ if we created another line that was $y=-x$, wouldn't that make it perpendicular?
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 A: Consider a vector ${\bf v}=(p,q)$,  attached at ${\bf 0}$ and pointing into the first quadrant. Turning this vector  $90^\circ$ counterclockwise produces the new vector ${\bf v}^*=(-q,p)$. These two vectors could serve as direction vectors of two perpendicular lines $g$ and $g^*$. The slope $m$ of $g$ is then the "slope" of ${\bf v}$, whence $$m={q\over p}\ .$$
The slope $m^*$ of $g^*$ is the "slope" of ${\bf v}^*$, whence
$$m^*={-p\over q}=-{1\over m}\ .$$
A: Remember that the slope of a given line is $m_1=\tan\theta$, where $\theta$ is the angle it makes with the positive X-axis. So if you just put a minus sign in front of the slope, you'll get a line with slope $m_2=-m_1=-\tan\theta=\tan(\pi-\theta)$. 

This line clearly makes angle $\theta$ with the negative X-axis, and clearly, the angle between the two lines would be $\pi-2\theta$ (or simply $2\theta$). So, these lines won't be perpendicular except in the case where $\theta=\pm\dfrac{\pi}{4}$.
The line perpendicular to the given line clearly makes angle $\dfrac{\pi}{2}+\theta$ with the positive X-axis, so its slope is $m_3=\tan\left(\dfrac{\pi}{2}+\theta\right)=-\cot\theta=\dfrac{-1}{\tan\theta}=\dfrac{-1}{m_1}$
A: If both perpendicular lines have a slope, neither of them is vertical.  If a line is not vertical, then the line perpendicular to it cannot be horizontal.  Thus, neither line has slope equal to zero.
We can choose our coordinate system so that the non-vertical perpendicular lines intersect at the origin, as shown in the diagram below.
If one line has slope $m$ and the line perpendicular to it has slope $m'$, then they have equations $y = mx$ and $y = m'x$, respectively (since the $y$-intercept is $0$).  
The line with slope $m$ intersects the vertical line $x = 1$ at the point $(1, m)$, while the line with slope $m'$ intersects the vertical line $x = 1$ at the point $(1, m')$.  
The three lines form a right triangle with vertices $(0, 0)$, $(1, m)$, and $(1, m')$.  
We can use the Distance Formula to determine the lengths of the legs and the hypotenuse.  The Distance Formula states that the distance between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is 
$$d(P_1, P_2) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$
The length of the leg from the origin to the point $(1, m)$ is $\sqrt{1 + m^2}$.  The length of the leg from the origin to the point $(1, m')$ is $\sqrt{1 + m'^2}$.  The length of the hypotenuse is the distance from $(1, m)$ to the point $(1, m')$, which is $|m - m'|$.  By the Pythagorean Theorem,
\begin{align*}
\sqrt{1 + m^2}^2 + \sqrt{1 + m'^2}^2 & = |m - m'|^2\\
1 + m^2 + 1 + m'^2 & = (m - m')^2\\
2 + m^2 + m'^2 & = m^2 - 2mm' + m'^2\\
2 & = -2mm'\\
-\frac{1}{m} & = m'\\
\end{align*} 
A: Consider two generic lines passing from the origin of the plane $x-y$:
$$y = mx, y = nx$$
Fix $x_1$ and $x_2$ and evaluate the two lines in these points:
$$y_1 = mx_1, y_2 = nx_2$$
The couples $[x1, y1]$ and $[x2, y2]$ are vectors that represents these points in the plane $x-y$. These points are perpendicular if their scalar product is null. In other words:
$$x_1x_2 + y_1y_2 = 0 \Rightarrow\\
\Rightarrow x_1x_2 + mx_1nx_2 = 0 \Rightarrow \\
\Rightarrow x_1x_2 (1 + mn) = 0$$
The previous is satisfied if:


*

*$x_1x_2 = 0$

*$1+mn = 0$


We want that the two lines are perpendicular, and this means that these vectors must be perpendicular for any $x_1$ and $x_2$. For any $x_1$ and $x_2$ we can't state that $x_1x_2 = 0$ . Then, we have to solve only:
$$1+mn = 0 \Rightarrow m = -\frac{1}{n}$$
The lines are:
$$y = mx, y = -\frac{1}{m}x$$
A: We can assume the given line $r$ has equation $y=mx$, so the point $A(0,1)$ does not belong to the line.
We want now to find the smallest distance from $A$ to $r$. Points $P$ on $r$ have coordinates $(t,mt)$, so the square of the distance from $A$ to $P$ is
$$
f(t)=t^2+(mt-1)^2=(1+m^2)t^2-2mt+1
$$
This function is a simple quadratic, so it has a minimum for $t=m/(1+m^2)$; it's easy to prove that the point at minimum distance is
$$
\left(\frac{m}{1+m^2},\frac{m^2}{1+m^2}\right)
$$
(see later if you don't know why).
The slope of the line connecting this point to $(0,1)$ is
$$
\frac{\dfrac{m^2}{1+m^2\mathstrut}-1}{\dfrac{m\mathstrut}{1+m^2}-0}=-\frac{1}{m}
$$

The minimum of a quadratic $f(t)=at^2+bt+c$ (with $a\ne0$) can be found by completing the square:
$$
f(t)=\frac{1}{4a}(4a^2t^2+4bt+4ac)=\frac{1}{4a}((2at+b)^2-(b^2-4ac))
$$
and since a square is nonnegative, the minimum value is attained when $2at+b=0$, that is, for
$$
t=-\frac{b}{2a}
$$
A: Suppose we have two blue lines (one vertical, the other horizontal). Next we rotate them, using the intersection as a pivot point, but when we do this, the lines magically become red. Suppose one of the red lines has a slope of $m$.
Try to take it from here. :)

