Proof that two straight lines are perpendicular and the product of their slopes is -1

When I was learning elementary functions a long time ago, if the two straight lines in the linear function y = m0x + b0, y = m1x + b1 are vertical, it is concluded that the slope product of the two straight lines is -1, that is, m0*m1 = -1, It was only used before, and it was not proved. When I was studying graphics recently, I suddenly remembered this point, so I took a note to prove it.

As shown in the figure above, there are two straight lines: $y_0 = m_0x + b_0$ and $y_1 = m_1x + b_1$ , which are perpendicular to each other. An implicit condition can be obtained here: $m_0 \neq m1$ (two straight lines with equal slopes and different y-axis intercepts are parallel, and if they are vertical, the slopes are not equal).

The coordinates of the intersection of the two straight lines in the figure can be obtained by solving the equation. The y of the intersection is the same, so we have:

$m_0x + b_0 = m_1x + b_1$

The x-coordinate of the intersection point obtained by solving is: $\frac{b_1 - b_0}{m_0 - m_1}$ ,

Substitute x into y0 and y1 respectively, and the y coordinates of the intersection points are:

$\frac{m_0(b_1 - b_0)}{m_0 - m_1} + b_0$ and $\frac{m_1(b_1 - b_0)}{m_0 - m_1} + b_1$ , the two values are equal

Therefore, the coordinates of the three key points in the figure are as follows:

The coordinates of the intersection point A of the straight line y0 on the y-axis are (0, b0)

The coordinates of the intersection point B of the straight line y1 on the y-axis are (0,b1)

The C coordinate of the intersection point of the two straight lines is that $(\frac{b_1 - b_0}{m_0 - m_1}, \frac{m_0(b_1 - b_0)}{m_0 - m_1} + b_0) , (\frac{b_1 - b_0}{m_0 - m_1}, \frac{m_1(b_1 - b_0)}{m_0 - m_1} + b_1)$ the two coordinates correspond to the same point.

Since the two straight lines are perpendicular, it can be known from the Pythagorean theorem that the square of the distance between the hypotenuse AB = the square of the distance between the right-angled side AC + the square of the distance between the right-angled side BC.

According to the distance formula between two points, the following equation can be obtained:

The square of the distance of AB = $(b_1 - b_0)^2$

The square of the distance of AC = $(\frac{m_0(b_1 - b_0)}{m_0 - m_1} )^2 + (\frac{b_1 - b_0}{m_0 - m_1})^2$ (Using the first form of C for distance calculation, b0 can be subtracted)

The square of the distance of BC = $(\frac{m_1(b_1 - b_0)}{m_0 - m_1} )^2 + (\frac{b_1 - b_0}{m_0 - m_1})^2$ (Using the second form of C for distance calculation, b1 can be subtracted)

According to the definition of Pythagorean shares, we can get:

$(b_1 - b_0)^2 = (\frac{m_0(b_1 - b_0)}{m_0 - m_1} )^2 + (\frac{b_1 - b_0}{m_0 - m_1})^2 + (\frac{m_1(b_1 - b_0)}{m_0 - m_1} )^2 + (\frac{b_1 - b_0}{m_0 - m_1})^2$

Sort it out and get:

$(b_1 - b_0)^2 = \frac{m_0^2(b_1 - b_0)^2}{(m_0 - m_1)^2} + 2\frac{(b_1 - b_0)^2}{(m_0 - m_1)^2}) + \frac{m_1^2(b_1 - b_0)^2}{(m_0 - m_1)^2}$

Subtract (b1- b0)^2 and get:

$(m_0 - m_1)^2 = m_0^2 + 2 + m_1^2$

Expand the difference of squares:

$m_0^2 + m_1^2 - 2m_0m_1 = m_0^2 + m_1^2 + 2$

sorted out

$-2m_0m_1 = 2$ , therefore $m_0m_1 = -1$

Proof that two straight lines are perpendicular and the product of their slopes is -1

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