# What is the equation of the line perpendicular to y=-9/7x  that passes through  (3,7) ?

Jan 6, 2016

Hi, here a "pretty long answer" but don't be afraid ! it's only logic, if you are able to do that, you are able to rules the world, promise ! draw it on a paper and everything will be ok (draw it without axis you don't need it, it's only geometry : ) ) what you need to know : Basic trigonometry, pythagore, determinant, polar coordinate and scalar product

I will explain how it work behind the scene

First you need to search two point of the line

take $x = 2$ you have $y = - \frac{18}{7}$
take $x = 1$ y you have $y = - \frac{9}{7}$

Ok you have two point $A = \left(2 , - \frac{18}{7}\right)$ and $B \left(1 , - \frac{9}{7}\right)$ those points are on the line

Now you want the vector formed by those points

$\vec{A B} = \left(1 - 2 , - \frac{9}{7} + \frac{18}{7}\right) = \left(- 1 , \frac{9}{7}\right)$

Let's call the point $\left(3 , 7\right)$ $P$

Ok now imagine the line you want which is perpendicular to our one, they intersect in one point, let's call this point $H$ we don't know what is $H$ and we want to know.

we know two things :

$\vec{A P} = \vec{A H} + \vec{H P}$

and $\vec{H P} \bot \vec{A B}$

$\det \left(\vec{A P} , \vec{A B}\right) = \det \left(\vec{A H} , \vec{A B}\right) + \det \left(\vec{H P} , \vec{A B}\right)$

Now consider that $\det \left(\vec{a} , \vec{b}\right) = a \cdot b \cdot \sin \left(\theta\right)$

where $a$ and $b$ are the norm and $\theta$ the angle between the two vector

Obviously $\det \left(\vec{A H} , \vec{A B}\right) = 0$ because $\vec{A H}$ and $\vec{A B}$ are on the same line ! so $\theta = 0$ and $\sin \left(0\right) = 0$

$\det \left(\vec{A P} , \vec{A B}\right) = \det \left(\vec{H P} , \vec{A B}\right)$

Now you wanted a line perpendicular to our one so

$\det \left(\vec{H P} , \vec{A B}\right) = H P \cdot A B \cdot \sin \left(\frac{\pi}{2}\right) = H P \cdot A B$

Finally do some calculation

$\det \left(\vec{A P} , \vec{A B}\right) = H P \cdot A B$

$\det \frac{\vec{A P} , \vec{A B}}{A B} = H P$

$\vec{A P} = \left(3 - 2 , 7 + \frac{18}{7}\right) = \left(1 , \frac{67}{7}\right)$

$\vec{A B} = \left(1 - 2 , - \frac{9}{7} + \frac{18}{7}\right) = \left(- 1 , \frac{9}{7}\right)$

$\det \left(\vec{A P} , \vec{A B}\right) = \frac{76}{7}$

$A B = \sqrt{{\left(- 1\right)}^{2} + {\left(\frac{9}{7}\right)}^{2}} = \frac{\sqrt{130}}{7}$

$H P = \frac{\frac{76}{7}}{\frac{\sqrt{130}}{7}} = \frac{76}{\sqrt{130}}$

Ok now we use pythagore to have $A H$

${\left(\frac{\sqrt{4538}}{7}\right)}^{2} = {\left(\frac{76}{\sqrt{130}}\right)}^{2} + A {H}^{2}$

$A H = \frac{277 \sqrt{\frac{2}{65}}}{7}$

Use trigonometry to have the angle formed by $\vec{A B}$ and the axis then have the angle formed by $\vec{A H}$ and the axis

You find $\cos \left(\theta\right) = - \frac{7}{\sqrt{130}}$

You find $\sin \left(\theta\right) = \frac{9}{\sqrt{130}}$

$x = r \cos \left(\theta\right)$

$y = r \sin \left(\theta\right)$

Where $r$ is the norm so :

$x = - \frac{277}{65}$

$y = \frac{2493}{455}$

$\vec{A H} = \left(- \frac{277}{65} , \frac{2493}{455}\right)$

$H = \left(- \frac{277}{65} + 2 , \frac{2493}{455} - \frac{18}{7}\right)$

$H = \left(- \frac{147}{65} , \frac{189}{65}\right)$

Now you have this point you can say "AAAAAAAAAAAAAH" because you finished soon

Just need to imagine one more point $M = \left(x , y\right)$ which can be anywhere

$\vec{H M}$ and $\vec{A B}$ are perpendicular if and only if $\vec{H M} \cdot \vec{A B} = 0$

It's only because $\vec{a} \cdot \vec{b} = a \cdot b \cdot \cos \left(\theta\right)$ if they are perpendicular $\theta = \frac{\pi}{2}$ and $\cos \left(\theta\right) = 0$

$\vec{H M} = \left(x + \frac{147}{65}\right) , \left(y - \frac{189}{65}\right)$

$\vec{H M} \cdot \vec{A B} = - \left(x + \frac{147}{65}\right) + \frac{9}{7} \left(y - \frac{189}{65}\right)$

$- \left(x + \frac{147}{65}\right) + \frac{9}{7} \left(y - \frac{189}{65}\right) = 0$ is your line

Point red is $H$
Point black is $P$
Line blue is $\vec{A B}$

You can see the two line