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 = -18/7#

take #x = 1# y you have #y = -9/7#

Ok you have two point #A = (2,-18/7)# and #B(1,-9/7)# those points are on the line

Now you want the vector formed by those points

#vec(AB) = (1-2,-9/7+18/7) = (-1,9/7)#

Let's call the point #(3,7)# #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(AP) = vec(AH) + vec(HP)#

and # vec(HP) _|_ vec(AB) #

add the determinant both side

#det(vec(AP),vec(AB)) = det(vec(AH),vec(AB)) + det(vec(HP),vec(AB))#

Now consider that #det(vec(a),vec(b)) = a*b*sin(theta)#

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

Obviously #det(vec(AH),vec(AB)) = 0# because #vec(AH)# and #vec(AB)# are on the same line ! so #theta = 0# and #sin(0) = 0 #

#det(vec(AP),vec(AB)) = det(vec(HP),vec(AB))#

Now you wanted a line perpendicular to our one so

#det(vec(HP),vec(AB)) = HP*AB*sin(pi/2) = HP*AB#

Finally do some calculation

#det(vec(AP),vec(AB)) = HP*AB#

#det(vec(AP),vec(AB))/(AB)= HP#

#vec(AP) = (3-2,7+18/7) = (1,67/7)#

#vec(AB) = (1-2,-9/7+18/7) = (-1,9/7)#

#det(vec(AP),vec(AB)) = 76/7#

#AB = sqrt((-1)^2+(9/7)^2) = sqrt(130)/7#

#HP = (76/7)/(sqrt(130)/7) = 76/sqrt(130)#

Ok now we use pythagore to have #AH#

#(sqrt(4538)/7)^2 = (76/sqrt(130))^2+AH^2#

#AH = (277 sqrt(2/65))/7#

Use trigonometry to have the angle formed by #vec(AB)# and the axis then have the angle formed by #vec(AH)# and the axis

You find #cos(theta) = -7/sqrt(130)#

You find #sin(theta) = 9/sqrt(130)#

#x = rcos(theta)#

#y = rsin(theta)#

Where #r# is the norm so :

#x = -277/65#

#y = 2493/455#

#vec(AH) = (-277/65, 2493/455)#

#H = (-277/65 + 2, 2493/455 - 18/7)#

#H = (-147/65, 189/65)#

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

Just need to imagine one more point #M = (x,y)# which can be anywhere

#vec(HM)# and #vec(AB)# are perpendicular if and only if #vec(HM)*vec(AB) = 0#

It's only because #vec(a)*vec(b) = a*b*cos(theta)# if they are perpendicular #theta = pi/2# and #cos(theta) = 0#

#vec(HM) = (x+147/65),(y-189/65)#

#vec(HM)*vec(AB) = -(x+147/65)+9/7(y-189/65)#

#-(x+147/65)+9/7(y-189/65)=0# is your line

Point red is #H#

Point black is #P#

Line blue is #vec(AB)#

You can see the two line