let's say that C is "x" units farther from B, that means that CB = x, and therefore AC = 1.5x.
![\bf \underset{\leftarrow ~~30~~\to}{\boxed{A}\stackrel{1.5x}{\rule[0.35em]{18em}{0.25pt}}C\stackrel{x}{\rule[0.35em]{10em}{0.25pt}}\boxed{B}} \\\\\\ AB=AC+CB\implies \stackrel{AB}{30}=\stackrel{AC}{1.5x}+\stackrel{CB}{x}\implies 30=2.5x \\\\\\ \cfrac{30}{2.5}=x\implies 12=x \\\\[-0.35em] ~\dotfill\\\\ AC=1.5(12)\implies AC=18~\hspace{8em} CB=x\implies CB=12](https://tex.z-dn.net/?f=%5Cbf%20%5Cunderset%7B%5Cleftarrow%20~~30~~%5Cto%7D%7B%5Cboxed%7BA%7D%5Cstackrel%7B1.5x%7D%7B%5Crule%5B0.35em%5D%7B18em%7D%7B0.25pt%7D%7DC%5Cstackrel%7Bx%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D%5Cboxed%7BB%7D%7D%0A%5C%5C%5C%5C%5C%5C%0AAB%3DAC%2BCB%5Cimplies%20%5Cstackrel%7BAB%7D%7B30%7D%3D%5Cstackrel%7BAC%7D%7B1.5x%7D%2B%5Cstackrel%7BCB%7D%7Bx%7D%5Cimplies%2030%3D2.5x%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B30%7D%7B2.5%7D%3Dx%5Cimplies%2012%3Dx%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0AAC%3D1.5%2812%29%5Cimplies%20AC%3D18~%5Chspace%7B8em%7D%20CB%3Dx%5Cimplies%20CB%3D12)
First, find the slope (m) =
=
=
= 
Now plug in ONE of the points and the slope into the point-slope equation:
y - y₁ = m(x - x₁); where (x₁, y₁) is the chosen point.
y - 5 =
(x - 1) (I used (1,5) as the chosen point)
3(y - 5) = x - 1
3y - 15 = x - 1
3y -14 = x
-14 = x - 3y → x - 3y = -14
Answer: x - 3y = -14
10 - 2x = -5x -5
+2x +2x
10 = -3x -5
+5 +5
15= -3x
X= -5
Answer:
-3/2
Explanation:
Slopes of perpendicular lines are opposite reciprocals, which means you flip them and switch them from positive to negative, or the other way around.
Answer: 5/4
Step-by-step explanation: