The points #(n-2, n - 3)#, #(n + 1, n+9)# , and #(70, n -31)# are collinear. What is the value of #n#?

1 Answer

#n = 79#

Explanation:

This line can't be vertical, because it's impossible
#n -2 = n + 1 = 70#.

For non-vertical lines we use #f(x) = ax + b# three times, so that we can find #a(n), b(n), n#.

#n - 3 = a (n - 2) + b Rightarrow a = frac{n - 3 - b}{n - 2}# // Eq 1

#n + 9 = a(n+1) + b# // Eq 2

#n - 31 = 70a + b Rightarrow b = n - 31 - 70 cdot frac{n - 3 - b}{n - 2}#

We solve for b:

#b - (70b)/(n-2) = n - 31 - 70 cdot (n-3)/(n-2)# // times (n-2)

#b(n-2 -70) = n(n-2) -31(n-2) - 70(n - 3)#

#b(n) = frac{n^2 -103n + 272}{n - 72}#

From Eq. 1 #Rightarrow a = frac{n - 3 - frac{n^2 -103n + 272}{n - 72}}{n - 2}#

#a(n) = frac{n(n - 72) - 3(n - 72) -n^2 + 103n -272}{(n-2)(n-72)}#

Finally from Eq. 2,

#n+ 9 = frac{28n - 56}{(n-2)(n-72)} cdot (n + 1) + frac{n^2 -103n + 272}{n - 72}#

#Rightarrow n(n-72) + 9(n-72) = frac{28(n - 2)}{n-2} cdot (n + 1) + n^2 -103n + 272#

#Rightarrow n^2 - 72 n + 9n - 648 = n^2 -75n + 300#

#Rightarrow - 63 n - 648 = -75n + 300#

#Rightarrow (75- 63) n = 300 + 648#