Two towns are located at points A(2,-2) and B (8,5). A new school is to be built on a straight road with equation -x+7y=-4. Find the location of the school so that it is equaidistant from the two towns

The location is (8.88, -1.84)

Step-by-step explanation:

We know that the school is on the line:

x + 7*y  = -4

or we can rewrite this as:

x = -4 - 7*y

Then, the points on this line can be written as:

(-4 - 7*y, y).

Then we can assume that the school is located at the point:

(-4 - 7*a, a)

and we need to find the value of a.

Now, we have two towns in the points:

A = (2, - 2)

B = (8, 5)

The distance between two points (a, b) and (c, d) is:

D = √( (a- c)^2 + (b - d)^2)

And we want that the school to be equidistant between the two cities, then if the distance to the citie A is:

Da = √( (2 + 4 + 7*a)^2 + (-2 - a)^2)

And the distance to the city B is:

Db = √( (8 + 4 + 7*a)^2 + (5 - a)^2)

Then we must have that:

Da = Db

√( (2 + 4 + 7*a)^2 + (-2 - a)^2) = √( (8 + 4 + 7*a)^2 + (5 - a)^2)

then:

(2 + 4 + 7*a)^2 + (-2 - a)^2 =  (8 + 4 + 7*a)^2 + (5 - a)^2

Let's solve this for a.

(6 + 7*a)^2 + (-2 - a)^2 =  (12 + 7*a)^2 + (5 - a)^2

36 + 84*a + 49*a^2 + 4 + 4*a + a^2 = 144 + 168*a + 49*a^2 + 25 - 10*a + a^2

let's simplify this:

36 + 84*a + 4 + 4*a = 144 + 168*a + 25 - 10*a

-129 = 70*a

-129/70 = a = -1.84

Then the location of the school is:

y = -1.84

x = -4 - 7*-1.84 = 8.88  

(8.88, -1.84)