Step-by-step explanation:
distance AB is (using Pythagoras with the coordinate differences between both points as sides and the distance as Hypotenuse = baseline of the triangle, the side opposite of the 90 degree angle):
distance² = (5 - -10)² + (7-2)² = 15² + 5² = 225 + 25 = 250
distance = sqrt(250)
to split this distance into a 3/2 ratio, we need actually 3+2=5 equal parts. and then AM gets 3 off these parts, and MB gets 2 of these parts.
so,
distance/5 = sqrt(250)/5 = sqrt(250/25) = sqrt(10)
so,
AM = 3×sqrt(10) = sqrt(90),
and
MB = 2×sqrt(10) = sqrt(40)
now, we need to calculate back using the same Pythagoras approach (calling the coordinates of M xm and ym)
AM² = (xm - -10)² + (ym - 2)² = (xm+10)² + (ym-2)² = 90
90 = xm² + 20xm + 100 + ym² - 4ym + 4
MB² = (5 - xm)² + (7 - ym)² = 40
40 = 25 - 10xm + xm² + 49 - 14ym + ym²
as a first approach we calculate AM² - MB²
90 = xm² + 20xm + ym² - 4ym + 104
- 40 = xm² - 10xm + ym² - 14ym + 74
----------------------------------------------------
50 = 0 + 30xm + 0 + 10ym + 30
20 = 30xm + 10ym
2 = 3xm + ym
ym = 2 - 3xm
this we use now e.g. in the first equation for AM.
90 = xm² + 20xm + (2-3xm)² - 4×(2-3xm) + 104
-14 = xm² + 20xm + 4 - 12xm + 9xm² - 8 + 12xm
-10 = 10xm² + 20xm
-1 = xm² + 2xm
xm² + 2xm + 1 = 0
solving such a squared equation
xm = (-b ± sqrt(b² - 4ac))/(2a)
a = 1
b = 2
c = 1
xm = (-2 ± sqrt(4 - 4))/2 = -1
only one combined solution (as a squared equation usually has 2 solutions).
ym = 2 - 3xm = 2 - 3×-1 = 2 + 3 = 5
so, M = (-1, 5)
