find the distance between S and T
[tex]d=\sqrt{(-2-18)^2+(-6-9)^2}[/tex]
[tex]d=25[/tex]
so i distance is ratio o 3:8
3+8=11
25/11=25/11
3*25/11=75/11
8*25/11=200/11
the distance from S to Q is 75/11 and distance rom Q to T is 200/11
if (x,y) is point Q
[tex]\frac{75}{11}=\sqrt{(x+2)^2+(y+6)^2}[/tex] is a circle made up of points where Q is 75/11 units away from S
if we find the equation of the line ST and find the intersection of the line and this circle, we can ind the point Q
or (-2,-6) and (18,9), slope is (-6-9)/(-2-18)=-15/-20=3/4
y=3/4x+b, if we put (-2,-6) in we get that b=-9/2
y=3/4x-9/2
if we subsitute for y we get
[tex]\frac{75}{11}=\sqrt{(x+2)^2+(\frac{3}{4}x-\frac{9}{2}+6)^2}[/tex]
if we solve this by squareing both sides and doing some algebraeic stuff we get 2 points (because the line crosses the circle at 2 points):
(-82/11,-111/11) and (38/11,-21/11)
we have to pick the one between S and T, check the y value
the irst point has a y value of -10.0909 which is below S, so this isn't the point
the 2n point has a vaue of -1.90909 and this is between -6 and 9 so this is the point
Q is [tex](\frac{38}{11},\frac{-21}{11})[/tex]
