Part A
Refer to the diagram below. Specifically, refer to figure 1.
Point P is added to form a right triangle.
J is at (-7,2) and P is at (11,2). Segment JP is 18 units long.
We split segment JP into 4 equal pieces to get 18/4 = 4.5; this result is added onto the x coordinate of J to get -7+4.5 = -2.5 which is the x coordinate of point Q. So Q is at (-2.5, 2)
The value -2.5 is the x coordinate of point S as well.
Notice how the ratio JQ:QP is equal to 1:3.
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The distance from P to H is 6 units, which divided into four pieces means 6/4 = 1.5 is added onto the y coordinate of P to get 2+1.5 = 3.5; this is the y coordinate of points R and S as the diagram indicates. Like in the previous section, the ratio PR:RH is equal to 1:3
So ultimately point S is located at (-2.5, 3.5)
If we used the distance formula to compute the lengths of segments JS and SH, then we should find that they are in the ratio 1:3. In other words, segment SH is 3 times longer compared to segment JS. This would help confirm the answer. I'll let you do this confirmation part.
Answer: (-2.5, 3.5)
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Part B
Refer to figure 2 this time.
We'll have a similar thing going on here as done in part A, though the points Q and R are in different locations.
To find where Q must be, we take 5/6 of the distance from J to P (which we found was 18 units). So (5/6)*18 = 15. This 15 is added onto the x coordinate of J to get -7+15 = 8. This is the x coordinate of point Q and point T.
Similarly, (5/6)*(length of segment PH) = (5/6)*6 = 5 is added onto the y coordinate of P to form the point R(11,7). The y coordinate of this helps form the y coordinate of the point T.
In short, point T is located at (8,7)
You should find that JS/SH is equal to 5/6 when you apply the distance formula to compute the lengths of JS and SH. I'll let you do this confirmation.
Answer: (8,7)
