Answer:
To find the point H that lies two-thirds of the way along the line segment FG, you can use the following formula:
H(x,y) = ([tex]\frac{2}{3} . x_{2} +\frac{1}{3} . x_{1} , \frac{2}{3} . y_{2} + \frac{1}{3} . y_{1}[/tex])
where ([tex]x_{1} , y_{1}[/tex]) are the coordinates of point F and ([tex]x_{2} , y_{2}[/tex]) are the coordinates of point G.
For the given points F(3,4) and G(9,13), plug these values into the formula:
H(x,y) = ([tex]\frac{2}{3} . 9 +\frac{1}{3} . 3 , \frac{2}{3} . 13 + \frac{1}{3} .4[/tex])
Now calculate:
H(x,y) = (6 + 1, [tex]\frac{26}{3}[/tex] + [tex]\frac{4}{3}[/tex])
Simplify:
H(x,y) = (7,10)
Therefore, the coordinates of point H are indeed (7, 10), confirming that it lies two-thirds of the way along the line segment FG.
Answer:
Method #1:
m = (13 - 4)/(9 - 3) = 9/6 = 3/2
From F, go up 3 units, then right 2 units, to (5, 7). From (5, 7), go up 3 units, then right 2 units, to (7, 10). From (7, 10). go up 3 units, then right 2 units, to (9, 13).
So H has coordinates (7, 10).
Method #2:
(9 - 3)/3 = 2, (13 - 4)/3 = 3
The coordinates of H are (3 + 2(2), 4 + 3(2)) = (3 + 4, 4 + 6) = (7, 10).
