Answer:
a) (8,12) is the point [tex]\frac{3}{5}[/tex] of the way from R to S, b) (0, 6) is the point [tex]\frac{1}{5}[/tex] of the way from R to S.
Step-by-step explanation:
Let be [tex]X[/tex] a point in the line between endpoints [tex]R[/tex] and [tex]S[/tex]. The ratio of the line segment [tex]RX[/tex] to the line segment [tex]RS[/tex] is given by:
[tex]r = \frac{RX}{RS}[/tex]
Where:
[tex]RX = \sqrt{(x_{X}-x_{R})^{2}+(y_{X}-y_{R})^{2}}[/tex]
[tex]RS = \sqrt{(x_{S}-x_{R})^{2}+(y_{S}-y_{R})^{2}}[/tex]
Let be [tex]x_{R} = -4[/tex], [tex]y_{R} = 3[/tex], [tex]x_{S} = 16[/tex] and [tex]y_{S} = 18[/tex].
a) If we know that [tex]x_{X} = 8[/tex] and [tex]y_{X} = 12[/tex], the ratio of the line segment [tex]RX[/tex] to the line segment [tex]RS[/tex] is:
[tex]RX =\sqrt{[8-(-4)]^{2}+(12-3)^{2}}[/tex]
[tex]RX = 15[/tex]
[tex]RS =\sqrt{[16-(-4)]^{2}+(18-3)^{2}}[/tex]
[tex]RS = 25[/tex]
Then,
[tex]r = \frac{15}{25}[/tex]
[tex]r = \frac{3}{5}[/tex]
(8,12) is the point [tex]\frac{3}{5}[/tex] of the way from R to S.
b) If we know that [tex]x_{X} = 0[/tex] and [tex]y_{X} = 6[/tex], the ratio of the line segment [tex]RX[/tex] to the line segment [tex]RS[/tex] is:
[tex]RX =\sqrt{[0-(-4)]^{2}+(6-3)^{2}}[/tex]
[tex]RX = 5[/tex]
[tex]RS =\sqrt{[16-(-4)]^{2}+(18-3)^{2}}[/tex]
[tex]RS = 25[/tex]
Then,
[tex]r = \frac{5}{25}[/tex]
[tex]r = \frac{1}{5}[/tex]
(0, 6) is the point [tex]\frac{1}{5}[/tex] of the way from R to S.
