Answer:
The location of point T is [tex]\vec T = (7.5, 2.25)[/tex]. Graph is included at the end of the explanation as attachment.
Step-by-step explanation:
The SQ : QT ratio means that there are 4 units of SQ for each 3 units of QT. Vectorially speaking, we notice that:
[tex]\overrightarrow {SQ} = \frac{4}{7}\cdot \overrightarrow {ST}[/tex] (Eq. 1)
[tex]\overrightarrow{QT} = \frac{3}{7}\cdot \overrightarrow {ST}[/tex] (Eq. 2)
Now, we eliminate [tex]\overrightarrow{ST}[/tex] by equalizing (Eq. 1) and (Eq. 2):
[tex]\frac{7}{3} \cdot \overrightarrow{QT} = \frac{7}{4}\cdot \overrightarrow{SQ}[/tex] (Eq. 3)
From Linear Algebra, we know that:
[tex]\frac{7}{3}\cdot (\vec T - \vec Q) = \frac{7}{4}\cdot (\vec Q-\vec S)[/tex] (Eq. 3b)
And we clear [tex]\vec T[/tex]
[tex]\frac{7}{3}\cdot \vec T -\frac{7}{3}\cdot \vec Q = \frac{7}{4}\cdot \vec Q -\frac{7}{4}\cdot \vec S[/tex]
[tex]\frac{7}{3}\cdot \vec T = \frac{49}{12}\cdot \vec Q - \frac{7}{4}\cdot \vec S[/tex]
[tex]\vec T = \frac{3}{7}\cdot \left(\frac{49}{12}\cdot \vec Q-\frac{7}{4}\cdot \vec S\right)[/tex]
[tex]\vec T = \frac{7}{4}\cdot \vec Q-\frac{3}{4}\cdot \vec S[/tex]
If we know that [tex]Q(x,y) = (3, 0)[/tex] and [tex]S(x, y) = (-3,-3)[/tex], then we have that [tex]T(x,y)[/tex] is:
[tex]\vec T = \frac{7}{4}\cdot (3, 0) - \frac{3}{4}\cdot (-3, -3)[/tex]
[tex]\vec T = \left(\frac{21}{4}, 0 \right) + \left(\frac{9}{4},\frac{9}{4}\right)[/tex]
[tex]\vec T = \left(\frac{21+9}{4},\frac{0+9}{4} \right)[/tex]
[tex]\vec T = \left(\frac{15}{2}, \frac{9}{4}\right)[/tex]
[tex]\vec T = (7.5, 2.25)[/tex]
At last we plot the segment with all points described on statement. We can check the result in the attachment set below.
