Question

Point Q lies on segment ST, where point S is located at (- 3, - 3) and location of point T is unknown. If SQ : QT = 4 : 3, and point Q is located at (3, 0) on the segment ST, select a place on the coordinate grid to plot point T and name its coordinate.

Answer 1

Answer:

The location of point T is $\vec T = (7.5, 2.25)$. 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:

$\overrightarrow {SQ} = \frac{4}{7}\cdot \overrightarrow {ST}$ (Eq. 1)

$\overrightarrow{QT} = \frac{3}{7}\cdot \overrightarrow {ST}$ (Eq. 2)

Now, we eliminate $\overrightarrow{ST}$ by equalizing (Eq. 1) and (Eq. 2):

$\frac{7}{3} \cdot \overrightarrow{QT} = \frac{7}{4}\cdot \overrightarrow{SQ}$ (Eq. 3)

From Linear Algebra, we know that:

$\frac{7}{3}\cdot (\vec T - \vec Q) = \frac{7}{4}\cdot (\vec Q-\vec S)$ (Eq. 3b)

And we clear $\vec T$

$\frac{7}{3}\cdot \vec T -\frac{7}{3}\cdot \vec Q = \frac{7}{4}\cdot \vec Q -\frac{7}{4}\cdot \vec S$

$\frac{7}{3}\cdot \vec T = \frac{49}{12}\cdot \vec Q - \frac{7}{4}\cdot \vec S$

$\vec T = \frac{3}{7}\cdot \left(\frac{49}{12}\cdot \vec Q-\frac{7}{4}\cdot \vec S\right)$

$\vec T = \frac{7}{4}\cdot \vec Q-\frac{3}{4}\cdot \vec S$

If we know that $Q(x,y) = (3, 0)$ and $S(x, y) = (-3,-3)$, then we have that $T(x,y)$ is:

$\vec T = \frac{7}{4}\cdot (3, 0) - \frac{3}{4}\cdot (-3, -3)$

$\vec T = \left(\frac{21}{4}, 0 \right) + \left(\frac{9}{4},\frac{9}{4}\right)$

$\vec T = \left(\frac{21+9}{4},\frac{0+9}{4} \right)$

$\vec T = \left(\frac{15}{2}, \frac{9}{4}\right)$

$\vec T = (7.5, 2.25)$

At last we plot the segment with all points described on statement. We can check the result in the attachment set below.