Question

Find the point, Q, along the directed line segment AB that

Answer 1

Answer:

Point Q is at a distance of 4.7 units from A.

Step-by-step explanation:

From the graph, AC = 10 units and BC = 6 units. Applying the Pythagoras theorem,

$AB^{2}$ = $AC^{2}$ + $BC^{2}$

      = $10^{2}$ + $6^{2}$

      = 100 + 36

     = 136

AB = $\sqrt{136}$

AB = 11.6619

AB = 11.66

     ≅ 11.7 units

But point Q divides AB into ratio 2:3. Therefore:

AQ = $\frac{2}{5}$ × AB

     =  $\frac{2}{5}$ × 11.66

     = 4.664

AQ = 4.664

AQ ≅ 4.7 units

QB = $\frac{3}{5}$ × AB

     =  $\frac{3}{5}$ × 11.66

     = 6.996

QB  ≅ 7.0 units

So that point Q is at a distance of 4.7 units from A.