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,
[tex]AB^{2}[/tex] = [tex]AC^{2}[/tex] + [tex]BC^{2}[/tex]
= [tex]10^{2}[/tex] + [tex]6^{2}[/tex]
= 100 + 36
= 136
AB = [tex]\sqrt{136}[/tex]
AB = 11.6619
AB = 11.66
≅ 11.7 units
But point Q divides AB into ratio 2:3. Therefore:
AQ = [tex]\frac{2}{5}[/tex] × AB
= [tex]\frac{2}{5}[/tex] × 11.66
= 4.664
AQ = 4.664
AQ ≅ 4.7 units
QB = [tex]\frac{3}{5}[/tex] × AB
= [tex]\frac{3}{5}[/tex] × 11.66
= 6.996
QB ≅ 7.0 units
So that point Q is at a distance of 4.7 units from A.
