Question

P, Q, R are three collinear points. The coordinates of P and R are (3, 4) and (11, 10) respectively and PQ is equal to 2.5 units. Coordinates of Q are-

• A. (5, 11/2)
• B. (11, 5/2)
• C. (5, -11/2)
• D. (-5, 11/2)

Answer 1

The correct option is A (5, 11/2)

Given that PQR are three collinear points and
PQ=2.5. Let us now find the distance of PR

Recall Distance Formula is =$ \sqrt { { \left( { x}_{ 2 }{ { x }_{ 1 } } \right) }^{ 2}+{ \left( { y }_{ 2 }{ { y }_{ 1 } } \right) }^{ 2 } } \\$

Distance PR=$\sqrt{\left(113{)}^2\right)+(104{)}^2}=\sqrt{100}=10$

Let us now consider S is the midpoint of PR,
then we have the co-ordinates of S as $\frac{11+3}{2},\frac{10+4}{2}=(7,7)$.

$\therefore$ it is clear that Q is the

midpoint of PS as we have know PQ=2.5 units.

Thus, to get the co-ordinates of Q take
midpoint of PS which is $\frac{3+7}{2},\frac{4+7}{2}=(5,11/2)$.