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-

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=$\sqrt { (113)^{ 2 })+( 104) ^{ 2 } }\\=\sqrt{100}=10$

Let us now consider S is the midpoint of PR,

then we have the co-ordinates of S as $\dfrac{11+3}{2},\dfrac{10+4}{2}=(7, 7)$.

$\therefore$ 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 $\dfrac{3+7}{2},\dfrac{4+7}{2}=(5,11/2)$.