Points P, Q, R are collinear such that d(P, Q) = 13cm, d(R,P) =4cm then find d(Q, R)
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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=
(113)
2
)+(104)
2
=
100
=10
Let us now consider S is the midpoint of PR,
then we have the co-ordinates of S as
2
11+3
,
2
10+4
=(7,7).
∴ 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
2
3+7
,
2
4+7
=(5,11/2).
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=
(113)
2
)+(104)
2
=
100
=10
Let us now consider S is the midpoint of PR,
then we have the co-ordinates of S as
2
11+3
,
2
10+4
=(7,7).
∴ 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
2
3+7
,
2
4+7
=(5,11/2).
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