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A (2, 5), B ( – 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.

Show that PQ = 1/3 BC

Answer:

**Solution:**

The Euclidean distance formula is used to calculate the separation between two points on a two-dimensional plane.

By distance formula, d(PQ) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

**Points are P(1,4) and Q(3,6).**

**So x _{1}= 1, y_{1}** **= 4**

**x _{2}** **= 3, y

d(PQ) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

d(PQ) = √[(3-1)^{2}+(6-4)^{2}]

d(PQ) = √[(2)^{2}+(2)^{2}]

d(PQ) = √(4+4)

d(PQ) = √8 = 2√2 ..(i)

By distance formula, d(BC) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

**Points are B(-1,2) and C(5,8).**

**So x _{1}= -1, y_{1}** **= 2**

**x _{2}** **= 5, y

d(BC) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

d(BC) = √[(5-(-1))^{2}+(8-2)^{2}]

d(BC) = √[(6)^{2}+(6)^{2}]

d(BC) = √(36+36)

d(BC) = √72 = √(36×2) = 6√2 ..(ii)

BC/3 = 6√2/3 = 2√2 = PQ

**PQ = 1/3 BC.**

**Hence proved.**

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