Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Asked by Abhisek | 1 year ago |  67

##### Solution :-

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the

line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

x$$\dfrac{(1×(-2) + 2×4)}{3}$$$$\dfrac{(-2 + 8)}{3}$$$$\dfrac{6}{3}$$ = 2

y1 = $$\dfrac{ (1×(-3) + 2×(-1))}{(1 + 2)}= \dfrac{ (-3 – 2)}{3}= \dfrac{-5}{3}$$

Therefore: P (x1, y1) = P(2, $$\dfrac{-5}{3}$$)

Point Q divides AB internally in the ratio 2:1.

x2 = $$\dfrac{(2×(-2) + 1×4)}{(2 + 1)}= \dfrac{(-4 + 4)}{3}=0$$

y2 = $$\dfrac{(2×(-3) + 1×(-1))}{(2 + 1)}= \dfrac{(-6 – 1)}{3}= \dfrac{-7}{3}$$

The coordinates of the point Q is (0,$$\dfrac{-7}{3}$$)

Answered by Pragya Singh | 1 year ago

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