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Mathematics

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

Coordinate Geometry

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Answer

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Let A and B be the points of trisection.

Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). NCERT Class 10 Mathematics CBSE Solutions.

So, A divides line segment joining (4, –1) and (–2, –3) in the ratio 1 : 2.

Let co-ordinates of A be (a, b).

⇒(a,b)=(1×−2+2×41+2,1×−3+2×−11+2)=(−2+83,−3−23)=(63,−53)=(2,−53).\Rightarrow (a, b) = \Big(\dfrac{1 \times -2 + 2 \times 4}{1 + 2}, \dfrac{1 \times -3 + 2 \times -1}{1 + 2}\Big) \\[1em] = \Big(\dfrac{-2 + 8}{3}, \dfrac{-3 - 2}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{-5}{3}\Big) \\[1em] = \Big(2, -\dfrac{5}{3}\Big).

B divides line segment joining (4, –1) and (–2, –3) in the ratio 2 : 1.

Let co-ordinates of B be (c, d).

⇒(c,d)=(2×−2+1×42+1,2×−3+1×−12+1)=(−4+43,−6−13)=(03,−73)=(0,−73).\Rightarrow (c, d) = \Big(\dfrac{2 \times -2 + 1 \times 4}{2 + 1}, \dfrac{2 \times -3 + 1 \times -1}{2 + 1}\Big) \\[1em] = \Big(\dfrac{-4 + 4}{3}, \dfrac{-6 - 1}{3}\Big) \\[1em] = \Big(\dfrac{0}{3}, \dfrac{-7}{3}\Big) \\[1em] = \Big(0, -\dfrac{7}{3}\Big).

Hence, points of intersection are (2,−53),(0,−73)\Big(2, -\dfrac{5}{3}\Big), \Big(0, -\dfrac{7}{3}\Big).

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