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Find the co-ordinates of the points of tri-section of the line joining the points (-3, 0) and (6, 6).

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Answer

From figure,

Find the co-ordinates of the points of tri-section of the line joining the points (-3, 0) and (6, 6). Section and Mid-Point Formula, Concise Mathematics Solutions ICSE Class 10.

Let A and B be the points of tri-section of the line joining the points (-3, 0) and (6, 6).

So, A and B divides the segment in three equal parts.

A divides the line segment in ratio 1 : 2. Let co-ordinates of A be (a, b).

By formula,

x=m1x2+m2x1m1+m2a=1×6+2×31+2a=663a=0y=m1y2+m2y1m1+m2b=1×6+2×01+2b=63=2.x = \dfrac{m1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow a = \dfrac{1 \times 6 + 2 \times -3}{1 + 2} \\[1em] \Rightarrow a = \dfrac{6 - 6}{3} \\[1em] \Rightarrow a = 0 \\[1em] y = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] \Rightarrow b = \dfrac{1 \times 6 + 2 \times 0}{1 + 2} \\[1em] \Rightarrow b = \dfrac{6}{3} = 2.

A = (a, b) = (0, 2).

B divides the line segment in ratio 2 : 1. Let co-ordinates of B be (c, d).

By formula,

x=m1x2+m2x1m1+m2c=2×6+1×31+2c=1233c=93=3.y=m1y2+m2y1m1+m2d=2×6+1×02+1d=123=4.x = \dfrac{m1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow c = \dfrac{2 \times 6 + 1 \times -3}{1 + 2} \\[1em] \Rightarrow c = \dfrac{12 - 3}{3} \\[1em] \Rightarrow c = \dfrac{9}{3} = 3. \\[1em] y = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] \Rightarrow d = \dfrac{2 \times 6 + 1 \times 0}{2 + 1} \\[1em] \Rightarrow d = \dfrac{12}{3} = 4.

B = (c, d) = (3, 4).

Hence, points of tri-section are (0, 2) and (3, 4).

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