Question

Using section formula, show that the points A(7.-5), B(9,-3) and C(13,1) are collinear.​

Answer 1

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3:2

Step-by-step explanation:

Given Using the section formula, show that the points A(7, -5), B(9, -3), and C(13,1) are collinear.

Let point C(13, 1) divide the other ratio k:1. So we can take as m = k and n = 1

So x1 = 7, y1 = - 5, x2 = 9, y2 = - 3

We know that section formula is given by

(mx2 + nx1 /m + n , my2 + ny1 /m + n)

(13 , 1) = (9k + 7(1) / k +1 , -3k + (-5)1 / k + 1)

Equating the coordinates to the respective numbers we get

9k + 7 / k + 1 = 13 ,  -3k - 5 / k + 1 = 1

- 4k = 6                      -4k = 6

 k = - 3/2                   k = - 3/2

Since k is negative, it divides externally in the ratio 3:2. Thus the points A, B, and C are collinear.

Answer 2

Answer:

Given Using the section formula, show that the points A(7, -5), B(9, -3), and C(13,1) are collinear.

Let point C(13, 1) divide the other ratio k:1. So we can take as m = k and n = 1

So x1 = 7, y1 = - 5, x2 = 9, y2 = - 3

We know that section formula is given by

(mx2 + nx1 /m + n , my2 + ny1 /m + n)

(13 , 1) = (9k + 7(1) / k +1 , -3k + (-5)1 / k + 1)

Equating the coordinates to the respective numbers we get

9k + 7 / k + 1 = 13 , -3k - 5 / k + 1 = 1

- 4k = 6 -4k = 6

k = - 3/2 k = - 3/2

Since k is negative, it divides externally in the ratio 3:2. Thus the points A, B, and C are collinear.