Question

show that the point (0, 3 )(6,0) and (4,1) lie on a straight line using the Section formula

Answer 1

since AB = BC the mid point is B so it all lies in a straight line

Answer 2

The points (0,3) (6,0) and (4,1) lie on a straight line and the point C lies internally with ratio 2:1

Step-by-step explanation:

Given Data

$(x_1, y_1) = (0, 3 ) \Rightarrow A$

$(x_2, y_2) = (6, 0) \Rightarrow B$

$(x_3, y_3) = (4, 1) \Rightarrow C$

Show that the above mentioned points are lie on a straight line using section formula

Let us consider the point c lies between the points A and B

The section formula is

$C = \frac{mx_2 + nx_1}{m+n} , \frac{my_2 + ny_1}{m+n}$

The point C (4,1) divides A (0, 3 ) and B (6,0) in the ratio of k : 1

Substitute the respective values in above formula, where m = k and n=1

$(4,1) = \frac{k (6) + (1)(0)}{k+1} , \frac{k(0) + (1)(3)}{k+1}$

$(4,1) = \frac{k (6) + 0}{k+1} , \frac{0 + 3}{k+1}$

$(4,1) = \frac{6k}{k+1} , \frac{3}{k+1}$

Equate the x and y terms

=>   $4 = \frac{6k}{k+1}$          =>     $1 = \frac{3}{k+1}$

     $6k = 4k + 4$                 $k + 1 = 3$

     2k = 4                         k = 3 -1

     k =2                             k = 2

The point C (4,1) divides A (0, 3 ) and B (6,0) in the ratio of 2 : 1

Therefore the points (0, 3 )(6,0) and (4,1) are lie in a straight line and the point C lies internally ( Between A and B)

To Learn more ....

1)Show by Section formula that the points ( 3, - 2 ) ,( 5, 2) and (8,8) are collinear

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

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