For what value of k will make the three points of (1, 4), (-3, 16) and (k, -2) collinear ?

Answer 1

Collinear points lie on the same straight line.

Slope of straight line:

m = ( y1 - y2 ) / ( x1 - x2 )

In this case:

x1 = 1 , y1 = 4 , x2 = - 3 , y2 = 16

m = ( y1 - y2 ) / ( x1 - x2 ) =

( 4 - 16 ) / [ 1 - ( - 3 ) ] =

- 12 / ( 1 + 3 ) = - 12 / 4 = - 3

m = - 3

also:

m = ( y2 - y3 ) / ( x2 - x3 )

In this case:

x2 = - 3 , y2 = 16 , x3 = k , y3 = - 2

m = - 3 = ( y2 - y3 ) / ( x2 - x3 ) =

[ 16 - ( - 2 ) ] / ( - 3 - k ) =

( 16 + 2) / ( - 3 - k ) = 18 / ( - 3 - k )

- 3 = 18 / ( - 3 - k )

Cross multiply

( - 3 ) * ( - 3 ) - 3 * ( - k ) = 18

9 + 3 k = 18

3 k = 18 - 9 = 9

k = 9 / 3 = 3

k = 3

Answer 2

(1,4), (-3,16), (k,-2).

m1 = (16-4)/(-3-1) = 12/-4 = -3.
The slope should be -3 at all points:
m2 = (-2-16)/k-(-3)) = (-2-16/(k+3) = -18/(k+3) = -3,
-18/(k+3) = -3,
-3k - 9 = -18,
K = 3.