Question

The points (3,-2) (-2,8) and (0,4) are three points in a plane. Show that these points are collinear

Answer 1

Answer:

soln

let a(3,-2),b(-2,8)and c(0,4)be the given points

let b(-2,8) divides the line segment joining the points a(3,-2)and c(0,4) in the ratio r:1

by section formula

B(mx2+nx1/m+n,my2+ny1/m+n)=b(-2,8)

(r×0+1×3/r+1,r×4+1×-2/r+1)= (-2,8)

(0+3/r+1,4r+(-2)/r+1)= (-2,8)

by equating corresponding coordinates

3/r+1=-2;4r-2/r+1=8

3=-2(r+1);r-2/r+1=2

3=-2r-2;r-2=2(r+1)

3+2r+2=0;r-2=2r+2

5+2r=0;r-2-2r-2=0

r=-5/2;1r-4=0

;r=4/1

therefore r=-5/2;r=4

they are not collinear

Answer 2

Answer:-

✈︎ By using area of the triangle formula

∆=

$= \frac{1}{2} |3(8 - 4) + ( - 2)(4 - ( - 2)) + 0(( - 2) - 8|$

$= > \frac{1}{2} |12 - 12| = 0$

☯︎ The area of the triangle is 0 . Hence the three points are collinear i.e., they lie on the same line.☯︎

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