Question

prove that the points (0,3),(6,0) and (4,1) are collinear​

Answer 1

$\huge\underline\mathcal\red{Question}$

prove that the points (0,3),(6,0) and (4,1) are collinear.

$\huge\underline\mathcal\red{Solution}$

Here for proving the lines collinear we have to proof that the area of triangle is 0.

A = (0,3 )=> (x1 , y1)

B = (6,0) => (x2 , y2)

C = (4,1) => (x3 , y3)

As we know that ,

Area of Triangle =$\frac{1}{2}(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))$

$= \frac{1}{2} (0(0 - 1) + 6(1 - 3) + 4(3 - 0))$

$= \frac{1}{2} (0 ( - 1) + 6( - 2) + 4(3)$

$= \frac{1}{2} (0 - 12 + 12)$

$= \frac{1}{2} (0)$

$= 0$

Area of triangle = 0

Hence , the given points are collinear .

Answer 2

Solution:-

To Proof:-

The points (0,3),(6,0) and (4,1) are collinear.

Proof:-

Let ( x1, y1) be ( 0,3)

( x2, y2) be ( 6,0)

& ( x3, y3) be ( 4, 1).

Now,

If the Points are Collinear , then the Area of Triangle will be equal to 0.

=) Area of Triangle = 1/2 [x1 ( y2 - y3) + x2( y3 - y1) + x3 ( y1 - y2) ]

=) 0 = [ 0(......) + 6( 1-3) + 4(3-0) ]

=) 0 = [ 0 + 6(-2) + 12]

=) 0 = 0 - 12 + 12

=) 0 = 0

Hence,

The Points (0,3),(6,0) and (4,1) are collinear.