Question

find the value of a for which points are collinear (2,3) (4,a)and (6,-3)​

Answer 1

Step-by-step explanation:

ANSWER

Points A(2,3),B(4,k) and C(6,−3) are collinear.

Area of triangle having vertices A, B and C=0

Area of a triangle =

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

Area of given ΔABC=0

⇒

2

1

[2(k−(−3))+4(−3−3)+6(3−k))]=0

⇒2k+6−24+18−6k=0

⇒−4k=0

or k=0

The value of k is zero.

Answer 2

Answer:

Hey buddy here's your answer ! Hope it helps you Formula for area of a triangle in co-ordinate Geometry

1/2 ( x1 ( y2 - y3) + x2 ( y3 - y1) + x3 ( y1 - y2)

Since they are collinear , it is equal to 0.

Here ,

x1 = 2 ; x2 = 4 and x3 = 6

y1 =3 ; y2 = a and y3 = -3

So,

1/2( 2(a-(-3)) + 4( -3-3) + 6(3-a)) = 0

1/2 ( 2a + 6 + (-24) + 18 - 6a) =0

1/2 ( -4a) =0

-4a = 0

a = 0