Question

if 1+2+3+4...+k , then find​

Answer 1

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.