Question

If A(1,3) B(3,0) and C(0,K) are collinear, find K​

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:

$k = \dfrac{9}{2}$

Step-by-step explanation:

If three points are collinear , the slope of first two points is equal to slope of next two points.

Slope of two points = $\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Slope of first two points = $\dfrac{0-3}{3-1}$

Slope of first two points = $\dfrac{-3}{2}$

Slope of next two points = $\dfrac{k-0}{0-3}$

Slope of next two points = $\dfrac{k}{-3}$

Since, the two slopes are equal.

$\dfrac{-3}{2}\:=\:\dfrac{k}{-3}\\-3 \times -3\:=\:2 \times k\\k\:=\:\dfrac{9}{2}$

Hope it helps u

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