Question

if (2,3), (0,2)&(-2,k) are collinear find k​

Answer 1

Step-by-step explanation:

Given -

• (2,3), (0,2), (-2,k) are collinear

To Find -

Value of k

Method 1 :-

As we know that :-

• x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)

» (2,3) = (x1,y1)

(0,2) = (x2,y2)

(-2,k) = (x3,y3)

Now,

» 2(2 - k) + 0(k - 3) - 2(3 - 2) = 0

» 4 - 2k + 0 - 6 + 4 = 0

» 2 = 2k

• » k = 1

Method 2 :-

As we know that :-

• Slope = y2 - y1/x2 - x1

Now,

Let a = (2,3), b = (0,2), c = (-2,k)

Then,

Slope of line AB =

(2,3) = (x1,y1)

(0,2) = (x2, y2)

» 0-2/2-3

» -2/-1

• » 2

And

Slope of line BC =

(0,2) = (x1,y1)

(-2, k) = (x2,y2)

» -2-0/k - 2

• » -2/k-2

Now,

» 2 = -2/k-2

» 2(k - 2) = -2

» 2k - 4 = -2

» 2k = 4 - 2

» 2k = 2

• » k = 1

Hence,

The value of k is 1

Answer 2

$\huge \mathfrak{Answer:-}$

$\implies$ k = 1

$\large\underline\mathrm{Given:-}$

• (2,3), (0,2)&(-2,k) are collinear.

$\large\underline\mathrm{To \: find}$

• value of k ?

$\large\underline\mathrm{Using \: formula:-}$

• x1(y2 - y3) + x2(y3 - y1) + x3(y1 -y2)

$\implies$ (x1, y1) = (2, 3)

$\implies$ (x2, y2) = (0, 2)

$\implies$ (x3, y3) = (-2, k)

$\large\underline\mathrm{Solution}$

$\implies$ x1(y2 - y3) + x2(y3 - y1) + x3(y1 -y2)

$\implies$ 2(2 - k) + 0(k - 3) - 2(3 - 2) = 0

$\implies$ 4 - 2k + 0 - 6 + 4 = 0

$\implies$ 2 = 2k

$\implies$ k = 1

$\large\underline\mathrm{hence,}$

$\large\underline\mathrm{the \: value \: of \: k \: is \: 1.}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$