Question

The value of k if the points A (2,3) B(4,k) c(6,-3)
are collinear is
options are
a)1 b)-1 c)2 d)0
$plz \: answer \: fast \: i \: will \: mark \: as \: bainlist \: if \: ans \: is \: correct$
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Answer 1

Answer:

So Easy we can solve it by graphing it ! and there are infinitely many equations to equate this....

Step-by-step explanation:

k=0

Answer 2

$\large\underline{\sf{Given- }}$

The following points are collinear

• A (2, 3)

• B (4, k)

• C (6, - 3)

$\large\underline{\sf{To\:Find - }}$

• The value of k

$\large\underline{\sf{Solution-}}$

Given that the points are collinear.

• A (2, 3)

• B (4, k)

• C (6, - 3)

We know that,

Three points are collinear iff

$\bf [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] = 0$

Here,

• x₁ = 2

• x₂ = 4

• x₃ = 6

• y₁ = 3

• y₂ = k

• y₃ = - 3

On substituting the values, we get

$\rm :\longmapsto\:2(k + 3) + 4( - 3 - 3) + 6(3 - k) = 0$

$\rm :\longmapsto\:2k + 6 - 24 + 18 - 6k = 0$

$\rm :\longmapsto\: - 4k + 24 - 24 = 0$

$\rm :\longmapsto\: - 4k = 0$

$\bf\implies \:k = 0$

Additional Information :-

Distance Formula :-

$\rm :\longmapsto\:Distance = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

Midpoint Formula :-

$\rm :\longmapsto\: \:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg)$

Section Formula :-

$\rm :\longmapsto\: \:( x, y) = \bigg(\dfrac{nx_1+mx_2}{m + n} , \dfrac{ny_1+my_2}{m + n} \bigg)$

Area of triangle :-

$\rm :\longmapsto\: \ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$