Question

9. Find the value of K, if the points A(2,3) B(4,K) & C(6.3) are collinear.
​

Answer 1

Answer:

k = 3

Step-by-step explanation:

Three points are collinear when they are in the same straight line. It means that the area of the triangle formed by these three points is zero.

We know that the area of a triangle formed by three points X(x1, y1); Y(x2, y2); Z(x3, y3) is given by

$Area = \dfrac{1}{2} \{x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\}$

Now, given points

A(2, 3)

B(4, k)

C(6, 3)

According to the question,

$ar(\Delta ABC) = \dfrac{1}{2} \{2(k - 3) + 4(3 - 3) + 6(3 - k)\}\\\\\\\implies \{2k - 6 + 4(0) + 18 - 6k\} = 0 \times 2\\\\\\\implies -4k + 12 + 0 = 0\\\\\\\implies -4k = -12\\\\\\\implies 4k = 12\\\\\\\implies k = \dfrac{12}{4}\\\\\\\implies \boxed{\boxed{\underline{\boxed{\bold{k = 3}}}}}$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\:\:$

$\green{\underline \bold{Given :}}$

$\:\:$

• The points A(2,3) B(4,K) & C(6,3) are collinear.

$\:\:$

$\red{\underline \bold{To \: Find:}}$

$\:\:$

• The value of K

$\:\:$

Firstly let us understand what is collinear points.

$\:\:$

Three or more points are said to be collinear if they lie on a single straight line.

$\:\:$

Also we know that,

$\:\:$

Three points are collinear if the value of area of triangle formed by the three points is zero.

$\:\:$

$\underline{\bold{\texttt{Area of triangle with 3 points :}}}$

$\:\:$

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

$\:\:$

$\underline{\bold{\texttt{Putting the values ,}}}$

$\:\:$

$\sf \longmapsto \dfrac{1}{2} [2(k - 3) + 4(3 - 3) + 6(3 - k)] = 0$

$\:\:$

$\sf \longmapsto [2k - 6 + 4(0) + 18 - 6k] = 0 \times 2$

$\:\:$

$\sf \longmapsto -4k + 12 + 0 = 0$

$\:\:$

$\sf \longmapsto -4k = -12$

$\:\:$

$\sf \longmapsto 4k = 12$

$\:\:$

$\sf \longmapsto k = \dfrac{12}{4}$

$\:\:$

$\bf \dashrightarrow k = 3$

$\:\:$

Hence,

$\:\:$

The value of 'k' is 3

$\rule{200}5$