Question

Find the value of K, if the points A(7, -2), B(5, 1) and C(3, K) are collinear.​

Answer 1

Question : -

Find the value of K, if the points A(7,-2),B(5,1) & C(3,K) are collinear ?

ANSWER

Given : -

The points A(7,-2),B(5,1) & C(3,K) are collinear

Required to find : -

• Value of K ?

Formula used : -

Area of the triangle (co-ordinate geometry)

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

Concept used : -

Q) What do we mean by collinearity ?

A) 2 or more points said to be collinear if they all line of the same line !

Similarly,

Q) How can we say if 2 points or more are collinear or not ?

A) 2 points or more points can be proved to be as collinear if the length of first 2 parts is equal to the length of the whole part

Eg : AC = 6 in which AB = 4 cm and BC = 2 cm so, we can say A,B,C are collinear.

Since, AB + BC = aC

Q) Shortcut to find the collinearity ?

A) Using the area of the triangle formula of co-ordinate geometry

• The area of the collinear points will be zero (0)
• since, no triangle is formed .

Solution : -

Given that;

The points of the line segment AC are

A(7,-2) B(5,1) C(3,K)

We need to find the value of K ?

So,

Here,

$\large{ \left \lgroup \begin{array} \: \sf x_1 = 7 \: , \: x_2 = 5 \: , \: x_3 = 3 \\ \\ \sf y_1 = -2 \: , \: y_2 = 1 \: , \: y_3 = k \end{array} \right\rgroup}$

Using the formula;

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

Area = (1)/(2)| 7(1-k) + 5(k-(-2)) + 3(-2-1)|

Area = (1)/(2)|7-7k + 5(k+2) + 3(-3)|

Area = (1)/(2)|7-7k+5k+10-9|

Area = (1)/(2)|-2k+8|

Area = (2k+8)/(2)

since,

Area of collinear points is 0

(2k+8)/(2) = 0

2k+8 = 0

2k = - 8

k = -8/2

k = -4

k = ±4 (since, we took modulus in the formula)

Therefore,

Value of k = +4 or -4