Question

Find the value of r for which the points (x, 1), (2, 1) and (4, 5) are collinear.

Answer 1

Answer:

these points can not be collinear

Step-by-step explanation:

(x , 1) , (2 , 1)  and ( 4 , 5) are collinear

if slope of (x , 1) , (2 , 1)  &  (2 , 1) ( 4 , 5) are equal

iff ( 1 - 1) /(2 - x)  = ( 5 - 1)/(4 - 2)

iff 0/(2 -x) = 4/2

=> 0 = 2

but 0 ≠ 2

=> for any values of x these points can not be collinear

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The value of r for which the points (x, 1), (2, 1) and (4, 5) are collinear.

CONCEPT TO BE IMPLEMENTED

If three points are collinear then the area of the triangle formed by the points are zero

CALCULATION

Three given points are (x, 1), (2, 1) and (4, 5)

Since the points are collinear

Hence the area of the triangle formed by

the points = 0

$\therefore \: \displaystyle \: \sf{ \frac{1}{2} \times \begin{vmatrix} x & 1 & 1\\ 2 & 1 & 1 \\ 4 & 5 & 1 \end{vmatrix} = 0}$

$\implies \: \displaystyle \: \begin{vmatrix} x & 1 & 1\\ 2 & 1 & 1 \\ 4 & 5 & 1 \end{vmatrix} = 0$

$\implies \: \displaystyle \: \begin{vmatrix} x - 2 & 0 & 0\\ 2 & 1 & 1 \\ 4 & 5 & 1 \end{vmatrix} = 0 \: \: \: \: \: \: \sf{( \: Using \: \: \: R'_1=R_1 - R_2 \: )\: }$

On Expansion

$(x - 2) \times (1 - 5) - 0 + 0 = 0$

$\implies \: - 4 (x - 2)= 0$

$\implies \: x - 2 = 0$

$\implies \:x = 2$

RESULT

$\boxed{\sf{ \: \: x = 2\: \: \: }}$

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