Coordinate Geometry Question 59
Question: If the points $ (x+1,2),\ (1,x+2),\ ( \frac{1}{x+1},\frac{2}{x+1} ) $ are collinear, then x is
[RPET 2002]
Options:
A) 4
B) 0
C) -4
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ A\equiv (x+1,2),B\equiv (1,x+2),C\equiv ( \frac{1}{x+1},\frac{2}{x+1} ) $ then A, B, C are collinear if area of $ \Delta ABC=0 $
$ \Rightarrow \begin{vmatrix} x+1 & 2 & 1 \\ 1 & x+2 & 1 \\ \frac{1}{x+1} & \frac{2}{x+1} & 1 \\ \end{vmatrix} =0 $
$ \Rightarrow \begin{vmatrix} x & -x & 0 \\ 1 & x+2 & 1 \\ \frac{1}{x+1} & \frac{2}{x+1} & 1 \\ \end{vmatrix} =0 $
$ (R_1\to R_1-R_2) $
$ \Rightarrow \begin{vmatrix} x & 0 & 0 \\ 1 & x+3 & 1 \\ \frac{1}{x+1} & \frac{3}{x+1} & 1 \\ \end{vmatrix} =0 $
$ (C_2\to C_2+C_1) $
$ \Rightarrow x( x+3-\frac{3}{x+1} )=0\Rightarrow x(x^{2}+3+4x-3)=0 $
$ \Rightarrow x^{2}(x+4)=0\Rightarrow x=0,-4 $ .
BETA
