Question

The lines (a + b-2c)x + (b +c-2a)y +(c + a-2b) = 0, (b + c-2a)x +(c +a-2b)y + (a + b-2c) = 0,(c + a- 2b)x + (a + b-2c)y + (b + c- 2a) = 0where a, b, c are real numbers

A)Form an equilateral triangle
B)Are concurrence
C)Form a right angled triangle
D)None of these​

Answer 1

your answer is

1. none of these

i hope it will help you

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Answer 2

Lines

B) Are concurrent

L1: $(a + b-2c)x + (b +c-2a)y +(c + a-2b) = 0$

L2: $(b + c-2a)x +(c +a-2b)y + (a + b-2c) = 0$

L3: $(c + a- 2b)x + (a + b-2c)y + (b + c- 2a) = 0$

There is  a condition for checking concurrency of lines, the determinant should be equal to zero.

$\det A = \begin{vmatrix}a+b-2c & b+c-2a & c+a-2b \\b+c-2a & c+a-2b & a+b-2c \\c+a-2b & a+b-2c & b+c-2a\end{vmatrix}=0$

$R1->R1+R2+R3$

Then,

$\det A = \begin{vmatrix}0 & 0 & 0 \\b+c-2a & c+a-2b & a+b-2c \\c+a-2b & a+b-2c & b+c-2a\end{vmatrix}=0$

$\det A = 0,$ as all element of row R1 is 0.

Hence, these lines are concurrent lines.

• For equilateral triangle , the distance between all lines should be similar.
• For right angled triangle, it should follow hypotenuse rule by finding the distance between them.