
The lines $(p - q)x + (q - r)y + (r - p) = 0,(q - r)x + (r - p)y + (p - q) = 0,(r - p)x + (p - q)y + (q - r) = 0$ are
A. parallel
B. Perpendicular
C. Concurrent
D. None of these
Answer
163.8k+ views
Hint: To solve this type of question we need to have knowledge of the types of lines. The answer will start with knowledge of concurrent lines. Lines are called concurrent if two or more two lines pass through a common point. To check whether the lines are concurrent or not we define a determinant that contains coefficients of line variables and set them equal to zero. In this question, we will equate the determinant with zero.
Complete step by step Solution:
The question asks us to find the nature of line if the three lines $\
(p - q)x + (q - r)y + (r - p) = 0 \\
(q - r)x + (r - p)y + (p - q) = 0 \\
(r - p)x + (p - q)y + (q - r) = 0 \\
\ $
The steps that should be used to solve this question is that, as we know that for lines to be concurrent the determinant should be equal to zero. On solving the determinants of the matrix which we get from the coefficient of $x,y$ . The matrix formed using the three lines will be:
$\left( {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}}&{{c_1}} \\
{{a_2}}&{{b_2}}&{{c_2}} \\
{{a_3}}&{{b_3}}&{{c_3}}
\end{array}} \right) = 0$
The above is the formula for the coefficient of $x,y$ present in the line. On substituting the value, we get:
$\left( {\begin{array}{*{20}{c}}
{p - q}&{q - r}&{r - p} \\
{q - r}&{r - p}&{p - q} \\
{r - p}&{p - q}&{q - r}
\end{array}} \right)$
${R_1} \to {R_1} + {R_2} + {R_3}$ we get,
$\left( {\begin{array}{*{20}{c}}
0&0&0 \\
{q - r}&{r - p}&{p - q} \\
{r - p}&{p - q}&{q - r}
\end{array}} \right) = 0$
Hence these lines are concurrent.
Hence, the correct option is C.
Note: Sometimes it can happen that the line equation given to us would have only a single term either of X or Y. Well, in that case, we should put the coefficient of that term as zero.
For example: if a line $x + c = 0$ is given, it means that the coefficient of y will be written as zero in the matrix.
Complete step by step Solution:
The question asks us to find the nature of line if the three lines $\
(p - q)x + (q - r)y + (r - p) = 0 \\
(q - r)x + (r - p)y + (p - q) = 0 \\
(r - p)x + (p - q)y + (q - r) = 0 \\
\ $
The steps that should be used to solve this question is that, as we know that for lines to be concurrent the determinant should be equal to zero. On solving the determinants of the matrix which we get from the coefficient of $x,y$ . The matrix formed using the three lines will be:
$\left( {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}}&{{c_1}} \\
{{a_2}}&{{b_2}}&{{c_2}} \\
{{a_3}}&{{b_3}}&{{c_3}}
\end{array}} \right) = 0$
The above is the formula for the coefficient of $x,y$ present in the line. On substituting the value, we get:
$\left( {\begin{array}{*{20}{c}}
{p - q}&{q - r}&{r - p} \\
{q - r}&{r - p}&{p - q} \\
{r - p}&{p - q}&{q - r}
\end{array}} \right)$
${R_1} \to {R_1} + {R_2} + {R_3}$ we get,
$\left( {\begin{array}{*{20}{c}}
0&0&0 \\
{q - r}&{r - p}&{p - q} \\
{r - p}&{p - q}&{q - r}
\end{array}} \right) = 0$
Hence these lines are concurrent.
Hence, the correct option is C.
Note: Sometimes it can happen that the line equation given to us would have only a single term either of X or Y. Well, in that case, we should put the coefficient of that term as zero.
For example: if a line $x + c = 0$ is given, it means that the coefficient of y will be written as zero in the matrix.
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