If the given lines y = m_1x + c_1,y = m_2x + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is

Vedclass · Accepted Answer

$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$

Vedclass · Answer

(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero: $\begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = 0$ Rewriting the given lines as $m_1x - y + c_1 = 0$,$m_2x - y + c_2 = 0$,and $m_3x - y + c_3 = 0$,the condition for concurrency is: $\begin{vmatrix} m_1 & -1 & c_1 \ m_2 & -1 & c_2 \ m_3 & -1 & c_3 \end{vmatrix} = 0$ Expanding along the second column: $-(-1) \begin{vmatrix} m_2 & c_2 \ m_3 & c_3 \end{vmatrix} + (-1) \begin{vmatrix} m_1 & c_1 \ m_3 & c_3 \end{vmatrix} - (-1) \begin{vmatrix} m_1 & c_1 \ m_2 & c_2 \end{vmatrix} = 0$ $(m_2c_3 - m_3c_2) - (m_1c_3 - m_3c_1) + (m_1c_2 - m_2c_1) = 0$ Rearranging the terms: $m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$.

If the given lines $y = m_1x + c_1$,$y = m_2x + c_2$,and $y = m_3x + c_3$ are concurrent,then:

Similar Questions

If the lines $4x + 3y - 1 = 0$,$x - y + 5 = 0$,and $kx + 5y - 3 = 0$ are concurrent,then $k=$

The lines $15x - 18y + 1 = 0$,$12x + 10y - 3 = 0$ and $6x + 66y - 11 = 0$ are

The family of straight lines $4ax + 3by + c = 0$ such that $a + b + c = 0$ (where $a, b, c$ are real constants) are concurrent at the point...

If the lines $ax + by + c = 0$,$bx + cy + a = 0$,and $cx + ay + b = 0$ are concurrent,then:

Find the equation of the line passing through the intersection of the lines $4x - 3y - 1 = 0$ and $5x - 2y - 3 = 0$ and parallel to the line $2x - 3y + 2 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module