Equation of line in Straight lines

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If the lines $2 x + y - k = 0$ , $x - y + 1 = 0$ and $- 2 x - 3 y + 8 = 0$ are concurrent then k=

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Solution

The condition that the lines $a_{1} x + b_{1} y + c_{1} = 0$ , $a_{2} x + b_{2} y + c_{2} = 0$ and $a_{3} x + b_{3} y + c_{3} = 0$ are to be concurrent is $|\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}| = 0$
Hence,
$|\begin{matrix} 2 & 1 & - k \\ 1 & - 1 & 1 \\ - 2 & - 3 & 8 \end{matrix}| = 0$
Expand the determinant
$\begin{matrix} 2 (- 8 + 3) - 1 (8 + 2) - k (- 3 - 2) = 0 \\ - 10 - 10 + 5 k = 0 \\ k = \frac{20}{5} \\ = 4 \end{matrix}$
Therefore, $k = 4$

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Equation of line in Straight lines

Remember concepts with our Masterclasses.

If the lines 2x+y−k=0, x−y+1=0 and −2x−3y+8=0 are concurrent then k=

The condition that the lines a1x+b1y+c1=0 , a2x+b2y+c2=0 and a3x+b3y+c3=0 are to be concurrent is a1b1c1a2b2c2a3b3c3=0Hence, 21−k1−11−2−38=0 Expand the determinant 2−8+3−18+2−k−3−2=0−10−10+5k=0k=205=4Therefore, k=4

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If the lines $2 x + y - k = 0$ , $x - y + 1 = 0$ and $- 2 x - 3 y + 8 = 0$ are concurrent then k=