If the lines x+y-2=0,3x-4y+1=0 and 5x+ky-7=0 | vedclass

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(D) Step $1$: Find the point of concurrency $(\alpha, \beta)$ by solving the first two equations:$x+y=2$  $(i)$$3x-4y=-1$ (ii)Multiply $(i)$ by $4$: $

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$x-2y=-1$

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(D) Step $1$: Find the point of concurrency $(\alpha, \beta)$ by solving the first two equations:$x+y=2$  $(i)$$3x-4y=-1$ (ii)Multiply $(i)$ by $4$: $4x+4y=8$ (iii)Adding (ii) and (iii): $7x=7 \implies x=1$.Substitute $x=1$ into $(i)$: $1+y=2 \implies y=1$.So,the point of concurrency is $(1, 1)$.Step $2$: Find $k$ using the third line $5x+ky-7=0$ passing through $(1, 1)$:$5(1)+k(1)-7=0 \implies 5+k-7=0 \implies k=2$.Step $3$: Find the equation of the line passing through $(1, 1)$ and perpendicular to $kx+y-k=0$ (i.e.,$2x+y-2=0$):The slope of $2x+y-2=0$ is $m_1 = -2$.The slope of the required line $m_2$ is $\frac{-1}{m_1} = \frac{-1}{-2} = \frac{1}{2}$.Step $4$: Use the point-slope form $y-y_1 = m_2(x-x_1)$:$y-1 = \frac{1}{2}(x-1) \implies 2y-2 = x-1 \implies x-2y = -1$.

If the lines $x+y-2=0$,$3x-4y+1=0$ and $5x+ky-7=0$ are concurrent at $(\alpha, \beta)$,then the equation of the line concurrent with the given lines and perpendicular to $kx+y-k=0$ is

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