If x+2y-3=0,3x+4y-7=0,2x+3y-4=0,and 4x+5y-6=0 | vedclass

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(D) The given lines are: $x+2y-3=0$ $\dots(i)$ $3x+4y-7=0$ $\dots(ii)$ $2x+3y-4=0$ $\dots(iii)$ $4x+5y-6=0$ $\dots(iv)$ Solving equations $(i)$ and $(

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not all of them are concurrent

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(D) The given lines are: $x+2y-3=0$ $\dots(i)$ $3x+4y-7=0$ $\dots(ii)$ $2x+3y-4=0$ $\dots(iii)$ $4x+5y-6=0$ $\dots(iv)$ Solving equations $(i)$ and $(ii)$: From $(i)$,$x = 3-2y$. Substituting in $(ii)$: $3(3-2y) + 4y - 7 = 0$ $9 - 6y + 4y - 7 = 0$ $-2y + 2 = 0 \implies y = 1$. Then $x = 3 - 2(1) = 1$. The point of intersection is $P(1, 1)$. Now,check if $P(1, 1)$ satisfies $(iii)$: $2(1) + 3(1) - 4 = 2 + 3 - 4 = 1 
eq 0$. Since the point of intersection of the first two lines does not lie on the third line,the lines are not concurrent.

If $x+2y-3=0$,$3x+4y-7=0$,$2x+3y-4=0$,and $4x+5y-6=0$ are the equations of four lines,then

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