If the lines x+3y-5=0,5x+2y-12=0,and 3x-ky-1= | vedclass

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(C) Three lines do not form a triangle if they are concurrent or if any two of them are parallel. First,find the intersection point of the first two l

Vedclass · Accepted Answer

$\frac{-6}{5}$

Vedclass · Answer

(C) Three lines do not form a triangle if they are concurrent or if any two of them are parallel. First,find the intersection point of the first two lines: $x+3y=5$ $(i)$ $5x+2y=12$ (ii) Multiplying $(i)$ by $5$: $5x+15y=25$ (iii) Subtracting (ii) from (iii): $13y=13 \Rightarrow y=1$. Substituting $y=1$ in $(i)$: $x+3(1)=5 \Rightarrow x=2$. The intersection point is $(2, 1)$. For the lines to be concurrent,this point must satisfy the third line $3x-ky-1=0$: $3(2)-k(1)-1=0$ $\Rightarrow 6-k-1=0$ $\Rightarrow k=5$. However,checking the options,we must also consider the case where the lines are parallel. If $3x-ky-1=0$ is parallel to $x+3y-5=0$,then $\frac{3}{1} = \frac{-k}{3} \Rightarrow k=-9$. If $3x-ky-1=0$ is parallel to $5x+2y-12=0$,then $\frac{3}{5} = \frac{-k}{2} \Rightarrow k=-\frac{6}{5}$. Comparing with the given options,the correct value is $k=-\frac{6}{5}$.

If the lines $x+3y-5=0$,$5x+2y-12=0$,and $3x-ky-1=0$ do not form a triangle,then a value of $k$ is

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