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If the straight lines $x-2 y=0$ and $k x+y=1$ intersect at the
point $\left(1, \frac{1}{2}\right)$, then what is the value of $k ?$
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point $\left(1, \frac{1}{2}\right)$, then what is the value of $k ?$
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Verified Answer
The correct answer is:
$1 / 2$
Since. the straight lines $x-2 y=0$ and $k x+y=1$ intersect at the point $\left(1, \frac{1}{2}\right)$.
$\therefore \quad$ The point $\left(1, \frac{1}{2}\right)$ satisfies the equation $k x+y=1$
$\therefore \quad$ Put $x=1$, and $y=\frac{1}{2}$ in $\mathrm{eq}^{\mathrm{n}} k x+y=1$
we get $k .1+\frac{1}{2}=1 \Rightarrow k=\frac{1}{2}$
$\therefore \quad$ The point $\left(1, \frac{1}{2}\right)$ satisfies the equation $k x+y=1$
$\therefore \quad$ Put $x=1$, and $y=\frac{1}{2}$ in $\mathrm{eq}^{\mathrm{n}} k x+y=1$
we get $k .1+\frac{1}{2}=1 \Rightarrow k=\frac{1}{2}$
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