The distance of the point (1, 2) from the lin | vedclass

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(A) Let the point be $P(1, 2)$. The line passing through $P$ and parallel to $3x - y = 2$ has the equation $3x - y = k$. Since it passes through $(1,

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$\frac{3 \sqrt{10}}{4}$ units

Vedclass · Answer

(A) Let the point be $P(1, 2)$. The line passing through $P$ and parallel to $3x - y = 2$ has the equation $3x - y = k$. Since it passes through $(1, 2)$,we have $3(1) - 2 = k$,so $k = 1$. The equation of the line is $3x - y = 1$,or $y = 3x - 1$. To find the intersection point $Q$ of this line with $x + y = 0$,substitute $y = 3x - 1$ into $x + y = 0$: $x + (3x - 1) = 0 \implies 4x = 1 \implies x = \frac{1}{4}$. Then $y = -x = -\frac{1}{4}$. So $Q = (\frac{1}{4}, -\frac{1}{4})$. The distance $PQ$ is $\sqrt{(1 - \frac{1}{4})^2 + (2 - (-\frac{1}{4}))^2} = \sqrt{(\frac{3}{4})^2 + (\frac{9}{4})^2} = \sqrt{\frac{9}{16} + \frac{81}{16}} = \sqrt{\frac{90}{16}} = \frac{3 \sqrt{10}}{4}$ units.

The distance of the point $(1, 2)$ from the line $x + y = 0$ measured parallel to the line $3x - y = 2$ is

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