The distance of the line 2x - 3y = 4 from the | vedclass

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(A) The slope of the line $x + y = 1$ is $m = -1$. This line makes an angle of $	heta = 135^\circ$ with the positive $x$-axis. The equation of the li

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$\sqrt{2}$

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(A) The slope of the line $x + y = 1$ is $m = -1$. This line makes an angle of $	heta = 135^\circ$ with the positive $x$-axis. The equation of the line passing through $(1, 1)$ and making an angle of $135^\circ$ is given by $\frac{x - 1}{\cos 135^\circ} = \frac{y - 1}{\sin 135^\circ} = r$. Substituting the values,we get $\frac{x - 1}{-1/\sqrt{2}} = \frac{y - 1}{1/\sqrt{2}} = r$. Thus,any point on this line is of the form $(1 - \frac{r}{\sqrt{2}}, 1 + \frac{r}{\sqrt{2}})$. Since this point lies on the line $2x - 3y = 4$,we substitute these coordinates into the equation: $2(1 - \frac{r}{\sqrt{2}}) - 3(1 + \frac{r}{\sqrt{2}}) = 4$. $2 - \frac{2r}{\sqrt{2}} - 3 - \frac{3r}{\sqrt{2}} = 4$. $-1 - \frac{5r}{\sqrt{2}} = 4$. $-\frac{5r}{\sqrt{2}} = 5$. $r = -\sqrt{2}$. Since distance is the magnitude,the distance is $|r| = \sqrt{2}$.

The distance of the line $2x - 3y = 4$ from the point $(1, 1)$ measured parallel to the line $x + y = 1$ is

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