Find the distance of the point (2, 3) from th | vedclass

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(B) The slope of the line $x - y + 1 = 0$ is $m = 1$. Thus,it makes an angle of $	heta = 45^{\circ}$ with the $x$-axis.The equation of a line passing

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$4\sqrt{2}$

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(B) The slope of the line $x - y + 1 = 0$ is $m = 1$. Thus,it makes an angle of $	heta = 45^{\circ}$ with the $x$-axis.The equation of a line passing through $(2, 3)$ with an angle of $45^{\circ}$ is given by $\frac{x - 2}{\cos 45^{\circ}} = \frac{y - 3}{\sin 45^{\circ}} = r$.Any point on this line can be represented as $(2 + r \cos 45^{\circ}, 3 + r \sin 45^{\circ}) = (2 + \frac{r}{\sqrt{2}}, 3 + \frac{r}{\sqrt{2}})$.Since this point lies on the line $2x - 3y + 9 = 0$,we substitute the coordinates:$2(2 + \frac{r}{\sqrt{2}}) - 3(3 + \frac{r}{\sqrt{2}}) + 9 = 0$.$4 + \frac{2r}{\sqrt{2}} - 9 - \frac{3r}{\sqrt{2}} + 9 = 0$.$4 - \frac{r}{\sqrt{2}} = 0 \implies \frac{r}{\sqrt{2}} = 4 \implies r = 4\sqrt{2}$.Thus,the required distance is $4\sqrt{2}$.

Find the distance of the point $(2, 3)$ from the line $2x - 3y + 9 = 0$ measured along the direction of the line $x - y + 1 = 0$.

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