A line passing through the point A(2, 3) make | vedclass

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Question

(D) The equation of a line passing through $(x_1, y_1)$ with angle $	heta$ is given by $\frac{x - x_1}{\cos 	heta} = \frac{y - y_1}{\sin 	heta} = r

Vedclass · Accepted Answer

$3\sqrt{2}$

Vedclass · Answer

(D) The equation of a line passing through $(x_1, y_1)$ with angle $	heta$ is given by $\frac{x - x_1}{\cos 	heta} = \frac{y - y_1}{\sin 	heta} = r$.Here,$(x_1, y_1) = (2, 3)$ and $	heta = 45^{\circ}$.So,$\frac{x - 2}{\cos 45^{\circ}} = \frac{y - 3}{\sin 45^{\circ}} = r$.This gives $x = 2 + r \cos 45^{\circ} = 2 + \frac{r}{\sqrt{2}}$ and $y = 3 + r \sin 45^{\circ} = 3 + \frac{r}{\sqrt{2}}$.Since point $P$ lies on the line $x + y + 1 = 0$,we substitute these coordinates into the equation:$(2 + \frac{r}{\sqrt{2}}) + (3 + \frac{r}{\sqrt{2}}) + 1 = 0$$6 + \frac{2r}{\sqrt{2}} = 0$$6 + r\sqrt{2} = 0$$r\sqrt{2} = -6$$r = -\frac{6}{\sqrt{2}} = -3\sqrt{2}$.The distance $AP = |r| = |-3\sqrt{2}| = 3\sqrt{2}$.

$A$ line passing through the point $A(2, 3)$ makes an angle of $45^{\circ}$ with the $x$-axis and intersects the line $x + y + 1 = 0$ at point $P$. Find the distance $AP$.

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