The slope of a line L passing through the poi | vedclass

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(A) The line $L$ passes through $(-2, -3)$ and has an undefined slope,so its equation is $x = -2$ (a vertical line).The angle between $x = -2$ and the

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$\pi - 	an^{-1}\left(\frac{1}{2}ight)$

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(A) The line $L$ passes through $(-2, -3)$ and has an undefined slope,so its equation is $x = -2$ (a vertical line).The angle between $x = -2$ and the line $ax - 2y + 3 = 0$ is $45^{\circ}$.The slope of the line $ax - 2y + 3 = 0$ is $m_1 = \frac{a}{2}$.The angle $	heta$ between a vertical line and a line with slope $m_1$ is given by $|	an(90^{\circ} - 	heta)| = |\frac{1}{m_1}|$.Since the angle is $45^{\circ}$,we have $|\frac{1}{a/2}| = 	an(45^{\circ}) = 1$.Thus,$\frac{2}{a} = 1$,which gives $a = 2$.Now,substitute $a = 2$ into the equation $x + ay - 4 = 0$ to get $x + 2y - 4 = 0$.The slope of this line is $m = -\frac{1}{2}$.The angle $	heta$ made with the positive $X$-axis is given by $	an 	heta = -\frac{1}{2}$.Since the slope is negative,the angle is in the second quadrant,so $	heta = \pi - 	an^{-1}\left(\frac{1}{2}ight)$.

The slope of a line $L$ passing through the point $(-2, -3)$ is not defined. If the angle between the lines $L$ and $ax - 2y + 3 = 0$ $(a > 0)$ is $45^{\circ}$,then the angle made by the line $x + ay - 4 = 0$ with the positive $X$-axis in the anticlockwise direction is

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