If M is the foot of the perpendicular drawn f | vedclass

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(A) The given line is $x-2y+3=0$. To find the $X$-intercept $A$,set $y=0$: $x+3=0 \implies x=-3$. So,$A = (-3, 0)$. To find the $Y$-intercept $B$,set

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$\frac{6 \sqrt{5}}{5}$

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(A) The given line is $x-2y+3=0$. To find the $X$-intercept $A$,set $y=0$: $x+3=0 \implies x=-3$. So,$A = (-3, 0)$. To find the $Y$-intercept $B$,set $x=0$: $-2y+3=0 \implies y=3/2$. So,$B = (0, 3/2)$. The foot of the perpendicular $M(x_1, y_1)$ from the origin $(0, 0)$ to the line $ax+by+c=0$ is given by $\frac{x_1-0}{a} = \frac{y_1-0}{b} = -\frac{ax_0+by_0+c}{a^2+b^2}$. Here,$a=1, b=-2, c=3$. $\frac{x_1}{1} = \frac{y_1}{-2} = -\frac{1(0)-2(0)+3}{1^2+(-2)^2} = -\frac{3}{5}$. So,$x_1 = -3/5$ and $y_1 = 6/5$. Thus,$M = (-3/5, 6/5)$. Now,calculate the distance $AM$ where $A = (-3, 0)$ and $M = (-3/5, 6/5)$: $AM = \sqrt{(-3/5 - (-3))^2 + (6/5 - 0)^2} = \sqrt{(-3/5 + 15/5)^2 + (6/5)^2} = \sqrt{(12/5)^2 + (6/5)^2} = \sqrt{144/25 + 36/25} = \sqrt{180/25} = \frac{\sqrt{36 	imes 5}}{5} = \frac{6 \sqrt{5}}{5}$.

If $M$ is the foot of the perpendicular drawn from the origin to the line $x-2y+3=0$ which meets the $X$ and $Y$-axes at $A$ and $B$ respectively,then $AM=$

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