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(A) The equation of the line passing through $(4, -9)$ with slope $m$ is given by $y + 9 = m(x - 4)$.To find the $x$-intercept $A$,set $y = 0$: $9 = m

Vedclass · Accepted Answer

$25$

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(A) The equation of the line passing through $(4, -9)$ with slope $m$ is given by $y + 9 = m(x - 4)$.To find the $x$-intercept $A$,set $y = 0$: $9 = m(x - 4) \Rightarrow x = 4 + \frac{9}{m}$. Thus,$A = (4 + \frac{9}{m}, 0)$.To find the $y$-intercept $B$,set $x = 0$: $y + 9 = m(-4) \Rightarrow y = -9 - 4m$. Since $B$ is a point on the axis and we consider distance,we look at the magnitude. Given $m > 0$,the $y$-coordinate is negative,so $B = (0, -(9 + 4m))$.The sum of the distances from the origin is $S = OA + OB = |4 + \frac{9}{m}| + |-(9 + 4m)|$.Since $m > 0$,$S = 4 + \frac{9}{m} + 9 + 4m = 13 + 4m + \frac{9}{m}$.Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$: $\frac{4m + \frac{9}{m}}{2} \geq \sqrt{4m 	imes \frac{9}{m}} = \sqrt{36} = 6$.Therefore,$4m + \frac{9}{m} \geq 12$.Thus,$S \geq 13 + 12 = 25$.

Let a variable line of slope $m > 0$ passing through the point $(4, -9)$ intersect the coordinate axes at the points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is

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