The length of the segment of the straight lin | vedclass

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(B) Given points are $(3,3)$ and $(7,6)$.The slope of the line $m = \frac{6-3}{7-3} = \frac{3}{4}$.The equation of the line is $(y-3) = \frac{3}{4}(x-

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$\frac{5}{4}$

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(B) Given points are $(3,3)$ and $(7,6)$.The slope of the line $m = \frac{6-3}{7-3} = \frac{3}{4}$.The equation of the line is $(y-3) = \frac{3}{4}(x-3)$,which simplifies to $4y - 12 = 3x - 9$,or $3x - 4y = -3$.To find the $y$-intercept,set $x=0$: $3(0) - 4y = -3 \Rightarrow y = \frac{3}{4}$. The point is $(0, \frac{3}{4})$.To find the $x$-intercept,set $y=0$: $3x - 4(0) = -3 \Rightarrow x = -1$. The point is $(-1, 0)$.The length of the segment between $(0, \frac{3}{4})$ and $(-1, 0)$ is given by the distance formula:$d = \sqrt{(0 - (-1))^2 + (\frac{3}{4} - 0)^2} = \sqrt{1^2 + (\frac{3}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

The length of the segment of the straight line passing through $(3,3)$ and $(7,6)$ cut off by the coordinate axes is

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