If the lines 3x - 4y + 4 = 0 and 6x - 8y - 7 | vedclass

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(B) The given lines are $L_1: 3x - 4y + 4 = 0$ and $L_2: 6x - 8y - 7 = 0$.We can rewrite $L_2$ as $3x - 4y - \frac{7}{2} = 0$.Since the lines are para

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$\frac{3}{4}$ units

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(B) The given lines are $L_1: 3x - 4y + 4 = 0$ and $L_2: 6x - 8y - 7 = 0$.We can rewrite $L_2$ as $3x - 4y - \frac{7}{2} = 0$.Since the lines are parallel and tangent to the same circle,the distance between them is equal to the diameter $D$ of the circle.$D = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|4 - (-7/2)|}{\sqrt{3^2 + (-4)^2}} = \frac{|4 + 3.5|}{5} = \frac{7.5}{5} = 1.5$.Since the diameter $D = 1.5$,the radius $r = \frac{D}{2} = \frac{1.5}{2} = 0.75 = \frac{3}{4}$ units.

If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangent to a circle,then the radius of the circle is

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