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QuestionFind values of aa for lines L1L_{1} and L2L_{2} with slopes m1=6a46m_{1}=\frac{-6 a-4}{-6} and m2=4a+85m_{2}=\frac{-4 a+8}{-5} such that: (i) L1L_{1} is parallel to L2L_{2}: a=a= (ii) L1L_{1} is perpendicular to L2L_{2}: a=a=

Studdy Solution
implify the square root to find the exact values of aa.
a=43±53a = \frac{4}{3} \pm \frac{\sqrt{5}}{3}So, when a=43+53a = \frac{4}{3} + \frac{\sqrt{5}}{3} or a=4353a = \frac{4}{3} - \frac{\sqrt{5}}{3}, the lines 1_{1} and _{} are perpendicular.
(i) 1_{1} parallel to :a=343_{}: a = -\frac{34}{3} (ii) 1_{1} perpendicular to :a=43+53,4353_{}: a = \frac{4}{3} + \frac{\sqrt{5}}{3}, \frac{4}{3} - \frac{\sqrt{5}}{3}

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