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How do you find the angle between the lines 3x+4y=12 and 4x-3y=12?

1 Answer

May 17, 2018

$90^{\circ}$ $90^@$

Explanation:

Rearrange so $3 x + 4 y = 12 \Rightarrow 4 y = - 3 x + 12 \Rightarrow y = - \frac{3}{4} x + 3$ $3x+4y=12 =>4y=-3x+12=>y=-3/4x+3$

$4 x - 3 y = 12 \Rightarrow - 3 y = - 4 x + 12 \Rightarrow y = \frac{4}{3} x - 4$ $4x-3y=12=>-3y=-4x+12=>y=4/3x-4$

the two gradients are $- \frac{3}{4} and \frac{4}{3}$ $-3/4 and 4/3$

$- \frac{3}{4} \times \frac{4}{3} = - \frac{12}{12} = - 1$ $-3/4xx4/3=-12/12=-1$

so they are perpendicular.

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