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Question

Find the distance between the parallel lines $$3x-4y+7=0$$ and $$3x-4y+5=0$$.


Solution


$$A=3, B=-4, C_{1}=7, C_{2}=5$$
$$d=\dfrac{|C_{1}-C_{2}|}{\sqrt{a^2+b^2}}$$

$$d=\dfrac{|7-5|}{\sqrt{25}}=\dfrac{2}{5}$$

Mathematics

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