Answer:
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2≡−−7=0L2≡qx−py−7=0(3,1) is their point of intersection then(3,1) is their point of intersection then⟹3−1+4=0⟹3p−1+4=0=−1p=−1⟹(3)−(−1)(1)−7=0⟹q(3)−(−1)(1)−7=0=2
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