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Miscellaneous Exercise · Q18

Q.If the lines y=3x+1y = 3x + 1 and 2y=x+32y = x + 3 are equally inclined to the line y=mx+4y = mx + 4, find the value of mm.

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Equating the tangents of the angles each line makes with y=mx+4y=mx+4 gives 7m2−2m−7=07m^2-2m-7=0, so m=1±527m=\dfrac{1\pm5\sqrt{2}}{7}.

Concept: "equally inclined"

Two lines are equally inclined to a third line when they make equal-magnitude angles with it. The angle θ\theta between lines of slopes m1m_1 and m2m_2 satisfies

tan⁡θ=∣m1−m21+m1m2∣.\tan\theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|.

So the angle between y=mx+4y=mx+4 and the first line equals the angle between y=mx+4y=mx+4 and the second line.

Step-by-step solution

Slopes. y=3x+1y=3x+1 has slope 33; 2y=x+3⇒y=12x+322y=x+3\Rightarrow y=\tfrac12 x+\tfrac32 has slope 12\tfrac12; y=mx+4y=mx+4 has slope mm.

Equal-angle condition.

∣3−m1+3m∣=∣12−m1+12m∣=∣1−2m2+m∣.\left|\frac{3-m}{1+3m}\right|=\left|\frac{\tfrac12-m}{1+\tfrac12 m}\right|=\left|\frac{1-2m}{2+m}\right|.

Case 1 (same sign):

(3−m)(2+m)=(1−2m)(1+3m)  ⇒  6+m−m2=1+m−6m2  ⇒  5m2=−5,(3-m)(2+m)=(1-2m)(1+3m)\;\Rightarrow\;6+m-m^2=1+m-6m^2\;\Rightarrow\;5m^2=-5,

which has no real solution.

Case 2 (opposite sign): …

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