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Exercise 6.4 · Q10

Q.A △OPQ\triangle OPQ is formed by the pair of straight lines x2−4xy+y2=0x^2 - 4xy + y^2 = 0 and the line PQPQ. The equation of PQPQ is x+y−2=0x + y - 2 = 0. Find the equation of the median of the triangle △OPQ\triangle OPQ drawn from the origin OO.

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Split the pair x2−4xy+y2=0x^2-4xy+y^2=0 into its two lines OA,OBOA,OB; intersect each with PQ:x+y−2=0PQ:x+y-2=0 to get P,QP,Q; the median from OO passes through their midpoint.

The median from OO in △OPQ\triangle OPQ is the line joining OO to the midpoint of PQPQ, so we first need the coordinates of PP and QQ, which are where the two component lines of the pair meet PQPQ.

Step 1. Split the pair into two lines through the origin. For x2−4xy+y2=0x^2-4xy+y^2=0, treat as a quadratic in xx: x2−4yx+y2=0x^2-4yx+y^2=0, so

x=4y±16y2−4y22=4y±12 y2=(2±3)y.x=\frac{4y\pm\sqrt{16y^2-4y^2}}{2}=\frac{4y\pm\sqrt{12}\,y}{2}=(2\pm\sqrt3)y.

So the two lines are x−(2+3)y=0x-(2+\sqrt3)y=0 and x−(2−3)y=0x-(2-\sqrt3)y=0.

Step 2. Intersect the first line with PQ:x+y−2=0PQ:x+y-2=0 (i.e. x=2−yx=2-y). Setting (2+3)y=2−y(2+\sqrt3)y=2-y:

(2+3)y+y=2 ⟹ (3+3)y=2 ⟹ y=23+3=2(3−3)(3+3)(3−3)=2(3−3)6=3−33.(2+\sqrt3)y+y=2\ \Longrightarrow\ (3+\sqrt3)y=2\ \Longrightarrow\ y=\frac{2}{3+\sqrt3}=\frac{2(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}=\frac{2(3-\sqrt3)}{6}=\frac{3-\sqrt3}{3}.

Then x=2−y=2−3−33=6−3+33=3+33x=2-y=2-\dfrac{3-\sqrt3}{3}=\dfrac{6-3+\sqrt3}{3}=\dfrac{3+\sqrt3}{3}. So

P=(3+33, 3−33).P=\left(\frac{3+\sqrt3}{3},\ \frac{3-\sqrt3}{3}\right).

Step 3. Intersect the second line with PQPQ. Setting (2−3)y=2−y(2-\sqrt3)y=2-y: …

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