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Exercise 12.1 · Q7

Q.Reduce the equation x+y−2=0x + y - 2 = 0 to the normal form.

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Convert x+y−2=0x+y-2=0 to normal form xcos⁡ω+ysin⁡ω=px\cos\omega+y\sin\omega=p by dividing through by a2+b2\sqrt{a^2+b^2}.

For a line ax+by+c=0ax+by+c=0, the normal form is xcos⁡ω+ysin⁡ω=px\cos\omega+y\sin\omega=p, obtained by dividing by a2+b2\sqrt{a^2+b^2} and choosing the sign so that p>0p>0. Here ω\omega is the angle the perpendicular from the origin makes with the x-axis, and pp is the length of that perpendicular.

  1. Write the line as x+y=2x+y=2, so here a=1, b=1, c=−2a=1,\ b=1,\ c=-2 (constant on RHS =2>0=2>0, so we keep the sign as is).
  2. Compute a2+b2=12+12=2\sqrt{a^2+b^2}=\sqrt{1^2+1^2}=\sqrt2.
  3. Divide both sides of x+y=2x+y=2 by 2\sqrt2: x2+y2=22=2\frac{x}{\sqrt2}+\frac{y}{\sqrt2}=\frac{2}{\sqrt2}=\sqrt2 …

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