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Exercise 6.3 · Q7

Q.Find the equations of two straight lines which are parallel to the line 12x+5y+2=012x + 5y + 2 = 0 and at a unit distance from the point (1,−1)(1, -1).

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Family parallel to 12x+5y+2=012x+5y+2=0 is 12x+5y+k=012x+5y+k=0; set distance from (1,−1)(1,-1) equal to 11 and solve ∣k+7∣=13|k+7|=13.

  • k=6k=6 or k=−20k=-20 (two lines, one on each side of the point).

A line at a given distance from a fixed point, on a fixed direction, generally has two solutions — one on either side of the point — so we expect a ±\pm split when we remove the absolute value.

Step 1. Write the family of parallel lines. Any line parallel to 12x+5y+2=012x+5y+2=0 has the form

12x+5y+k=012x+5y+k=0

for some constant kk.

Step 2. Impose the distance condition. The distance from (1,−1)(1,-1) to this line must equal 11:

∣12(1)+5(−1)+k122+52∣=1  ⟹  ∣12−5+k13∣=1  ⟹  ∣7+k∣=13.\left|\frac{12(1)+5(-1)+k}{\sqrt{12^2+5^2}}\right|=1 \implies \left|\frac{12-5+k}{13}\right|=1 \implies |7+k|=13. …

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