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Exercise 6.5 · Q10

Q.The equation of the line with slope 22 and length of the perpendicular from the origin equal to 5\sqrt5 is

(1) x+2y=5x+2y=\sqrt5
(2) 2x+y=52x+y=\sqrt5
(3) 2x+y=52x+y=5
(4) x+2y−5=0x+2y-5=0
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Write the line as 2x−y+c=02x-y+c=0 and use the perpendicular-distance-from-origin formula ∣c∣/5=5|c|/\sqrt5=\sqrt5 to get c=±5c=\pm5; matching the official answer 2x+y=52x+y=5 shows the intended slope is −2-2 (the source text's "slope 22" is a likely dropped minus sign).

Step 1. General approach — line of given slope mm at a given perpendicular distance pp from the origin. Write the line as mx−y+c=0mx-y+c=0, then use

distance from origin=∣c∣m2+1=p.\text{distance from origin} = \frac{|c|}{\sqrt{m^2+1}} = p.

Step 2. Check each option's distance from the origin, for the given p=5p=\sqrt5. For a line ax+by=kax+by=k, distance =∣k∣a2+b2=\dfrac{|k|}{\sqrt{a^2+b^2}}.

  • Option (1) x+2y=5x+2y=\sqrt5: distance =51+4=55=1=\dfrac{\sqrt5}{\sqrt{1+4}}=\dfrac{\sqrt5}{\sqrt5}=1. Not 5\sqrt5 — rejected.
  • Option (2) 2x+y=52x+y=\sqrt5: distance =55=1=\dfrac{\sqrt5}{\sqrt5}=1. Not 5\sqrt5 — rejected.
  • Option (3) 2x+y=52x+y=5: distance =55=5=\dfrac{5}{\sqrt5}=\sqrt5. ✓ Satisfies the distance condition.
  • Option (4) x+2y−5=0x+2y-5=0, i.e. x+2y=5x+2y=5: distance =55=5=\dfrac{5}{\sqrt5}=\sqrt5. ✓ Also satisfies the distance condition.

Step 3. Distinguish (3) and (4) using the slope. Option (3) 2x+y=5⇒y=−2x+52x+y=5 \Rightarrow y=-2x+5 has slope −2-2. Option (4) x+2y=5⇒y=−12x+52x+2y=5 \Rightarrow y=-\tfrac12x+\tfrac52 has slope −12-\tfrac12. Since a slope of magnitude 22 (not 12\tfrac12) is required, option (3) is the match — this shows the line's true slope is −2-2, i.e. the given data should read "slope −2-2", and the printed "slope 22" most likely dropped a minus sign in transcription. …

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