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Exercise 12.1 · Q1

Q.Find the equation of a line which is equidistant from the lines y=8y = 8 and y=−2y = -2.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est
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✓ Free question

A line equidistant from two parallel horizontal lines is the horizontal line exactly midway between them.

Midway line between y=k1y=k_1 and y=k2y=k_2: y=k1+k22y = \dfrac{k_1+k_2}{2}.

  1. Given lines. y=8y=8 and y=−2y=-2 — both horizontal, hence parallel.

  2. A line equidistant from both must itself be horizontal (parallel to them) and lie exactly midway between their yy-values:

y=8+(−2)2=62=3y = \frac{8+(-2)}{2} = \frac{6}{2} = 3

  1. Verify equal distances. Distance from y=3y=3 to y=8y=8 is ∣8−3∣=5|8-3|=5; distance from y=3y=3 to y=−2y=-2 is ∣3−(−2)∣=5|3-(-2)|=5. Both equal 55, confirming equidistance.

  2. Self-check. 33 is indeed the arithmetic mean of 88 and −2-2, and lies strictly between them, as expected for the equidistant line.

✓Final answer

y=3y = 3

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