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Exercise 6.5 · Q12

Q.The equation of the straight line perpendicular to the line x−y+5=0x-y+5=0, through the point of intersection of the yy-axis and the given line, is

(1) x−y−5=0x-y-5=0
(2) x+y−5=0x+y-5=0
(3) x+y+5=0x+y+5=0
(4) x+y+10=0x+y+10=0
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Find where x−y+5=0x-y+5=0 meets the yy-axis (set x=0x=0), then write the perpendicular line (slope −1-1) through that point.

Step 1. Find the intersection with the yy-axis. On the yy-axis, x=0x=0. Substituting into x−y+5=0x-y+5=0: −y+5=0  ⟹  y=5-y+5=0 \implies y=5. So the point is (0,5)(0,5).

Step 2. Slope of the given line. x−y+5=0  ⟹  y=x+5x-y+5=0 \implies y=x+5, slope =1=1.

Step 3. Slope of the required perpendicular. Negative reciprocal of 11 is −1-1.

Step 4. Equation through (0,5)(0,5) with slope −1-1. …

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