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Exercise 6.3 · Q13

Q.Find the family of straight lines

(i) perpendicular to
(ii) parallel to 3x+4y−12=03x + 4y - 12 = 0.
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For a line ax+by+c=0ax+by+c=0: every line perpendicular to it has the form bx−ay+k=0bx-ay+k=0, and every line parallel to it has the form ax+by+k1=0ax+by+k_1=0, for an arbitrary real parameter.

Here a=3, b=4a=3,\ b=4 for the given line 3x+4y−12=03x+4y-12=0. Use the two standard one-parameter families built from a,ba,b.

Step 1. Family of lines perpendicular to 3x+4y−12=03x+4y-12=0.

A line perpendicular to ax+by+c=0ax+by+c=0 has the roles of a,ba,b swapped with a sign flip: bx−ay+k=0bx-ay+k=0. With a=3, b=4a=3,\ b=4:

4x−3y+k=0,k∈R4x-3y+k=0,\qquad k\in\mathbb{R}

(Check: the slope of 3x+4y−12=03x+4y-12=0 is −34-\tfrac34; the slope of 4x−3y+k=04x-3y+k=0 is 43\tfrac43, and −34⋅43=−1-\tfrac34\cdot\tfrac43=-1, confirming perpendicularity for every kk.)

Step 2. Family of lines parallel to 3x+4y−12=03x+4y-12=0. …

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