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Exercise 6.3 · Q4

Q.Write the equation of the lines through the point (1,−1)(1, -1)

(i) parallel to x+3y−4=0x + 3y - 4 = 0
(ii) perpendicular to 3x+4y=63x + 4y = 6.
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Line through (x1,y1)(x_1,y_1) parallel to ax+by+c=0ax+by+c=0 is ax+by=ax1+by1ax+by=ax_1+by_1; perpendicular is bx−ay=bx1−ay1bx-ay=bx_1-ay_1.

  • (i) parallel to x+3y−4=0x+3y-4=0 through (1,−1)(1,-1): x+3y=1−3=−2x+3y=1-3=-2.

  • (ii) perpendicular to 3x+4y−6=03x+4y-6=0 through (1,−1)(1,-1): 4x−3y=4+3=74x-3y=4+3=7.

Both parts use the ready-made formulas for the parallel/perpendicular line through a fixed point (x1,y1)=(1,−1)(x_1,y_1)=(1,-1), applied to two different reference lines.

Step 1 (i). Parallel line — set up. For a line ax+by+c=0ax+by+c=0, the line through (x1,y1)(x_1,y_1) parallel to it is ax+by=ax1+by1ax+by=ax_1+by_1. Here the reference line is x+3y−4=0x+3y-4=0, so a=1, b=3a=1,\ b=3.

Step 2 (i). Substitute the point (1,−1)(1,-1).

x+3y=(1)(1)+(3)(−1)=1−3=−2  ⟹  x+3y+2=0.x+3y = (1)(1)+(3)(-1) = 1-3=-2 \implies x+3y+2=0.

Check: coefficients (1,3)(1,3) match the given line (parallel ✓), and at (1,−1)(1,-1): 1+3(−1)+2=1−3+2=01+3(-1)+2=1-3+2=0 ✓. …

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