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Exercise 6.4 · Q6

Q.Find the equation of the pair of straight lines passing through the point (1,3)(1, 3) and perpendicular to the lines 2x−3y+1=02x - 3y + 1 = 0 and 5x+y−3=05x + y - 3 = 0.

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Find the two individual lines through (1,3)(1,3) perpendicular to each given line (negative reciprocal slope), then multiply them together to get the combined equation.

We need the pair of lines through (1,3)(1,3), one perpendicular to each given line; find each separately, then combine.

Step 1. Line through (1,3)(1,3) perpendicular to 2x−3y+1=02x-3y+1=0. This line has slope 23\tfrac23, so the perpendicular has slope −32-\tfrac32. Through (1,3)(1,3):

y−3=−32(x−1) ⟹ 2(y−3)=−3(x−1) ⟹ 2y−6=−3x+3 ⟹ 3x+2y−9=0.y-3=-\tfrac32(x-1)\ \Longrightarrow\ 2(y-3)=-3(x-1)\ \Longrightarrow\ 2y-6=-3x+3\ \Longrightarrow\ 3x+2y-9=0.

Step 2. Line through (1,3)(1,3) perpendicular to 5x+y−3=05x+y-3=0. This line has slope −5-5, so the perpendicular has slope 15\tfrac15. Through (1,3)(1,3):

y−3=15(x−1) ⟹ 5(y−3)=x−1 ⟹ 5y−15=x−1 ⟹ x−5y+14=0.y-3=\tfrac15(x-1)\ \Longrightarrow\ 5(y-3)=x-1\ \Longrightarrow\ 5y-15=x-1\ \Longrightarrow\ x-5y+14=0.

Step 3. Combine the two lines.

(3x+2y−9)(x−5y+14)=0.(3x+2y-9)(x-5y+14)=0.

Expand:

3x(x−5y+14)+2y(x−5y+14)−9(x−5y+14)3x(x-5y+14)+2y(x-5y+14)-9(x-5y+14) …

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