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Miscellaneous Exercise 4 (Solve) · Q71

Q.Find the joint equation of lines: passing through (−1,2)(-1,2) and perpendicular to the lines x+2y+3=0x + 2y + 3 = 0 and 3x−4y−5=03x - 4y - 5 = 0.

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Perpendicular to x+2y+3=0x+2y+3=0 (slope −1/2-1/2) has slope 22; through (−1,2)(-1,2): 2x−y+4=02x-y+4=0. Perpendicular to 3x−4y−5=03x-4y-5=0 (slope 3/43/4) has slope −4/3-4/3; through (−1,2)(-1,2): 4x+3y−2=04x+3y-2=0. Combined: (2x−y+4)(4x+3y−2)=0(2x-y+4)(4x+3y-2)=0, e …

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