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Miscellaneous · Q25

Q.Find the equation of the line passing through the point (2,3)(2, 3) and perpendicular to the line joining the points (1,4)(1, 4) and (5,−2)(5, -2).

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Slope of the line joining (1,4)(1,4) and (5,−2)(5,-2): −2−45−1=−64=−32\dfrac{-2-4}{5-1}=\dfrac{-6}{4}=-\dfrac32. The required line is perpendicular to this, so its slope is the negative reciprocal: 23\dfrac{2}{3} (check: −32×23=−1-\tfrac32\times\tfrac23=-1 ✓). Using the point-slope form through (2,3)(2,3): y−3=23(x−2)y-3=\dfrac23(x-2). Multiply by 33: 3y−9=2x−43y-9=2x-4, i.e. 2x−3y+5=02x-3y+5=0. Check with (2,3)(2,3): 2(2)−3(3)+5=4−9+5=02(2)-3(3)+5=4-9+5=0 ✓. [!ANSWER] 2x−3y+5=02x-3y+5=0.

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