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Miscellaneous · Q26

Q.The line through the points (h,3)(h, 3) and (2,7)(2, 7) is perpendicular to the line 5x−2y−10=05x - 2y - 10 = 0. Find the value of hh.

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Slope of 5x−2y−10=05x-2y-10=0: solving for yy, y=52x−5y=\dfrac52x-5, so its slope is 52\dfrac52. For the line through (h,3)(h,3) and (2,7)(2,7) to be perpendicular to it, its slope must be the negative reciprocal, −25-\dfrac25. But this slope is also 7−32−h=42−h\dfrac{7-3}{2-h}=\dfrac{4}{2-h}. Setting these equal: 42−h=−25  ⟹  4×5=−2(2−h)  ⟹  20=−4+2h  ⟹  2h=24  ⟹  h=12\dfrac{4}{2-h}=-\dfrac25 \implies 4\times5 = -2(2-h) \implies 20=-4+2h \implies 2h=24 \implies h=12. Check: slope …

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