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anonymous

  • 5 years ago

Indicate the equation of the given line in standard form. The line through point (-3, 4) and perpendicular to a line that has slope 2/5

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  1. anonymous
    • 5 years ago
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    Right, you have a point, and you need a gradient so you may then use the point-gradient formula for a straight line. You're told that the line perpendicular to yours has slope \[m_1=\frac{2}{5}\]Two lines are perpendicular if their gradients satisfy the following condition,\[m_1m_2=-1\]This means your gradient must be\[m_2=\frac{-1}{m_1}=-\frac{1}{\frac{2}{5}}=-\frac{5}{2}\]

  2. anonymous
    • 5 years ago
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    The point-gradient formula then gives,\[y-y_1=m(x-x_1)\]which here becomes,\[y-4=-\frac{5}{2}(x-(-3)) \rightarrow 2(y-4)=-5(x+3)\]after multiplying both sides by 2. Expanding and collecting terms, you have\[2y-8=-5x-15 \rightarrow 5x+2y=-7\]

  3. anonymous
    • 5 years ago
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    The last equation is your equation in standard form:\[Ax+By=C\]

  4. anonymous
    • 5 years ago
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    Thank you so much!

  5. anonymous
    • 5 years ago
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    np. Just hope it's clear enough for you to do it for yourself next time :)

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