## 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

1. anonymous

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

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

The last equation is your equation in standard form:$Ax+By=C$

4. anonymous

Thank you so much!

5. anonymous

np. Just hope it's clear enough for you to do it for yourself next time :)

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