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Find the equation of line passing through (4,5) and perpendicular to the line 7x - 5y = 420

written 5.2 years ago by teamques10 ★ 68k

modified 2.6 years ago by pedsangini276 • 4.8k

engineering mathematics

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written 5.2 years ago by teamques10 ★ 68k

Solution:

$Point =\left(x_{1}, y_{1}\right)=(4,5)$

$Slope of the line 7 x-5 y=420 is,$

$m=-\frac{a}{b}=-\frac{7}{-5}=\frac{7}{5}$

$\therefore Slope of the required line is,$

$m_{1}=-\frac{1}{m}=\frac{-5}{7}$

$\therefore equation is,$

$\quad y-y_{1}=m_{1}\left(x-x_{1}\right)$ $\therefore y-5=\frac{-5}{7}(x-4)$

$\therefore 5 x+7 y-55=0$

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