If my tangent line at point (4,8) has the equation $y = 5 \frac{x}{6} - 9$ $y=5x/6 - 9$ , what is the equation of the normal line at the same point?

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Answer 1 · 2014-10-11T02:05:31

The normal line is the line that is perpendicular to the the tangent line.

If the slope of a line is $m$ $m$ then the slope of the perpendicular line is $- \frac{1}{m}$ $-1/m$ , this is also known as the negative reciprocal.

The given equation is $y = \frac{5}{6} x - 9$ $y=5/6x-9$ the slope is $\frac{5}{6}$ $5/6$ so the slope of the normal is $- \frac{6}{5}$ $-6/5$ .

The point $(x, y) \to (4, 8)$ $(x,y)->(4,8)$

$y = m x + b \to$ $y=mx+b ->$ Substitute in the values of $m$ $m$ , $x$ $x$ and $y$ $y$

$8 = - \frac{6}{5} (4) + b$ $8=-6/5(4)+b$

$8 = - \frac{24}{5} + b$ $8=-24/5+b$

$\frac{24}{5} + 8 = b$ $24/5+8=b$

$\frac{24}{5} + \frac{40}{5} = b$ $24/5+40/5=b$

$\frac{64}{5} = b$ $64/5=b$

The equation of the normal line is $\to y = - \frac{6}{5} x + \frac{64}{5}$ $-> y=-6/5x+64/5$

If my tangent line at point (4,8) has the equation y=5x/6 - 9y=5x6−9y=5x/6 - 9, what is the equation of the normal line at the same point?