Question

What is the equation of the line perpendicular to 3x+y=6 that passes through the point (0,8)​

Answer 1

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$3x + y = 6$

$y = - 3x + 6$

The coefficient of x is the slope of the line , So the slope of the above equation is -3 .

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Remember from now on ,

Product of multiplying the slopes of two lines which are perpendicular to each other , is -1 .

Thus ;

$( - 3) \times (m) = - 1$

m is the slope of the line which we want.

$- 3m = - 1$

Negatives simplifies

$3m = 1$

Divide sides by 3

$\frac{3m}{3} = \frac{1}{3}$

$m = \frac{1}{3}$

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We have following equation to find the point-slope form of the linear functions :

$y - y(0) = m(x - x(0))$

x(0) and y(0) are the coordinates of the point which the line passed through.

$y - 8 = \frac{1}{3} (x - 0)$

$y - 8 = \frac{1}{3} x$

Add sides 8

$y = \frac{1}{3} x + 8$

Done....

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