Question

Write an equation in standard form for the line that passes through the point (2,-3) and is perpendicular to the line y + 4 = -2/3(x-12)

Answer 1

For this case we have that by definition, the standard form of a linear equation is given by:

$ax + by = c$

By definition, if two lines are perpendicular then the product of their slopes is -1. That is to say:

$m_ {1} * m_ {2} = - 1$

We have the following point-slope equation of a line:

$y+4 = -\frac {2} {3}(x-12)$

The slope is:

$m_ {1} = - \frac {2} {3}$

We find the slope $m_ {2}$of a perpendicular line:

$m_ {2} = \frac {-1}{m_ {1}}\\m_ {2} = \frac {-1} {-\frac {2} {3}}\\m_ {2} = \frac{3} {2}$

Thus, the equation is of the form:

$y-y_ {0} = \frac {3} {2} (x-x_ {0})$

We have the point through which the line passes:

$(x_ {0}, y_ {0}) :( 2, -3)$

Thus, the equation is:

$y - (- 3) = \frac {3} {2} (x-2)\\y + 3 = \frac {3} {2} (x-2)$

We manipulate algebraically:

$y + 3 = \frac{3} {2} x- \frac {3} {2} (2)\\y + 3 = \frac{3} {2} x-3$

We add 3 to both sides of the equation:

$y + 3 + 3 = \frac {3} {2} x\\y + 6 = \frac {3} {2} x$

We multiply by 2 on both sides of the equation:

$2(y + 6) = 3x\\2y + 12 = 3x$

We subtract 3x on both sides:

$2y-3x + 12 = 0$

We subtract 12 from both sides:

$2y-3x = -12$

ANswer:

$-3x + 2y = -12$