Question

On a coordinate plane, a line goes through (negative 1, 1) and (0, negative 3). a point is at (negative 4, negative 3) and (0, negative 3). what is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the point (−4, −3)? y 3 = −4(x 4) y 3 = –one-fourth(x 4) y 3 = one-fourth(x 4) y 3 = 4(x 4)

Answer 1

The equation in point-slope form is $y + 3 = \frac{1}{4} (x+4)$

$m_{AB} = \frac{y_2 - y_1}{x_2-x_1}\\\\m_{AB} = \frac{-3-1}{0-(-1)}\\\\m_{AB} = -4$

Because AB is perpendicular to CD,

$m_{AB} * m_{CD} = -1\\-4 * m_{CD} = -1\\m_{CD} = \frac{1}{4}$

Equation of CD

$y - y_{1} = m(x - x_1) \\y + 3 = \frac{1}{4} (x+4)$

Thus  equation in point-slope form is $y + 3 = \frac{1}{4} (x+4)$

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