What is the equation of the line perpendicular to #y=2/15x # that passes through # (-4,4) #?

Question

What is the equation of the line perpendicular to #y=2/15x    # that passes through #  (-4,4)  #?

Eva Abramson · Accepted Answer

Equation of the line is #y = -15/2 x -26# Slope of the line, # y=2 /15 x ; [y=m x+c]# is #m_1= 2/15# [Compared with slope-intercept form of equation] The product of slopes of the pependicular lines is #m_1*m_2=-1# #:.m_2= -15/2#. The equation of line passing through #(x_1,y_1)# having slope of #m_2# is #y-y_1=m_2(x-x_1)#. The equation of line passing through #(-4,4)# having slope of #-15/2# is #y-4=-15/2(x+4) or y = -15/2 x +4-30#. or #y = -15/2 x -26#. Equation of the line is #y = -15/2 x -26# [Ans]

Kennedy Adams · Answer

undefined The equation of the line perpendicular to $y=\frac{2}{15}x$ that passes through (-4,4) can be found using the point-slope form of a linear equation.

First, determine the slope of the line perpendicular to $y=\frac{2}{15}x$. The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. The given line has a slope of $\frac{2}{15}$, so the perpendicular line will have a slope of $-\frac{15}{2}$.

Now, use the point-slope form with the point (-4,4) and the slope $-\frac{15}{2}$ to find the equation of the perpendicular line:

$y - y_1 = m(x - x_1)$

$y - 4 = -\frac{15}{2}(x - (-4))$

$y - 4 = -\frac{15}{2}(x + 4)$

$y - 4 = -\frac{15}{2}x - 30$

$y = -\frac{15}{2}x - 26$