Home > Algebra > Equations of Perpendicular Lines

How do you write the equation of a line passing through (-6, 2) and perpendicular to y=-3/5x+6?

Answer 1

Kennedy Adams

#5x-3y+36=0#

The slope of line: #y=-3/5 x+6# is #=-3/5#

Now, the slope #m# of line perpendicular to the given line: #y=-3/5x+6#

#m=-1/{-3/5}#

#=5/3#

Hence the equation of line passing through the point #(x_1, y_1)\equiv(-6, 2)# & having slope #m=5/3# is given by following formula

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

#y-2=5/3(x-(-6))#

#3y-6=5x+30#

#5x-3y+36=0#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Julia Adams

To write the equation of a line passing through the point ((-6, 2)) and perpendicular to the line (y = -\frac{3}{5}x + 6), you first find the slope of the perpendicular line, which is the negative reciprocal of (-\frac{3}{5}). The negative reciprocal of (-\frac{3}{5}) is (\frac{5}{3}). Then, you use the point-slope form of the equation of a line, which is (y - y_1 = m(x - x_1)), where (m) is the slope and ((x_1, y_1)) is the given point. Substituting (\frac{5}{3}) for (m) and ((-6, 2)) for ((x_1, y_1)), you get (y - 2 = \frac{5}{3}(x + 6)). Finally, simplify the equation to get the slope-intercept form, (y = \frac{5}{3}x + 12).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.