How do you write an equation in point slope and slope intercept form given (8, -2) and perpendicular to the line whose equation is x-5y-7=0?

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Answer 1 · 2018-06-05T12:05:31

Point slope form equation: #(y+2)=-5(x-8)#
Slope intercept form equation: #y= -5 x +38#

Slope of the line, # x-5 y -7=0 or y= 1/5 x -7/5; [y=m x+c]#

is #m_1=1/5# [Compared with slope-intercept form of equation]

The product of slopes of the pependicular lines is #m_1*m_2=-1#

#:.m_2=(-1)/(1/5)=-5#. In point slope form, equation of line

passing through #(x_1=8,y_1=-2)# having slope of #m_2=-5#

is #y-y_1=m_2(x-x_1):. y-(-2)= -5 (x-8) #or

#(y+2)=-5(x-8)#. Slope intercept form:

#(y+2)=-5(x-8) or y +2 =-5 x+40 # or

#y= -5 x +40-2 or y= -5 x +38# [Ans]