What is the equation of the line that passes through #(6,-1)# and is perpendicular to the line that passes through the following points: #(8,-3),(12,10) #?

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Answer 1 · 2016-03-18T19:19:19

#y=-4/13x+11/13#

#P_1(6,-1) #
#P_A(x,y) " any point on line pases trough (6,-1)"#
#m_1=(y-(-1))/(x-6)#
#m_1=(y+1)/(x-6)" slope of line"#

#m_2=(10-(-3))/(12-8)#
#m_2=13/4" slope of other line pases trough (8,-3)(12,10)"#
#m_1*m_2=-1" (if lines are perpendicular)"#

#(y+1)/(x-6)*13/4=-1#

#(13y+13)/(4x-24)=-1#

#13y+13=-4x+24#

#13y=-4x+24-13#

#13y=-4x+11#

#y=-4/13x+11/13#