Question

which is an equation of the line that is perpendicular to y+1=-3(x-5) and pass through the point (4,-6)

Answer 1

Answer:

Step-by-step explanation:

y+1=-3(x-5)

y + 1 = -3x + 15

     y = -3x  + 15 - 1

     y = -3x + 14

Slope m1 = -3

The required line is perpendicular to this y+1=-3(x-5)

so. m1 * m2 = -1

         -3*m2 = -1

                m2 = -1/-3 = 1/3

Slope = 1/3; pass through (4, -6)

y -y1 = m(x-x1)

y - [-6]  = 1/3 (x - 4)

y + 6   = 1/3 (x - 4)

 y + 6 = 1/3x - 4/3

        y = 1/3x - 4/3 +6      {-4/3 + 6 = -4/3 + 6*3/1*3 =-4/3+18/3 =(-4+18)/3 = 14/3}

         y= 1/3x - 14/3