find the equation of the line passing through (4,5) and :

Question

a. parallel to the line 2x - 3y + 4 =0

b. perpendicullar to the line 13x - 6y - 11 = 0

Answer 1

a.  Through (4,5) and parallel to the line 2x - 3y + 4 =0

 

     2x - 3y = -4

     For line in format ax + by = c, the slope is m = -a/b

     The slope of this line is -2/(-3) = 2/3

     Parallel lines have the same slope so you are looking

     for a line through (4,5) with slope m = 2/3

 

    y = mx + b

    y = (2/3)x + b

    Plug in the given point (4,5) for (x,y) and solve for b

    5 = (2/3)(4) + b

    5 = 8/3 + b

    5 - 8/3 = b

    7/3 = b

    y = (2/3)x + 7/3

 

    This can be simplified by multiplying through by 3

    to eliminate the fractions

    3y = 2x + 7

    -2x + 3y - 7 = 0

 

b.  Through (4,5) and perpendicular to 13x-6y-11 = 0

     13x - 6y = 11

     m = -a/b = -13/(-6) = 13/6

 

     Perpendicular lines have slopes that are negative

     reciprocals of each other.  m_⊥ = -1/m

     m_⊥ = -1/(13/6) = -6/13

 

     You now have the slope m = -6/13 and the point (4,5)

     y = (-6/13)x + b

 

     Follow the same steps as in part a to

     solve for b and finish the equation.