Equation of line solution15



In this page equation of line solution15 we are going to see solution of each problem with detailed explanation of the worksheet slope of the line.

(14) If the vertices of a triangle ABC are A (2,-4), B (3,3) and C (-1,5). Find the equation of the straight line along the altitude from vertex B.

Solution:

Slope of the line AC:

                        m = (y₂ - y₁)/(x₂ - x₁)

                            = (5-(-4))/(-1-2)

                            = (5+4)/(-3)

                            = 9/(-3)

                            = -3  

Slope of BD = -1/(-3)

                = 1/3

Equation of BD:

 (y - y₁) = m(x - x)

(y - 3) = (1/3) (x - 3)

3 (y - 3) = 1 (x - 3)

3 y - 9 = x - 3

x - 3 y - 3 + 9 = 0

x - 3 y + 6 = 0


(15) If the vertices of triangle ABC are (-4, 4) , B (8 ,4) and C (8,10). Find the equation of the straight line along the median from A.

Solution:

Slope of the line BC:

                        m = (y₂ - y₁)/(x₂ - x₁)

                            = (10-4)/(8-8)

                            = 6/0

                            = 0

Slope of AD = -1/0

Equation of AD:

 (y - y₁) = m(x - x)

(y - 4) = (-1/0) (x - (-4))

0 (y - 4) = -1 (x + 4)

0 = - x - 4

x + 4 = 0


(16) Find the coordinates of the foot from the origin on the straight line 3 x +2 y = 13.

Solution:

In this question we have to find the point "P". The line OP is perpendicular to the 3 x + 2 y = 13.

Equation of the line OP

2 x - 3 y + k = 0

The line OP is passing through the origin (0 , 0)

2 (0) - 3 (0) + k = 0

k = 0

Therefore the equation of the line OP

2 x - 3y + 0 = 0

2 x - 3 y = 0

Both lines are intersecting at the point P.

3 x + 2 y = 13   -------- (1)

2 x - 3 y = 0   -------- (2)

(1) x 3 = >  9 x + 6 y = 39

(2) x 2 =>   4 x - 6 y = 0

               --------------

                 13 x = 39

                     x = 39/13

                     x = 3

Substitute x = 3 in the second equation

     2 (3) - 3 y = 0

       6 - 3 y = 0

          - 3 y = -6

               y = (-6)/(-3)

               y = 2

Therefore the required point P is (3 , 2).

equation of line solution15  equation of line solution15