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 - x₁)

(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 - x₁)

(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