This page 10th grade math solution exercise 5.5 part4 is going to provide you solution for every problems that you find in the exercise no 5.5

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

Solution:

A line drawn from the vertex B is perpendicular to the line AC.So the product of their slopes will be equal to -1.

Slope of AC = (y₂ - y₁)/(x₂ - x₁)

x₁ = 2, y₁ = -4, x₂ = -1 , y₂ = 5

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

= (5 + 4)/(-3)

= -9/3

= -3

slope of line drawn from the vertex B = -1/m

= -1/(-3)

= 1/3

Equation of the line drawn from the vertex B:

(y - y₁) = m (x - x₁)

B (3,3) slope = 1/3

x₁ = 3, y₁ = 3

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

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

3 y - 9 = x - 3

x + 3y - 3 + 9 = 0

x + 3y + 6 = 0

In the page10th grade math solution exercise 5.5 part4 we are going to see the solution of next problem

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

Solution:

Since AD is median,it passes through the midpoint of the side BC.

x₁ = 8, y₁ = 4, x₂ = 8 , y₂ = 10

= [(x₁ + x₂)/2 , (y₁ + y₂)/2]

= [(8 + 8)/2 , (4 + 10)/2]

= 16/2 , 14/2

= (8,7)

Equation of the line AD

(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)

(-4,4) (8,7)

x₁ = -4, y₁ = 4, x₂ = 8 , y₂ = 7

(y -4)/(7-4) = (x-(-4))/(8-(-4))

(y -4)/3 = (x+4)/(8+4)

(y -4)/3 = (x+4)/12

12(y - 4) = 3(x + 4)

12 y - 48 = 3 x + 12

3 x - 12 y + 12 + 48 = 0

3 x - 12y + 60 = 0

dividing the whole equation by 3

x - 4 y + 20 = 0

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