# Equations of Lines and Intersections

The diagram below (see attachment) shows a triangle ABC whose vertices are at A (-1, 3), B (6, 5) and C (8, -3). The line BP is perpendicular to the line AC, and M is the midpoint of BC.

Note that BP is called an altitude of triangle ABC and that AM is called a median of triangle ABC.

a) Find the gradient of

i) The line AC

ii) The altitude BP

(Show answers as a fraction in its simplest form)

b) Find the equation of the altitude BP, expressing your answer in the form: y = mx + c

c) Find the coordinates of M.

d) Find the equation of the median AM, expressing your answer in the form: y = mx + c

e) Explain how you would find the coordinates of the point of intersection of the altitude BP and the median AM.

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#### Solution Preview

a) Find the gradient of

gradient of a line (also called slope ) = (y2-y1) /(x2-x1)

i) The line AC

Gradient of a line (also called slope ) = (y2-y1) /(x2-x1)

A (-1,3) , C (8,-3)

Therefore gradient ={ (-3)-(3) }/{ (8)-(-1)} = -6/9 = -2/3

Answer: slope of line AC= -2/3

ii) The altitude BP

Altitude BP is perpendicular to AC

The product of the gradients of perpendicular lines = -1

Therefore ,

Slope of line AC x slope of line BP = -1

Or ( -2/3) x slope of line BP = -1

Or slope of line BP = -1 / (-2/3) = 3 /2

Answer: slope of line BP= 3/2

(Show answers as a fraction in its simplest form)

b) Find the ...

#### Solution Summary

The expert solves equations of lines and intersections. The altitudes of triangles are provided.