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

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.

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