# 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 This solution is **FREE** courtesy of BrainMass!

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 equation of the altitude BP, expressing your answer in the form

y = mx + c

y = mx + c

But we have calculated m = 3/2 in part a above

Thus

y = 3/2 x + c

This line passes through B whose coordinates are (6,5)

Thus the coordinates of B must satisfy the equation of line BP

Therefore

5 = 3/2 (6 ) + c

or 5= 9 + c

or c= -4

Therefore the equation of line BP is y= 3/2 x - 4

Answer: Equation of altitude BP is y= 3/2 x - 4

c) Find the coordinates of M.

The coordinates of a point which divides a straight line internally in the ratio m1:m2 is given by

x coordinate = (m1x1 + m2x2 ) / (m1+ m2)

y coordinate = (m1y1 + m2y2 ) / (m1+ m2)

Here m1=m2=1 (M is the midpoint of BC. )

x1= 6 , y1=5

x2= 8 , y2=-3

Therefore coordinates of M are

x coordinate = (6 + 8 ) / (2)= 7

y coordinate = {5+ ( -3 )} / (2)= 1

Thus the coordinates of M is (7,1)

d) Find the equation of the median AM, expressing your answer in the form

y = mx + c

The coordinates of

A( -1,3)

M (7,1)

Therefore slope= (1-3) / { ( 7 ) - (-1) } = -2 / 8 = - Â¼

Thus the equation of the line AM is y = - Â¼ x + c

This line passes through A( -1,3)

Therefore

3 = -(1/4 ) x (-1) + c

or c = 3 - Â¼ = 11/4

or Equation of line AM is

y= - Â¼ x + 11/4

Answer: Equation of line AM is y= - Â¼ x + 11/4

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

We have two equations

Equation of line BP is y= 3/2 x - 4

Equation of line AM is y= - Â¼ x + 11/4

We will solve these equations simultaneously to find the coordinates of the point of intersection of the altitude BP and the median AM.

If we do so we get

3/2 x - 4 = - Â¼ x + 11/4

or 7 / 4 x = 27 /4

or x= 27/7

Substituting the value of x in y= 3/2 x - 4

y= 25/14

Therefore the coordinates of the point of intersection of the altitude BP and the median AM are ( 27/7 , 25/14)

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