c1 or c2 - i don't know :)

Triangle ABC has an angle of 90 degrees at B. Point A is on the y axis. AB is part of the line x - 2y + 8 = 0, and C is the point (6,2)

Find the equations AC and BC
Find the lengths of AB and BC and hence find the area of the triangle
Using your answer to the above, find the length of the perpendicular from B to AC

I have AC as u = 1/7x + 1 1/7.. no idea if this is right
I have BC so far as y = -2x + c
but can't find C..

any ideas?

Reply 1

samd

10

londongirl

Triangle ABC has an angle of 90 degrees at B. Point A is on the y axis. AB is part of the line x - 2y + 8 = 0, and C is the point (6,2)

(a) Find the equations AC and BC
(b) Find the lengths of AB and BC and hence find the area of the triangle
(c) Using your answer to the above, find the length of the perpendicular from B to AC

I have AC as u = 1/7x + 1 1/7.. no idea if this is right
I have BC so far as y = -2x + c
but can't find C..

any ideas?

Right. I'll give it a go from the beginning...

(a) Point A is on the y axis, and it also lies on x - 2y + 8 = 0.

So put x = 0 in this equation.

Giving: -2y + 8 = 0

Therefore: y = 4.

So: A is (0, 4)

AC's equation can then be found:

m = (y₂ - y₁) / (x₂ - x₁)
so: m = (2 - 4) / (6 - 0) = -1/3

Equation is therefore:

y - y₁ = m (x - x₁
y - 0 = -1/3 (x - 4)
y = -1/3x + 4/3

BC's equation can be found using the idea that -1/m is a gradient perpendicular to a gradient of m, as from the question BC and AB are perpendicular.

AB: x - 2y + 8 = 0

so: 2y = x - 8
y = x/2 -4
so: m_AB = 1/2

so: m_BC = -2

So equation of BC can be found using the coordinate C and m = -2:

y - y₁ = m (x - x₁)
y - 2 = -2(x - 6)
y = 14 - 2x

Answer to be can be found by the distance formula.

Reply 2

londongirl

OP

thanks, i'm still reading it over, but i get it :biggrin:

Reply 3

samd

10

Did you do the other parts of the question?

I'm sorry I really didn't have the energy to the other day.

Reply 4

steve2005

17

samd

Right. I'll give it a go from the beginning...

(a) Point A is on the y axis, and it also lies on x - 2y + 8 = 0.

So put x = 0 in this equation.

Giving: -2y + 8 = 0

Therefore: y = 4.

So: A is (0, 4)

AC's equation can then be found:

m = (y₂ - y₁) / (x₂ - x₁)
so: m = (2 - 4) / (6 - 0) = -1/3

Equation is therefore:

y - y₁ = m (x - x₁
y - 0 = -1/3 (x - 4)
y = -1/3x + 4/3

BC's equation can be found using the idea that -1/m is a gradient perpendicular to a gradient of m, as from the question BC and AB are perpendicular.

AB: x - 2y + 8 = 0

so: 2y = x - 8
y = x/2 -4
so: m_AB = 1/2

so: m_BC = -2

So equation of BC can be found using the coordinate C and m = -2:

y - y₁ = m (x - x₁)
y - 2 = -2(x - 6)
y = 14 - 2x

Answer to be can be found by the distance formula.

I have entered data from your answer into my graphing program and something is wrong: either I have entered the data incorrectly OR your solution contains an error.

See picture. The dots ought to be the vertices of the triangle ABC !

It may be my error .... I will investigate...

I attach picture 56 which contains part of the solution AND I will have to upload picture 59 which contains the rest of the solution because the pic are too large. Edit: I could have found length of AC more easily. AC=sqrt(6^2+2^2) Notice a diagram makes it all much easier.

last part of solution.

AC=6.324
Area=10
Perpendicular distance from B to AC =3.162

I have had problems uploading: the file was too large.

I hope I have posted the complete solution....got a bit confused...