Coordinate Geometry

[lusty lodge]

do you have part a and b?

[severe wagon]

yes

[lusty lodge]

yeh so the gradient of L will be the negative reciprocal of the gradient of AB

and it will pass through C

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but its a bisector

[lusty lodge]

yeh it bisects AB and is perpendicular to AB

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oh just cuts through

so gradient is -5/12

[lusty lodge]

make sure your gradient in part a is a fraction

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i havent drawn lines yet

but its 12/5

cuzz 11--1 / 7-2

so 12/5

cant i do m = y-y1 /x -x1

[lusty lodge]

oh i think you made a mistake

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HUH

HOW

[lusty lodge]

it should be (11-5)/7+1

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oh

i see

makes more sense

6/8

3/4

-4/3

[lusty lodge]

yes

[severe wagon]

-4/3 = y-8 / x-3

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

[lusty lodge]

use y-y_1 = m(x-x_1)

or y = mx + C

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that's what i used

[lusty lodge]

ohhhh

i see

you just divided

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cross multiply

[lusty lodge]

you should just expand

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yeah 3y-24 =-4x +12

4x+3y - 36

[lusty lodge]

yeh now take it all to the LHS

nice

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d i)

D( x , 0)

y= 0 sub into eqn

@lusty lodge hey i got down to iii

[lusty lodge]

on d?

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yes

im now confused

[lusty lodge]

use pythagoras' theorem

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but its not a right

angle

triangle

is it now

[lusty lodge]

it is

if they both lie on L

[severe wagon]

wheres H

[lusty lodge]

the hypoteneuse is CD

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wheres angle A

then

that majes hypotenuse Opposite

[lusty lodge]

you don't need any angles

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right

[lusty lodge]

the height is 8

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i drew triangle

yeah

width is 6

[lusty lodge]

in fact this is a very nice Pythagorean triple !

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fancy term that's cute

So Tan = 8 * A * 6

huh

[lusty lodge]

what are you doing

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so Tan /48 = A?

[lusty lodge]

you don't need trig here

oh

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what

[lusty lodge]

part iii

?

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yes

how to find

[lusty lodge]

oh in fact

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tan CÂD

huh

confused now

drew a triangle

its not light bulbing

what do i do

[lusty lodge]

since the line through CD is perpendicular to the line through AC

angle ACD is right angle

[severe wagon]

but i need A

AH

ur confusing me

[lusty lodge]

Well you know the length CD from part ii

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yeah 10

[lusty lodge]

Now you just need to find the length AC

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8?

you've actually lost me

[lusty lodge]

well tan(CAD) is going to be CD/AC

since it's the ratio of the opposite to the adjacent

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but A is unkown

[lusty lodge]

wdym

you know the point A

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(3,0)?

[lusty lodge]

yeh

so find the length of AC

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okay

gotchu

so (3-3) ^ 2 + (0-8)^2

its just 8

cuz (8)^2

64

square root

8

A id 90°

@lusty lodge

[lusty lodge]

is it not 5?

[severe wagon]

no?

cuz c is (3,8)

A (3,0)

[lusty lodge]

A is -1,5

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Why

[lusty lodge]

it says at the start

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thats for different

cuz this is line L

[lusty lodge]

it doesn't matter

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so

[lusty lodge]

you just want the length AC

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(3--1)2 + (8-5)2

[lusty lodge]

which is 5 p sure

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16 + 9

25

[lusty lodge]

yeh

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yh 5

[lusty lodge]

so now it's just CD/AC

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10/5

= 2

[lusty lodge]

bingo

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tan cad = 2

?

[lusty lodge]

i believe so

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why

[lusty lodge]

tan(x) = opposite/adjacent

take angle CAD to be x

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yeah

[lusty lodge]

then you have this triangle

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wouldnt it be tan - 1

to get x

[lusty lodge]

You want the value of tan(CAD) though

Not the actual angle CAD

[severe wagon]

ohh

lowkey will ask my teacher

to explain