Linear Functions Help D: (1 Viewer)

[mawissah]

Given the points A(6,0) and B(0,9) find, in general form, the equation of
a) the line AB
b) the line through A perpendicular to AB

 

[Smile12345]

Find the gradient and then use the point-gradient formula...

 

[Menomaths]

Equation of line AB
y(1) - y(2)/x(1)-x(2) =m
0-9/6-0 = -9/6 = -3/2
Therefore gradient is = -3/2
y-y(1)=m(x-x(1))
y-0 = -3/2(x-6)
2y=-3(x-6)
2y+3x-18 = 0

 

[Smile12345]

Equation of line AB
y(1) - y(2)/x(1)-x(2) =m
0-9/6-0 = -9/6 = -3/2
Therefore gradient is = -3/2
y-y(1)=m(x-x(1))
y-0 = -3/2(x-6)
2y=-3(x-6)
2y+3x-18 = 0

Awe you beat me!!

 

[Menomaths]

Awe you beat me!!

You can do the second one

 

[Smile12345]

You can do the second one

Thanks.

 

[mawissah]

Equation of line AB
y(1) - y(2)/x(1)-x(2) =m
0-9/6-0 = -9/6 = -3/2
Therefore gradient is = -3/2
y-y(1)=m(x-x(1))
y-0 = -3/2(x-6)
2y=-3(x-6)
2y+3x-18 = 0

thankyou! ^^

 

[Menomaths]

thankyou! ^^

No problem.

 

[Smile12345]

b) m1 x m2 = -1 (m is the gradient)
Since m1 (last answer's gradient) is -3/2... m2 = 2/3 .... -3/2 x 2/3 = -1

Now for the point-gradient formula...
y - 0 = 2/3 (x - 6)
y = 2/3 (x -6)
3y = 2x - 12
2x - 3y - 12 = 0

Hope this is correct after all that Menomaths!

 

[Menomaths]

b) m1 x m2 = -1 (m is the gradient)
Since m1 (last answer's gradient) is -3/2... m2 = 2/3 .... -3/2 x 2/3 = -1

Now for the point-gradient formula...
y - 0 = 2/3 (x - 6)
y = 2/3 (x -6)
3y = 2x - 12
2x - 3y - 12 = 0

Hope this is correct after all that Menomaths!

Ask OP she's got the answers

 

[Smile12345]

Ask OP she's got the answers

True!! I nearly made a mistake!! I just hoped I'd got it right after I'd asked to do it!

Hehe!

 

[Smile12345]

Mawissah is my answer right?

U there still???

 

[mawissah]

Mawissah is my answer right?

U there still???

yeah you're answer was right, thankyou so much :3

 

[Smile12345]

yeah you're answer was right, thankyou so much :3

Very good... No worries al all.