Help with math problem

[keishton]

So I take 4/-3 and -b/4 and graph it to find b?

for line 1: (7-3)/(-2-1) = 4/-3 = gradient
for line 2: (-1-b)/(8-4) = (-1-b)/4 = perpendicular gradient
perpendicular gradient = -1/gradient

solve for b

no idea how to use a graphics calculator anymore lol

[Snakeseye0]

for line 1: (7-3)/(-2-1) = 4/-3 = gradient
for line 2: (-1-b)/(8-4) = (-1-b)/4 = perpendicular gradient
perpendicular gradient = -1/gradient

solve for b

no idea how to use a graphics calculator anymore lol

I got -1 as b?

[Deleted]

it's been a while but my approach would be:

- set up an equation of a circle with the center being (8,-1) and the radius as the variable.
- look at l1 as the tangent line
- intersect l1 with the circle and you will get an intercept point
- now set up a new equation for the line passing through (let's call it l2) using the intercept point and (8,-1)
- then calculate y=k*x+d
- and finally put y as b

sorry for my english

[Lothrik]

It'd be a lot easier if you actually got a pencil and paper and tried to graph this. You draw a line that passes through (1,3) and (-2,7). You draw another line perpendicular to that first line that passes through (8,-1). You can find b by looking along the second line you drew and looking at the y value where it intersects x=4.

http://www.wolframalpha.com/input/?i...o+%288%2C-1%29

Do you understand what it means for two lines to be perpendicular to each other?

[Deleted]

I got -1 as b?

that can't be correct

- - - Updated - - -

it's been a while but my approach would be:

- set up an equation of a circle with the center being (8,-1) and the radius as the variable.
- look at l1 as the tangent line
- intersect l1 with the circle and you will get an intercept point
- now set up a new equation for the line passing through (let's call it l2) using the intercept point and (8,-1)
- then calculate y=k*x+d
- and finally put y as b

sorry for my english

edit: there is most likely an easier way but can't remember lul

[Snakeseye0]

It'd be a lot easier if you actually got a pencil and paper and tried to graph this. You draw a line that passes through (1,3) and (-2,7). You draw another line perpendicular to that first line that passes through (8,-1). You can find b by looking along the second line you drew and looking at the y value where it intersects x=4.

http://www.wolframalpha.com/input/?i...o+%288%2C-1%29

Do you understand what it means for two lines to be perpendicular to each other?

Not the same?

[Garnier Fructis]

Not the same?

[Snakeseye0]

Well thanks for the help guys, I still really can't get this so I guess I'll just skip it. It's extra credit anyways, so not that important I guess.

[Deleted]

that can't be correct

- - - Updated - - -



edit: there is most likely an easier way but can't remember lul

edit2: what I proposed makes in fact no sense at all. forget what I was saying.

edit3: shouldn't have replied to my own entry with a quote. sorry guys

[Jordaen]

Well thanks for the help guys, I still really can't get this so I guess I'll just skip it. It's extra credit anyways, so not that important I guess.

If you can't do the extra credit I doubt you'll be able to do the test. Just put some time into drawing the graph or asking your teacher for help tomorrow, unless it's due tomorrow in which case still ask because if your test is going to be like this problem, you should learn how to read it.

[Ulfhedinn]

for line 1: (7-3)/(-2-1) = 4/-3 = gradient
for line 2: (-1-b)/(8-4) = (-1-b)/4 = perpendicular gradient
perpendicular gradient = -1/gradient

solve for b

no idea how to use a graphics calculator anymore lol

This gives you everything you need to find the answer.

Also if you have any math questions Khan academy is your friend.

[Deleted]

it's been a while but my approach would be:

- set up an equation of a circle with the center being (8,-1) and the radius as the variable.
- look at l1 as the tangent line
- intersect l1 with the circle and you will get an intercept point
- now set up a new equation for the line passing through (let's call it l2) using the intercept point and (8,-1)
- then calculate y=k*x+d
- and finally put y as b

sorry for my english

edit4: it actually does make sense!

interception point = (q,w)

r = √[(8-q)^2 + (-1-w)^2]

now r is no longer a variable in the intersection of l1 and circle!


edit5: can some mathematics-genius confirm?

edit6: it's not correct after all