Answer Equations of Lines Tangent to xy+y=2 and xy+x=2

[ArmiAldi]

i have 2 simmilar questions
1) Find the equation of the line tangent to the curve at the point
xy + y = 2 ; (1, 1) and i did this

xy + y = 2
y + xy' + y' = 0
xy' + y' = - y
y'(x + 1) = -y
y' = -y/(x + 1)
y' = -1/2
y- y_1 = m(x - x_1) and i hope so this is OK
but how can i do this one?
2) xy + x = 2 ; (1, 1) please help have to answer in 3 hours

 

[rock.freak667]

y- y_1 = m(x - x_1) and i hope so this is OK
but how can i do this one?

Well you found y' at (1,1) [your "m"]. In that form [itex]y- y_1 = m(x - x_1)[/itex], [itex](x_1,y_1)[/itex] is a point on the line...which you have.

The same way...to find the equation of a line, you need (i) The gradient of the line and (ii) A point which lies on the line.

To find the gradient at a point, you find y' and then use the point given to find the gradient.

 

[HallsofIvy]

i have 2 simmilar questions
1) Find the equation of the line tangent to the curve at the point
xy + y = 2 ; (1, 1) and i did this

xy + y = 2
y + xy' + y' = 0
xy' + y' = - y
y'(x + 1) = -y
y' = -y/(x + 1)
y' = -1/2
y- y_1 = m(x - x_1) and i hope so this is OK

That's good so far but what you give in your final line is a general formula, not the answer to this problem. What is your answer?

but how can i do this one?
2) xy + x = 2 ; (1, 1) please help have to answer in 3 hours

?? You have these labled (1) and (2) so I thought they were different problems- but they are exactly the same. Are you just asking HOW to give the final answer? Use what you said above: y-y_1= m(x- x_1). (x_1, y_1) is, of course, the point you were given: (1, 1) and m is the slope you calculated: -1/2. Put those numbers in.