# Need quick help

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?

Homework Helper
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 $y- y_1 = m(x - x_1)$, $(x_1,y_1)$ is a point on the line...which you have.

ArmiAldi;17176 2) xy + x = 2 ; (1 said:
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.

Homework Helper
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?