Question

2x+3y=5 at (-2,3) find the equation of the tangent line

Answer 1

The slope of the tangent line to the line 2x+3y=5 can be found by differentiating 2x+3y=5.

Differentiating 2x+3y=5 with respect to x, we get

`\begin{gathered} 2+3(dy)/(dx)=0 \\ 3(dy)/(dx)=-2 \\ (dy)/(dx)=(-2)/(3) \end{gathered}`

m=dy/dx is the slope of tangent line.

Hence, slope, m=-2/3.

Now, the equation of the tangent line passing through point (x1, y1)=(-2, 3) with slope m=-2/3 can be found as,

`\begin{gathered} m=(y_1-y)/(x_1-x) \\ (-2)/(3)=(3-y)/(-2-x) \\ -2(-2-x)=3(3-y) \\ 4+2x=9-3y \\ 3y+2x=5 \end{gathered}`

Therefore, the equation of the tangent line is 3y+2x=5.