Find the equation of tangent line passing $(2,3)$ and perpendicular to $3x+4y=8$. I need to find the equation of tangent line passing $(2,3)$ and perpendicular to $3x+4y=8$. Need help in this and also show me how you got the answer. I will be very thankful.
 A: Your approach my depend on what you have seen about slopes and/or normal vectors and their relations with respect to being perpendicular.
Slope approach
A line through $(x_1,y_1)$ with slope $m$ has the following equation:
$$y = m(x-x_1)+y_1$$
If $m_1$ and $m_2$ are slopes corresponding to perpendicular lines, then $m_1m_2 = -1$. The slope of the given line is is $-3/4$, so a perpendicular line has slope $4/3$; filling in:
$$y = \tfrac{4}{3} (x-2)+3$$
Normal vector approach
The normal vector of a line $ax+by+c=0$ is $(a,b)$ so any line with normal vector $(b,-a)$, or any non-zero multiple of this vector, is perpendicular to it (since their dot product is zero); so you're looking for a line of the form:
$$4x-3y=C$$
Substitution of $(2,3)$ gives you $C$. Or you may know the standard form to go directly to:
$$a(x-x_1)+b(y-y_1) = 0 \longrightarrow 4(x-2)-3(y-3)=0$$
A: The equation is: $-4x + 3y = a$       
By taking $(-4,3)$ (in front of $x$ and $y$) you make the line
perpendicular to the given one which has $(3,4)$.       
You determine the $a$ by putting $(2,3)$ in there.
You get:  $-4x + 3y = 1$    
