Q.Using Calculus, find perpendicular distance from (0, 0) on 3x + 4y + 5 = 0.

Minimize the squared distance from the origin to a general point on the line using calculus (derivative $=0$), then take the square root.

Given line: $3x+4y+5=0$. A general point on this line can be written as $(x,\,y)$ with $y=-\dfrac{3x+5}{4}$.

The squared distance from the origin $(0,0)$ to this point is:

$D^2 = x^2+y^2 = x^2 + \left(\frac{3x+5}{4}\right)^2$

To minimize $D^2$ (equivalently, minimize $D$, since $D\ge0$), differentiate with respect to $x$ and set to zero:

$\frac{d(D^2)}{dx} = 2x + 2\left(\frac{3x+5}{4}\right)\cdot\frac34 = 0$

Multiply through by $8$: $16x + 3(3x+5) = 0 \ \Rightarrow\ 16x+9x+15=0 \ \Rightarrow\ 25x=-15 \ \Rightarrow\ x=-\dfrac35$

Then $y = -\dfrac{3(-3/5)+5}{4} = -\dfrac{-9/5+25/5}{4} = -\dfrac{16/5}{4} = -\dfrac45$

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