$d=\frac{\left|a x_{0}+b y_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}$

Step 1 ：Let \(x = \sqrt{a^2 + b^2}\). Then \(x^2 = a^2 + b^2\), so \(d = \frac{|ax_0 + by_0 + c|}{x}\)

Step 2 ：We can rewrite the equation as \(|ax_0 + by_0 + c| = dx\)

Step 3 ：Square both sides to get rid of the absolute value: \((ax_0 + by_0 + c)^2 = d^2x^2\)

Step 4 ：Expand the equation: \(a^2x_0^2 + 2ax_0by_0 + b^2y_0^2 + 2cx_0a + 2cy_0b + c^2 = d^2(a^2 + b^2)\)

Step 5 ：Substitute \(x^2\) back into the equation: \(a^2x_0^2 + 2ax_0by_0 + b^2y_0^2 + 2cx_0a + 2cy_0b + c^2 = d^2x^2\)

Step 6 ：Divide both sides by \(x^2\): \(\frac{a^2x_0^2 + 2ax_0by_0 + b^2y_0^2 + 2cx_0a + 2cy_0b + c^2}{x^2} = d^2\)

Step 7 ：Simplify the equation: \(\frac{a^2x_0^2}{x^2} + \frac{2ax_0by_0}{x^2} + \frac{b^2y_0^2}{x^2} + \frac{2cx_0a}{x^2} + \frac{2cy_0b}{x^2} + \frac{c^2}{x^2} = d^2\)

Step 8 ：Cancel out the common terms: \(a^2\left(\frac{x_0^2}{x^2}\right) + 2ab\left(\frac{x_0y_0}{x^2}\right) + b^2\left(\frac{y_0^2}{x^2}\right) + 2ca\left(\frac{x_0}{x^2}\right) + 2cb\left(\frac{y_0}{x^2}\right) + c^2\left(\frac{1}{x^2}\right) = d^2\)

Step 9 ：Simplify further: \(a^2\left(\frac{x_0^2}{a^2 + b^2}\right) + 2ab\left(\frac{x_0y_0}{a^2 + b^2}\right) + b^2\left(\frac{y_0^2}{a^2 + b^2}\right) + 2ca\left(\frac{x_0}{a^2 + b^2}\right) + 2cb\left(\frac{y_0}{a^2 + b^2}\right) + c^2\left(\frac{1}{a^2 + b^2}\right) = d^2\)

Step 10 ：Now, we can see that the equation is in the form of \(d^2\), so we can take the square root of both sides to get the final answer: \(d = \sqrt{a^2\left(\frac{x_0^2}{a^2 + b^2}\right) + 2ab\left(\frac{x_0y_0}{a^2 + b^2}\right) + b^2\left(\frac{y_0^2}{a^2 + b^2}\right) + 2ca\left(\frac{x_0}{a^2 + b^2}\right) + 2cb\left(\frac{y_0}{a^2 + b^2}\right) + c^2\left(\frac{1}{a^2 + b^2}\right)}\)

Step 11 ：\(\boxed{d = \sqrt{a^2\left(\frac{x_0^2}{a^2 + b^2}\right) + 2ab\left(\frac{x_0y_0}{a^2 + b^2}\right) + b^2\left(\frac{y_0^2}{a^2 + b^2}\right) + 2ca\left(\frac{x_0}{a^2 + b^2}\right) + 2cb\left(\frac{y_0}{a^2 + b^2}\right) + c^2\left(\frac{1}{a^2 + b^2}\right)}}\)