Find the missing value. The distance between -8-17i and 3+bi is 61 units. The positive value of b is

[tex]\bf \stackrel{a}{-8}\stackrel{b}{-17}i\qquad \qquad \stackrel{a}{3}\stackrel{b}{+b}i \\\\\\ \textit{so let's use those two points of (-8, -17) and (3,b)} \\\\\\ \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -8}}\quad ,&{{ -17}})\quad % (c,d) &({{ 3}}\quad ,&{{ b}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{61}{d}=\sqrt{[3-(-8)]^2+[b-(-17)]^2}\implies 61=\sqrt{(3+8)^2+(b+17)^2}[/tex]

[tex]\bf 61=\sqrt{11^2+(b^2+34b+17^2)} \\\\\\ 61=\sqrt{121+b^2+34b+289}\implies 61^2=b^2+34b+410 \\\\\\ 0=b^2+34b+410-\stackrel{61^2}{3721} \implies 0=b^2+34b-3311 \\\\\\ 0=(b+77)(b-43)\implies b= \begin{cases} -77\\ \boxed{43} \end{cases}[/tex]

bear in mind that the "i" in bi, is indicating is the coordinate value on the imaginary axis, however, the coordinate ordinal value is just "b" without the "i", when it comes to just using the coordinates in a point notation.