Answer
$(-14, -6), (16, -6)$
Work Step by Step
a) (Refer to image)
Using Pythagorean theorem on $\triangle ADC,$
$(8, 15, 17)$ is a Pythagorean triplet.
$\implies DC = 15.$
(The formula $DC^{2} = AC^{2}-AD^{2}$ will yield the same result.)
Doing the same steps for $\triangle ADB$, we get $BD = 15.$
$\implies$ Distance of C from the y-axis = Distance of C from D + 1
and,
$\implies$ Distance of B from the y-axis = Distance of B from D - 1
($\because D$ is shifted one unit to the right.)
$\therefore$ Co-ordinates of C = (16, -6) and Co-ordinates of B = (-14, -6)
b) Distance formula - $D = \sqrt{({x_{2}-x_{1}})^{2}+({y_{2}-y_{1}})^{2}}$
Here, $x_{1} = 1, x_{2} = x (To\space Find), y_{1} = 2, y_{2} = -6, D = 17$
Substituting in the formula,
$D = \sqrt{({x-1})^{2}+({(-6)-2})^{2}} = \sqrt{x^{2}+1-2x+64} = \sqrt{x^{2}-2x+65} = 17$
Squaring both sides,
$\implies x^{2}-2x+65 = 289 \implies x^{2}-2x -224 = 0$
This is a quadratic equation with a = 1, b = -2, c = -224.
Using the formula, $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$,
$\implies x = \frac{2 \pm \sqrt{(-2)^{2}-4\times1\times-224}}{2}$,
$\implies x = \frac{2 \pm \sqrt{300}}{2} \implies \frac{-2 \pm 30}{2}$
$\implies x$ can be either 16 or -14
$\therefore x \in \{16, -14\}$
$\therefore$ Co-ordinates of C = (16, -6) and Co-ordinates of B = (-14, -6)
