Answer
The x-coordinate of particle D's position must be $~~x = 0.716~d$
Work Step by Step
We can find the y-component of the gravitational force on A due to B:
$F_y = \frac{G~m_A~m_B}{d^2}$
$F_y = \frac{G~m_A~(2.00~m_A)}{d^2}$
$F_y = 2.00~\frac{G~m_A^2}{d^2}$
We can find the x-component of the gravitational force on A due to C:
$F_x = -\frac{G~m_A~m_C}{(1.5~d)^2}$
$F_x = -\frac{G~m_A~(3.00~m_A)}{2.25~d^2}$
$F_x = -1.33~\frac{G~m_A^2}{d^2}$
We can find the magnitude of the net gravitational force on A due to B and C:
$F = \sqrt{F_x^2+F_y^2}$
$F = \sqrt{(-1.33~\frac{G~m_A^2}{d^2})^2+(2.00~\frac{G~m_A^2}{d^2})^2}$
$F = 2.40~\frac{G~m_A^2}{d^2}$
We can find the direction of the net gravitational force on A due to B and C as an angle $\theta$ above the negative x axis:
$tan~\theta = \frac{\vert F_y \vert}{\vert F_x \vert}$
$tan~\theta = \frac{2.00~\frac{G~m_A^2}{d^2}}{1.33~\frac{G~m_A^2}{d^2}}$
$tan~\theta = \frac{2.00}{1.33}$
$\theta = tan^{-1}~(1.5)$
$\theta = 56.3^{\circ}$
The direction of the net gravitational force on A due to B and C is an angle of $~~56.3^{\circ}~~$ above the negative x axis.
For the net gravitational force on particle A to be zero, the magnitude of the gravitational force on A due to D must be $F = 2.40~\frac{G~m_A^2}{d^2}$ and the direction must be an angle of $~~56.3^{\circ}~~$ below the positive x axis.
We can find the required distance $r$ of particle D from particle A:
$F = 2.40~\frac{G~m_A^2}{d^2} = \frac{G~m_A~m_D}{r^2}$
$2.40~\frac{m_A}{d^2} = \frac{4.00~m_A}{r^2}$
$2.40~r^2 = 4.00~d^2$
$r^2 = \frac{4.00~d^2}{2.40}$
$r = \sqrt{\frac{4.00}{2.40}}~d$
$r = 1.29~d$
We can find the x-coordinate of particle D's position:
$x = (1.29~d)~cos~56.3^{\circ}$
$x = 0.716~d$
The x-coordinate of particle D's position must be $~~x = 0.716~d$
