$R_x = \Sigma F_x$
$R_x = 750 \sin 60^\circ + 250$
$R_x = 899.52 \, \text{ lb to the right}$
$R_y = \Sigma F_y$
$R_y = 750 \cos 60^\circ - 1250$
$R_y = -875 \, \text{ lb}$
$R_y = 875 \, \text{ lb downward}$
$R = \sqrt{{R_x}^2 + {R_y}^2}$
$R = \sqrt{899.52^2 + 875^2}$
$R = 1254.89 \, \text{ lb}$
$\tan \theta_x = \dfrac{R_y}{R_x}$
$\tan \theta_x = \dfrac{875}{899.52}$
$\theta = 44.21^\circ$
$M_{axle} = \Sigma M_{center}$
$M_{axle} = 250(1.25) + 1250(0.5) - 750(1.25)$
$M_{axle} = 0$
Thus, R = 1254.89 lb downward to the right at θx = 44.21° and passes through the axle.