Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).

Answer Details

To find the locus of points equidistant from two fixed points, we need to find the perpendicular bisector of the line segment joining the two points. This perpendicular bisector will be the locus of all points that are equidistant from the two given points. So, let's first find the midpoint of the line segment PQ joining P(4, 5) and Q(-6, -1). The midpoint M is given by: \begin{align*} M &= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\\ &= \left(\frac{4 + (-6)}{2}, \frac{5 + (-1)}{2}\right)\\ &= (-1, 2) \end{align*} Now, let's find the slope of the line segment PQ: \begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{-1 - 5}{-6 - 4}\\ &= \frac{-6}{-10}\\ &= \frac{3}{5} \end{align*} Since the perpendicular bisector of PQ will be perpendicular to PQ, its slope will be the negative reciprocal of the slope of PQ: \begin{align*} m_\perp &= -\frac{1}{m}\\ &= -\frac{1}{\frac{3}{5}}\\ &= -\frac{5}{3} \end{align*} Now we have the slope of the perpendicular bisector and a point on it, which is the midpoint M. We can use point-slope form to find the equation of the perpendicular bisector: \begin{align*} y - y_1 &= m_\perp(x - x_1)\\ y - 2 &= -\frac{5}{3}(x + 1) \end{align*} Simplifying this equation, we get: \begin{align*} 3y - 6 &= -5x - 5\\ 5x + 3y - 1 &= 0 \end{align*} Therefore, the locus of points equidistant from P(4, 5) and Q(-6, -1) is the line 5x + 3y - 1 = 0. Hence, the correct option is (d) 5x + 3y - 1 = 0.