The correct answer is **A) $(0, -1)$**. Here's a step-by-step solution: **1. Understand the properties of point P:** Let the coordinates of point $P$ be $(x_P, y_P)$. Since $P$ lies on the line $y = 2x-1$, its coordinates can be expressed as $(x_P, 2x_P-1)$. **2. Utilize the equidistant property:** Point $P$ is equidistant from $A(1, 2)$ and $B(3, 0)$. This means the distance $PA$ is equal to the distance $PB$. We can write this as $PA = PB$, or more conveniently, $PA^2 = PB^2$ to avoid square roots. **3. Apply the distance formula:** The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Calculate $PA^2$: $PA^2 = (x_P - 1)^2 + ((2x_P-1) - 2)^2$ $PA^2 = (x_P - 1)^2 + (2x_P - 3)^2$ $PA^2 = (x_P^2 - 2x_P + 1) + (4x_P^2 - 12x_P + 9)$ $PA^2 = 5x_P^2 - 14x_P + 10$ Calculate $PB^2$: $PB^2 = (x_P - 3)^2 + ((2x_P-1) - 0)^2$ $PB^2 = (x_P - 3)^2 + (2x_P - 1)^2$ $PB^2 = (x_P^2 - 6x_P + 9) + (4x_P^2 - 4x_P + 1)$ $PB^2 = 5x_P^2 - 10x_P + 10$ **4. Equate $PA^2$ and $PB^2$ and solve for $x_P$:** $5x_P^2 - 14x_P + 10 = 5x_P^2 - 10x_P + 10$ Subtract $5x_P^2$ from both sides: $-14x_P + 10 = -10x_P + 10$ Subtract $10$ from both sides: $-14x_P = -10x_P$ Add $14x_P$ to both sides: $0 = 4x_P$ Therefore, $x_P = 0$. **5. Find $y_P$ using the line equation:** Substitute $x_P = 0$ into the line equation $y = 2x-1$: $y_P = 2(0) - 1$ $y_P = -1$ So, the coordinates of point $P$ are $(0, -1)$. **Alternative Method (using perpendicular bisector):** **1. Identify the locus of points equidistant from A and B:** The locus of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of the line segment $AB$. **2. Find the midpoint of AB:** Midpoint $M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right)$ $M = \left(\frac{1+3}{2}, \frac{2+0}{2}\right) = \left(\frac{4}{2}, \frac{2}{2}\right) = (2, 1)$. **3. Find the slope of AB:** Slope $m_{AB} = \frac{y_B-y_A}{x_B-x_A} = \frac{0-2}{3-1} = \frac{-2}{2} = -1$. **4. Find the slope of the perpendicular bisector:** The slope of a line perpendicular to $AB$ is the negative reciprocal of $m_{AB}$. $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1$. **5. Write the equation of the perpendicular bisector:** Using the point-slope form $y - y_M = m_{\perp}(x - x_M)$ with $M(2, 1)$ and $m_{\perp}=1$: $y - 1 = 1(x - 2)$ $y = x - 2 + 1$ $y = x - 1$. **6. Find the intersection of the perpendicular bisector and the given line:** Point $P$ lies on both the given line ($y = 2x-1$) and the perpendicular bisector ($y = x-1$). So, we equate their $y$ values: $2x - 1 = x - 1$ $2x - x = -1 + 1$ $x = 0$ **7. Find the y-coordinate of P:** Substitute $x=0$ into either equation (e.g., $y = x-1$): $y = 0 - 1 = -1$. Thus, the coordinates of point $P$ are $(0, -1)$. Both methods yield the same result. **Checking the options:** A) $(0, -1)$: Lies on $y=2x-1$ ($-1 = 2(0)-1$, True). $PA = \sqrt{(0-1)^2 + (-1-2)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$. $PB = \sqrt{(0-3)^2 + (-1-0)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$. Equidistant. This is the correct option. B) $(1, 1)$: Lies on $y=2x-1$ ($1 = 2(1)-1$, True). $PA = \sqrt{(1-1)^2 + (1-2)^2} = \sqrt{0^2 + (-1)^2} = 1$. $PB = \sqrt{(1-3)^2 + (1-0)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4+1} = \sqrt{5}$. Not equidistant. C) $(-1, -3)$: Lies on $y=2x-1$ ($-3 = 2(-1)-1$, True). $PA = \sqrt{(-1-1)^2 + (-3-2)^2} = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4+25} = \sqrt{29}$. $PB = \sqrt{(-1-3)^2 + (-3-0)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$. Not equidistant. D) $(2, 3)$: Lies on $y=2x-1$ ($3 = 2(2)-1$, True). $PA = \sqrt{(2-1)^2 + (3-2)^2} = \sqrt{1^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$. $PB = \sqrt{(2-3)^2 + (3-0)^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1+9} = \sqrt{10}$. Not equidistant. The final answer is $\boxed{(0, -1)}$.