This solution provides a comprehensive, step-by-step approach to determine the area of a triangle formed by the intersection points of a parabola with the coordinate axes, by leveraging the properties of a circle that passes through these same points.

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Problem Statement Analysis

We are given a parabola $y=x^2+\mathrm{p} x-3$. It intersects the coordinate axes at three points, P, Q, and R. This implies:

• One point is the y-intercept (where $x=0$).
• Two points are the x-intercepts (where $y=0$). We are also given a circle C with its center at $(-1,-1)$, and this circle passes through P, Q, and R. Our goal is to find the area of $\triangle PQR$.

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1. Finding the Y-intercept: Point P

Concept: A point where a curve intersects the y-axis always has an x-coordinate of $0$.

Step-by-step working:

1. Substitute $x=0$ into the parabola's equation. The equation of the parabola is $y = x^2 + px - 3$. $y = (0)^2 + p(0) - 3$
2. Calculate the corresponding y-value. $y = -3$

Explanation (Why this step is taken): By definition, the y-intercept is the point where the graph crosses the y-axis. All points on the y-axis have an x-coordinate of zero. Substituting $x=0$ into the parabola's equation directly gives us the y-coordinate of this intersection point.

Result: The parabola intersects the y-axis at the point $\mathbf{P = (0, -3)}$. This is one of the vertices of $\triangle PQR$.

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2. Determining the Equation of Circle C

Concept: The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. The radius is the distance from the center to any point on the circle.

Step-by-step working:

1. Identify the given center of the circle. The center of circle C is $C_0 = (-1, -1)$. So, $h = -1$ and $k = -1$.
2. Use point P to find the radius. We know that point P $(0, -3)$ lies on the circle. The distance between the center $C_0(-1, -1)$ and point $P(0, -3)$ will be the radius $r$. We use the distance formula: $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting the coordinates of $C_0$ and $P$: $r = \sqrt{(0 - (-1))^2 + (-3 - (-1))^2}$ $r = \sqrt{(0+1)^2 + (-3+1)^2}$ $r = \sqrt{1^2 + (-2)^2}$ $r = \sqrt{1 + 4}$ $r = \sqrt{5}$
3. Write the equation of the circle. Substitute the center $(h,k) = (-1,-1)$ and radius $r = \sqrt{5}$ into the standard circle equation: $(x - (-1))^2 + (y - (-1))^2 = (\sqrt{5})^2$ $\mathbf{(x + 1)^2 + (y + 1)^2 = 5}$

Explanation (Why these steps are taken): The problem states that the circle passes through P, Q, and R. Since we have already found P, we can use its coordinates along with the given center to calculate the radius. Once the radius and center are known, the full equation of the circle can be determined. This equation is crucial because it will allow us to find the other two points (Q and R) that lie on the coordinate axes.

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3. Finding the X-intercepts: Points Q and R

Concept: A point where a curve intersects the x-axis always has a y-coordinate of $0$. Points Q and R are the x-intercepts of the parabola. Since the circle also passes through Q and R, these points must also be the x-intercepts of the circle.

Step-by-step working:

1. Substitute $y=0$ into the circle's equation. We use the equation of the circle we just found: $(x + 1)^2 + (y + 1)^2 = 5$. $(x + 1)^2 + (0 + 1)^2 = 5$
2. Solve for $x$. $(x + 1)^2 + 1^2 = 5$ $(x + 1)^2 + 1 = 5$ $(x + 1)^2 = 4$
3. Take the square root of both sides. $x + 1 = \pm\sqrt{4}$ $x + 1 = \pm 2$
4. Determine the two possible x-values.
   • Case 1: $x + 1 = 2 \implies x = 1$
   • Case 2: $x + 1 = -2 \implies x = -3$

Explanation (Why these steps are taken): The x-intercepts are points where the graph crosses the x-axis, meaning their y-coordinate is $0$. By substituting $y=0$ into the circle's equation, we find the x-coordinates of the points where the circle intersects the x-axis. These are precisely the points Q and R. Solving the resulting quadratic equation gives us the two distinct x-coordinates.

Result: The x-intercepts are $\mathbf{Q = (-3, 0)}$ and $\mathbf{R = (1, 0)}$. (The assignment of Q and R to specific coordinates doesn't affect the area calculation).

Self-Check (Optional - Finding 'p'): Although not required for the area, we can verify the consistency of our x-intercepts with the original parabola $y=x^2+px-3$. The x-intercepts are the roots of the quadratic equation $x^2+px-3=0$. Using Vieta's formulas:

• Sum of roots: $x_Q + x_R = (-3) + 1 = -2$. From the equation, sum of roots is $-p/1 = -p$. So, $-2 = -p \implies p=2$.
• Product of roots: $x_Q \cdot x_R = (-3) \cdot 1 = -3$. From the equation, product of roots is $-3/1 = -3$. This matches the constant term. This confirms that our x-intercepts are consistent with the parabola's form, and the value of $p=2$ could be deduced if needed.

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4. Calculating the Area of $\triangle PQR$

Concept: The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be calculated using the determinant formula: $\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$ However, if two vertices lie on a coordinate axis, a simpler formula, $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, can be used.

Step-by-step working:

1. List the vertices of $\triangle PQR$. $P = (0, -3)$ $Q = (-3, 0)$ $R = (1, 0)$

2. Identify the base of the triangle. Notice that points Q and R both have a y-coordinate of $0$, meaning they lie on the x-axis. We can choose the segment QR as the base of the triangle. Calculate the length of the base QR: $\text{Base (QR)} = |x_R - x_Q| = |1 - (-3)| = |1 + 3| = 4 \text{ units}$

3. Identify the height of the triangle. The height of the triangle, with base QR on the x-axis, is the perpendicular distance from the third vertex P to the x-axis. This distance is the absolute value of the y-coordinate of P. $\text{Height} = |y_P| = |-3| = 3 \text{ units}$

4. Calculate the area using the base-height formula. $\text{Area of } \triangle PQR = \frac{1}{2} \times \text{Base} \times \text{Height}$ $\text{Area} = \frac{1}{2} \times 4 \times 3$ $\text{Area} = \frac{1}{2} \times 12$ $\mathbf{\text{Area} = 6 \text{ square units}}$

Explanation (Why these steps are taken): Since two vertices (Q and R) lie on the x-axis, the segment connecting them forms a horizontal base. The height of the triangle relative to this base is simply the perpendicular distance from the third vertex (P) to the x-axis, which is the absolute value of P's y-coordinate. This method is generally simpler and less prone to calculation errors than the determinant formula when applicable.

Verification (Optional - using the determinant formula): For completeness, let's confirm the area using the general determinant formula: $P=(x_1, y_1) = (0, -3)$ $Q=(x_2, y_2) = (-3, 0)$ $R=(x_3, y_3) = (1, 0)$ $\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$ $\text{Area} = \frac{1}{2} \left| 0(0-0) + (-3)(0-(-3)) + 1(-3-0) \right|$ $\text{Area} = \frac{1}{2} \left| 0 + (-3)(3) + 1(-3) \right|$ $\text{Area} = \frac{1}{2} \left| 0 - 9 - 3 \right|$ $\text{Area} = \frac{1}{2} |-12|$ $\text{Area} = \frac{1}{2} (12) = 6 \text{ square units}$ Both methods yield the same result, confirming our calculation.

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5. Key Concepts Revisited & Tips for Success

• Intersection with Coordinate Axes:
  • Y-intercept: Set $x=0$ in the equation.
  • X-intercepts: Set $y=0$ in the equation.
• Connecting Multiple Curves: The crucial insight in this problem is that the points P, Q, and R are common to both the parabola and the circle. This allows us to use the properties of the simpler curve (the circle, once its equation is known) to find the intersection points that might be harder to determine directly from the more complex curve (the parabola, without knowing 'p').
• Equation of a Circle: Remember that knowing the center and any point on the circle is sufficient to determine its radius and thus its full equation.
• Area of a Triangle:
  • Always check if two vertices lie on a coordinate axis. If so, the $\frac{1}{2} \times \text{base} \times \text{height}$ formula is much more efficient.
  • The base length is the absolute difference of the relevant coordinates.
  • The height is the absolute value of the coordinate of the third vertex perpendicular to the base's axis.
• Common Mistake: Forgetting Absolute Values: Distances and areas are always non-negative. Remember to use absolute values when calculating lengths from coordinate differences (e.g., $|x_2 - x_1|$) or when extracting height from a coordinate (e.g., $|y_P|$).

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6. Key Takeaway

This problem beautifully illustrates how a multi-concept approach can simplify complex geometric problems. By strategically using