For the next three (3) items that follow:
What is the circumcenter of the triangle ABC?
(-2, -2)
The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. It is also the center of the circumscribed circle, which passes through all three vertices of the triangle. A key property of the circumcenter is that it is equidistant from each vertex of the triangle.
Let the coordinates of the vertices of triangle ABC be A(-2, 3), B(2, 1), and C(1, 2). Let the circumcenter be denoted by P(x, y).
Since the circumcenter P(x, y) is equidistant from A, B, and C, the distance from P to A is equal to the distance from P to B, and the distance from P to C. Mathematically, this means PA = PB = PC.
To avoid dealing with square roots, we can work with the square of the distances:
\(PA^2 = PB^2\) and \(PA^2 = PC^2\) (or \(PB^2 = PC^2\)).
Using the distance formula, the square of the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \((x_2 - x_1)^2 + (y_2 - y_1)^2\). Applying this for our vertices and the circumcenter P(x, y):
Now we set up equations using the equidistant property:
\((x+2)^2 + (y-3)^2 = (x-2)^2 + (y-1)^2\)
Expand both sides:
\((x^2 + 4x + 4) + (y^2 - 6y + 9) = (x^2 - 4x + 4) + (y^2 - 2y + 1)\)
\(x^2 + y^2 + 4x - 6y + 13 = x^2 + y^2 - 4x - 2y + 5\)
Subtract \(x^2 + y^2\) from both sides and move terms to one side:
\(4x - 6y + 13 = -4x - 2y + 5\)
\(4x + 4x - 6y + 2y + 13 - 5 = 0\)
\(8x - 4y + 8 = 0\)
Divide the equation by 4:
\(2x - y + 2 = 0\) (Equation 1)
\((x+2)^2 + (y-3)^2 = (x-1)^2 + (y-2)^2\)
Expand both sides:
\((x^2 + 4x + 4) + (y^2 - 6y + 9) = (x^2 - 2x + 1) + (y^2 - 4y + 4)\)
\(x^2 + y^2 + 4x - 6y + 13 = x^2 + y^2 - 2x - 4y + 5\)
Subtract \(x^2 + y^2\) from both sides and move terms to one side:
\(4x - 6y + 13 = -2x - 4y + 5\)
\(4x + 2x - 6y + 4y + 13 - 5 = 0\)
\(6x - 2y + 8 = 0\)
Divide the equation by 2:
\(3x - y + 4 = 0\) (Equation 2)
We now have a system of two linear equations with two variables x and y:
From Equation 1, we can express y in terms of x:
\(y = 2x + 2\)
Substitute this expression for y into Equation 2:
\(3x - (2x + 2) + 4 = 0\)
\(3x - 2x - 2 + 4 = 0\)
\(x + 2 = 0\)
\(x = -2\)
Now substitute the value of x back into the expression for y:
\(y = 2(-2) + 2\)
\(y = -4 + 2\)
\(y = -2\)
The coordinates of the circumcenter (x, y) are (-2, -2).
| Step | Description |
|---|---|
| 1 | Identify vertices A(-2, 3), B(2, 1), C(1, 2). Assume circumcenter P(x, y). |
| 2 | Use property: PA\(^2\) = PB\(^2\) = PC\(^2\). |
| 3 | Set up equations: (x+2)\(^2\) + (y-3)\(^2\) = (x-2)\(^2\) + (y-1)\(^2\) and (x+2)\(^2\) + (y-3)\(^2\) = (x-1)\(^2\) + (y-2)\(^2\). |
| 4 | Simplify equations to linear form: 2x - y + 2 = 0 and 3x - y + 4 = 0. |
| 5 | Solve the system of linear equations for x and y. |
| 6 | Obtain circumcenter coordinates (-2, -2). |
| Concept | Formula/Property |
|---|---|
| Distance between \((x_1, y_1)\) and \((x_2, y_2)\) | \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) |
| Squared Distance between \((x_1, y_1)\) and \((x_2, y_2)\) | \((x_2 - x_1)^2 + (y_2 - y_1)^2\) |
| Circumcenter Property | Equidistant from all vertices of the triangle. |
| Circumcenter Location | Intersection of perpendicular bisectors of sides. |
Besides the circumcenter, there are several other important centers in a triangle:
For a given triangle, the circumcenter, centroid, and orthocenter are collinear, lying on a line called the Euler line (unless the triangle is equilateral, in which case all four centers coincide).
Let ABC be a triangle. If D(2, 5) and E(5, 9) are the mid-points of the sides AB and AC respectively, then what is the length of the side BC?
Consider the following statements:
1. The third vertex has at least one irrational coordinate.
2. The area is irrational.
Which of the above statements is/are correct?
The difference of coordinates of the third vertex is
What is the centroid of the triangle ABC?
What is the foot of the altitude from the vertex A of the triangle ABC?
What is \(AB + BC\) equal to?
In the given figure, if \(RS = 3\sqrt{3} \text{ cm}\) and \(RPS\) is an equilateral triangle, then find the value of \(QR\)
For a pair of similar triangles shown in the figure, the angles made at $B$ and $R$ are same in both the triangles $BAC$ and $RAQ$. If the ratio of the areas of the two triangles is 2:1, and if the area of $BAC$ is $100\text{ cm}^2$ then what is the area of $RAQ$ in sq.cm.?
