Question

The equation of three sides of triangle are \(\mathrm{x}=2, \mathrm{y}+1=0\) and \(\mathrm{x}+2 \mathrm{y}=4\). The coordinates of the circumcentre of the triangle is (a) \((4,0)\) (b) \((2,-1)\) (c) \((0,4)\) (d) \((-1,2)\)

Short answer

The circumcenter of the triangle is \((3, 0)\), which is not in the given options.

Step-by-step solution

Find the vertices of the triangle

To find the vertices of the triangle, we need to solve the given equations in pairs. Let's denote the vertices as A, B, and C. For A, we'll solve equations (1) and (2): (1) \(x = 2\) (2) \(y + 1 = 0\) Substituting (1) into (2) gives \(y = -1\). Thus, vertex A is \((2, -1)\). For B, we'll solve equations (2) and (3): (2) \(y + 1 = 0\) (3) \(x + 2y = 4\) Substituting (2) into (3) gives \(x = 4\). Thus, vertex B is \((4, -1)\). For C, we'll solve equations (1) and (3): (1) \(x = 2\) (3) \(x + 2y = 4\) Substituting (1) into (3) gives \(y = 1\). Thus, vertex C is \((2, 1)\). The vertices of the triangle are A(2, -1), B(4, -1), and C(2, 1).

Find the midpoints of the sides of the triangle

To find the midpoints of the sides of the triangle, we use the midpoint formula: \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\). Midpoint of AB: \(M_{AB} = (\frac{2+4}{2}, \frac{-1+-1}{2}) = (3,-1)\) Midpoint of BC: \(M_{BC} = (\frac{4+2}{2}, \frac{-1+1}{2}) = (3, 0)\) Midpoint of AC: \(M_{AC} = (\frac{2+2}{2}, \frac{-1+1}{2}) = (2, 0)\)

Calculate the slopes of the sides and their perpendiculars

To calculate the slope of a side, we use the formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Once we have the slopes of the sides, we can find their perpendicular slopes as the negative reciprocal of the slope. Slope of AB: \(m_{AB} = \frac{-1 - (-1)}{4 - 2} = 0\) Perpendicular slope: \(m_{AB_\perp} = \frac{-1}{0}\) (undefined, it's a vertical line) Slope of BC: \(m_{BC} = \frac{1 - (-1)}{2 - 4} = -1\) Perpendicular slope: \(m_{BC_\perp} = \frac{-1}{-1} = 1\) Slope of AC: \(m_{AC} = \frac{1 - (-1)}{2 - 2} = \mathrm{Undefined}\) Perpendicular slope: \(m_{AC_\perp} = 0\) (horizontal line)

Find the equations of the perpendicular bisectors

We know the midpoints, and the perpendicular slopes. Now, we can use the point-slope form of a linear equation, \(y - y_1 = m(x - x_1)\), to find the equations of the perpendicular bisectors. Equation of perpendicular bisector of AB (using midpoint \(M_{AB}\) and slope \(m_{AB_{\perp}}\)): \(x = 3\) (vertical line) Equation of perpendicular bisector of BC (using midpoint \(M_{BC}\) and slope \(m_{BC_{\perp}}\)): \(y - 0 = 1(x - 3)\) \(y = x - 3\) Equation of perpendicular bisector of AC (using midpoint \(M_{AC}\) and slope \(m_{AC_{\perp}}\)): \(y = 0\) (horizontal line)

Find the intersection of the perpendicular bisectors

The circumcenter is the intersection point of the perpendicular bisectors. We need to find the coordinates of this point by solving the equations of the perpendicular bisectors. Solving the equations of the bisectors of AB and BC: \(x = 3\) \(y = x - 3\) Substituting \(x = 3\) into \(y = x - 3\) gives \(y = 3 - 3 = 0\). Thus, the intersection point is \((3, 0)\), which is the vertex opposite to the midpoint of BC. So, the circumcenter of the triangle is \((3, 0)\). The correct answer is not in the given options.

Key concepts

Coordinate Geometry

Coordinate geometry allows us to analyze geometric problems by using algebraic formulas and coordinate systems. Instead of working purely with angles and distances as in traditional geometry, coordinate geometry utilizes points positioned in a plane, each defined by a set of coordinates. These coordinates typically consist of two numbers: one for the x-coordinate (horizontal position) and one for the y-coordinate (vertical position).

In this exercise, you are given equations of lines in a coordinate plane. These lines form the sides of a triangle, which can be analyzed using the coordinates of their vertices. Coordinate geometry simplifies such problems by providing methods to calculate distances, slopes, and intersection points geometrically, using algebraic techniques.

Equations of Lines

Equations of lines describe how points in a geometric plane are connected. They are often written in a form like Ax + By = C, where A, B, and C are constants. The most straightforward line equation format is the vertical line, such as x = k, or the horizontal line y = k.

In this task, lines are part of the triangle’s sides, specifically given as x=2, y+1=0 (which can be rearranged as y = -1), and x + 2y = 4. Each line intersects with others to define the triangle's vertices. Using line equations, you can find these points of intersections, which are crucial for determining the circumcenter.

Intersection of Lines

When two lines intersect, they share a common point. This point can be found by solving the equations of the lines simultaneously.

In our scenario, the intersections are the vertices of the triangle. To find a vertex such as point A, you solve the equations x=2 and y+1=0. This yields the coordinates (2, -1). Repeating this process with other pairs of equations provides all three vertices of the triangle (points A, B, and C). Knowing how to find these intersections is key in geometry to further explore properties like circumcenters.

Midpoint Formula

The midpoint formula is a straightforward algebraic way to find the point that is exactly halfway between two other points on a line segment. It is calculated using the formula \((\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})\).

For the sides of a triangle, knowing midpoints is crucial, especially when determining the perpendicular bisectors, which leads us to the circumcenter. In this problem, the midpoints of each triangle side are calculated: for example, the midpoint of AB is found to be (3, -1) using this formula. Using midpoints helps establish the necessary perpendicular slopes needed for constructing perpendicular bisectors.