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Chapter 14: Problem 1277

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

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

Step by step solution

01

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).
02

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: \(M_{AC} = (\frac{2+2}{2}, \frac{-1+1}{2}) = (2, 0)\)
03

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)
04

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)
05

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.

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Most popular questions from this chapter

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