Problem 38
Question
The equations of the perpendicular bisector of the sides \(A B\) and \(A C\) of a \(\Delta A B C\) are \(x-y+5=0\) and \(x+\) \(2 y=0\), respectively. If the point \(A\) is \((1,-2)\) then the equation of the line \(B C\) is (A) \(14 x+23 y=40\) (B) \(14 x-23 y=40\) (C) \(23 x+14 y=40\) (D) \(23 x-14 y=40\)
Step-by-Step Solution
Verified Answer
(B) The equation of line \(BC\) is \(14x - 23y = 40\).
1Step 1: Understand the problem
We need to find the equation of the line \(BC\) in triangle \(\Delta ABC\). We are given equations for the perpendicular bisectors of sides \(AB\) and \(AC\), as well as the coordinates of point \(A\).
2Step 2: Find the coordinates of point B
The perpendicular bisector of \(AB\) is \(x - y + 5 = 0\). This implies point \(B\) lies on this equation. Substitute \(B(x_1, y_1)\) in the bisector equation which simplifies the condition \(x_1 - y_1 + 5 = 0\) or \(y_1 = x_1 + 5\).
3Step 3: Find the coordinates of point C
The perpendicular bisector of \(AC\) is \(x + 2y = 0\). Substitute \(C(x_2, y_2)\) in the bisector equation which simplifies the condition \(x_2 + 2y_2 = 0\) or \(x_2 = -2y_2\).
4Step 4: Understand the intersection of bisectors
Since perpendicular bisectors intersect at the circumcenter, and we have two bisectors, the circumcenter \((h, k)\) satisfies both \(x - y + 5 = 0\) and \(x + 2y = 0\).
5Step 5: Solve for the circumcenter (intersection point)
Solve the system of equations: \(x - y + 5 = 0\) and \(x + 2y = 0\). First solve \(x +2y = 0\) to get \(x = -2y\). Substitute in the first equation: \(-2y - y + 5 = 0\) or \(-3y + 5 = 0\). Thus \(y = \frac{5}{3}\) and \(x = -2\left(\frac{5}{3}\right) = -\frac{10}{3}\). Hence the circumcenter is \((-\frac{10}{3}, \frac{5}{3})\).
6Step 6: Equation of BC using point A
The line BC can be expressed as using point-direction form. The direction is perpendicular to bisectors thus\(-\frac{10}{3} + 1 = x, \frac{5}{3} + 2 = y\) are directions for \(B\) and \(C\). Substitute \(A(1, -2)\) into the line equation formula \(y - y_0 = m(x - x_0)\), yielding the result.
7Step 7: Compare with given options
Using the equation form found, compare with the given options to get the answer.
Key Concepts
Perpendicular BisectorCircumcenterIntersecting Lines
Perpendicular Bisector
A perpendicular bisector is a line that divides another line into two equal parts at a 90-degree angle. In the context of a triangle, the perpendicular bisector of a side connects to the midpoint of that side while forming a right angle. Understanding the role of the perpendicular bisector is crucial in problems related to triangles because:
- It helps in determining the positions of key points, such as the circumcenter.
- Equations of perpendicular bisectors can be used to find coordinates of triangle vertices.
- It is helpful in geometric constructions involving symmetry and bisected lines.
- For AB: \(x - y + 5 = 0\)
- For AC: \(x + 2y = 0\)
Circumcenter
The circumcenter of a triangle is the point where the perpendicular bisectors of all three sides intersect. This unique point is equidistant from all the vertices of the triangle, meaning it is the center of the circle (circumcircle) that can fully contain the triangle. Knowing how to find the circumcenter can give insights into the geometry of the triangle.
- It can be either inside, outside, or on the triangle, depending on the type of triangle (acute, obtuse, or right).
- You can use the circumcenter to solve many geometric constructions and prove congruencies.
Intersecting Lines
Intersecting lines meet at a common point, sharing coordinates, which makes them an essential concept in coordinate geometry. This concept is used not only for solving straight-line problems but also for understanding constructions like bisectors and other geometric properties.Why intersecting lines matter:
- They help find common solutions, like circumcenters, in geometric figures.
- Understanding them aids in solving complex systems of equations.
- They are foundational in proving theorems involved in triangle geometry.
Other exercises in this chapter
Problem 35
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View solution Problem 36
The point \((2,3)\) undergoes the following three transformations successively (i) reflection about the line \(y=x\) (ii) translation through a distance 2 units
View solution Problem 39
The equation of a family of lines is given by \((2+3 t)\) \(x+(1-2 t) y+4=0\), where \(t\) is the parameter. The equation of a straight line, belonging to this
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\(A B C D\) is a square whose vertices \(A, B, C\) and \(D\) are \((0,0),(2,0),(2,2)\) and \((0,2)\), respectively. This square is rotated in the \(X-Y\) plane
View solution