Mathematics
Equation of a Perpendicular Bisector
The equation of a perpendicular bisector is a mathematical formula that describes a line that passes through the midpoint of a line segment and is perpendicular to it. It is used to find the locus of points that are equidistant from the endpoints of the line segment. The equation can be derived using the slope-intercept form or the point-slope form.
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4 Key excerpts on "Equation of a Perpendicular Bisector"
eBook - PDF- Robert Bix(Author)
- 2014(Publication Date)
- Academic Press(Publisher)
Proof: Let / and m be the two given lines, and let p and q be the angle bisectors. One pair of vertical angles is divided by p into angles of equal measure 0, and the other pair of vertical angles is divided by q into angles of equal measure y/. The adjacent angles formed by / and m have measures 20 and 2y/, and so we have 20 + 2y/ = 180° (by Theorem 0.6). Dividing by 2 gives 0 + if/ = 90°. Since p and q form the angle 0 + y/, p and q are perpendicular. • We want to prove that the angle bisectors at the vertices of a triangle lie by threes on common points. The proof is analogous to the proof in Section 4 that the perpendicular bisectors of the three sides of a triangle meet at a point. By Theorem 4.2, the perpendicular bisector of two points A and B consists of exactly those points equidistant from A and B. Similarly, we show that the angle bisectors of two lines / and m consist of exactly those points equidistant from / and m. We use the following definition. DEFINITION 5.4. In the Euclidean plane, let P be a point and let / be a line. The foot of the perpendicular from P to / is the point F where / intersects the line on P perpendicular to /. The distance from P to / is the distance between the points P and F (Figure 5.8). • The next result is analogous to Theorem 4.2. THEOREM 5.5. In the Euclidean plane, let / and m be two lines on a point O. Then the angle bisectors consist of exactly those points equidistant from / and m (Figure 5.8). Proof: Let P be a point, and let F and G be the feet of the perpendiculars from P to I and m. First, assume that P doesn't lie on / or m and that neither F nor G equals O (Figure 5.8). P is equidistant from / and m if and only if it is equidistant from F and G (by Definition 5.4). FOP and GOP are right triangles that have a common side OP. Thus, P is equidistant from F and G if and only if triangles FOP and GOP are congruent (by the Pythagorean Theorem 0.9 and the SSS Property 0.1).
- Allen Ma, Amber Kuang(Authors)
- 2022(Publication Date)
- For Dummies(Publisher)
10. AF CF A median of a triangle is a line segment that connects the vertex of a triangle to the midpoint of the opposite side. 11. ∠ ADB and ∠ CDB are right angles. The altitude of a triangle is a segment that connects the vertex of a triangle perpendicular to the opposite side. 12. ∠ ∠ ABE CBE ≅ A bisector divides an angle into two congruent angles. 13. AF CF A midpoint divides a segment into two congruent segments. 14. Right A perpendicular bisector divides a segment into two congruent segments and also forms right angles at the point of intersection. 15. 12.5 A midpoint divides a segment into two congruent segments. Set the two segments equal and solve for x: CE BE x x x 2 25 50 2 2 5 12 5 . 16. 80° A bisector divides an angle into two congruent angles. Set the two angles equal and solve for x: m ADE m CDE x x x x x ∠ ∠ = -= + -= = = 3 5 25 2 5 25 2 30 15 CHAPTER 18 Answers and Explanations 159 ANSWERS 1–100 After finding the value of x, plug it in to find the value of the desired angle: m ADE m ADC m ADE ∠ ∠ ∠ = -= = = ( ) = 3 15 5 40 2 2 40 80 ( ) ( ) 17. 22 Perpendicular lines form right angles. Set the angle equal to 90° and solve for x: m BAD x x x ∠ = -= = = 90 5 2 0 9 0 5 110 22 18. 21 A bisector divides a segment into two congruent segments. This means that BE CE . Find the sum of BE and CE, set it equal to BC, and solve for x: BE CE BC x x x x x x x x ( ) ( ) 12 12 5 3 2 24 5 3 24 3 3 27 3 9 Now plug in the value of x to find BE: BE CE x 12 9 12 21 19. Linear pair A linear pair is a pair of adjacent angles whose sum is a straight angle. 20. Perpendicular Perpendicular lines are two lines that intersect to form right angles. 21. Acute An acute angle is an angle measuring more than 0° but less than 90°. 22. Isosceles An isosceles triangle is a triangle that has two congruent angles with the sides opposite them congruent as well.
eBook - PDFAnalytical Geometry
The Commonwealth and International Library of Science, Technology, Engineering and Liberal Studies: Mathematics Division
- Barry Spain, W. J. Langford, E. A. Maxwell, I. N. Sneddon(Authors)
- 2014(Publication Date)
- Pergamon(Publisher)
Hence the equations of the bisectors of the angles between u x = a 1 x+b 1 y+c 1 = 0 and u 2 ΞΞ a 2 x+b 2 y+c 2 = 0 are given by a 1 x+b 1 y+c 1 , a 2 x+b 2 y+c 2 It must be emphasised that the positive sign does not always correspond to the internal bisector. In fact, if c x c 2 is positive the perpendicular distances from the origin to u x = 0 and u 2 =0 have the same sign and so the positive sign corresponds to the bisector of the angle which contains the origin. If c Y c 2 is negative, we see similarly that the negative sign corresponds to the bisector of the angle which contains the origin. In a numerical example, it is advisable to draw a rough sketch and note the approximate gradients of the internal and external bisectors. Illustration: Obtain the bisectors of the angles between the lines 7x-y+6 = 0 and x+y+2 = 0. A rough sketch shows (Fig. 15) that the gradient of the bisector of the acute angle is large numerically, whilst that of the bisector of the obtuse angle is small. Both bisectors are given by lx-y+6 x+y+2 V(49+l) ^ 1 V(l + 1)· 30 ANALYTICAL GEOMETRY FIG. 15 That is, lx—y+6 = ± 5(x+y+2) and so we obtain the equations x—3y—2 = 0 and 3x+y+4 — 0. The first equation gives the bisector of the obtuse angle. EXAMPLES 54. Obtain the equation of the bisector of the acute angle between the pair of lines: (i) x+2y = 1, 2x +y+3=0; (ii) 3x-4y = 5, -5*+12j> = 2. 55. Obtain the coordinates of the centre of the circle inscribed in the triangle whose vertices are at (—7, —5), (17, 1) and (1,14). Further, calculate the radius of this inscribed circle. 56. Show that the point (3, — 1) is equidistant from the lines 3ΛΓ— Ay— 16=0 and 4χ+3>>-12 = 0. 57. The sides BC, CA and AB of a triangle have the! equations 5x- 2y-26 = 0, 3JC+4J>-10 = 0 and 4*+3j>+10 = 0 respectively. Find the coordinates of the incentre of the triangle and also the centre of the circle escribed to AC. x' 3x+y+4=0 .--_.--~ 7x-y+6=0 x-3y -2=0 .-'-.-.--.-'-x+y+2=0 y' X
eBook - PDF- Doug French(Author)
- 2004(Publication Date)
- Continuum(Publisher)
Many equations of standard curves can be obtained in this way as well as solutions to a variety of locus problems. Two simple examples are illustrated by Figure 10.7. In the first case the problem is to find the locus of a set of points that are equidistant from two given points. This is the perpendicular bisector of the line segment joining the points. Its equation can be found in a number of ways, but it is particularly instructive to find it by equating the distances of a general point denoted by (x,y) from the two given points. Pythagoras' theorem enables us to find expressions for the squares of the two distances. Simplification of the resulting equation then gives the equation of the required line as shown below: Figure 10.6 The properties of a quadrilateral Linking Geometry and AIgebra 25 Figure 10.7 Two locus problems In the second case the locus is the set of points at a fixed distance from a given point. That, of course, is the definition of the circle with the result being the equation of a circle. In the example the points are at a distance of 2 units from the point (3,2) and the simplest form of the equation of a circle is found immediately as a direct application of Pythagoras' theorem: An interesting problem where the nature of the locus is not immediately obvious from geometrical considerations is to find the locus of a point where the distances from two given points are in the ratio 2:1. In Figure 10.8 the points O and A, the origin and (6, 0), are taken as the two points and a general point P is such that OP = 2AP. The problem can be explored initially by constructing a set of concentric circles as shown in the diagram with centres at the points O and A. Points are identified at the intersections of appropriate circles to fulfil the ratio condition. The diagram shows that these points appear to lie on a circle.
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