fundamental to circles

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<p>Parallel and Perpendicular Lines Angle Between Two Lines Straight Lines</p> <p>Definition 1.6 The angle of inclination of a line or simply an inclination is the smallest angle, greater than or equal to 00, that the line makes with positive direction of the x-axis. If a line is inclined upward to the right, then the inclination E1 is between 00 to 900, that is 00 &lt; E1 &lt; 900. If the line is inclined upward to the left, then 900 &lt; E2 &lt; 1800. The inclination of a horizontal line is 00, that is, E3 = 00. The inclination of a vertical line is 900.</p> <p>L2 L1</p> <p>E2 E1</p> <p>(a)</p> <p>(b)L4</p> <p>L4</p> <p>L3</p> <p>90 9000</p> <p>(c)</p> <p>(d) Figure 10</p> <p>Example 5:1. Draw the line passing through (-2, 4) with E = 1200.</p> <p>1200</p> <p>1200</p> <p>2. What is the inclination E if tan E = -1? The trigonometric function tangent is negative on quadrant two and four. So, either E = 1350 or E = 3150. However inclination should be the smallest angle that the line makes with the positive x-axis. Thus, E must be 1350 .</p> <p>1350</p> <p>L</p> <p>(y E (x E</p> <p>Figure 11</p> <p>Figure 11 shows that the ratio of the differences between the y-coordinates and x-coordinates is actually the steepness of the line. So, the quotient (y/(x is the tangent of the inclination E. Thus, the slope is equal to tangent E.</p> <p>Definition 1.7 The slope of a line, denoted by m, is the tangent of the inclination. That is m = tan E. In Figure 9, observe that tangent of E1 is positive and tangent of E2 is negative. So, the slopes of lines L1 and L2 are positive and negative respectively; while the slopes of a horizontal line is 0 and the slope of a vertical line is undefined since tan 900 is undefined.Theorem 2.1 The slope of a line passing through two points P1(x1, y1)</p> <p>y 2 y1 . and P2(x2, y2) is m = x 2 x1</p> <p>.</p> <p>Example 6: 1. What is the slope of the line whose inclination is 600? Solution: The slope of this line is tangent of 600. So, m =3</p> <p>2. Find the slope of the line that passes through (-2, 4) and (3, -6).y (-2, 4)</p> <p>y (3, -6)</p> <p>Slopes of Parallel and Perpendicular LinesTheorem 1.2 Two nonvertical lines are parallel if and only if their slopes are equal. That is, lines L1 and L2 are parallel if and only if m1 = m2. Theorem .3 Two slant lines are perpendicular if and only if the product of their slopes is -1. That is, lines L1 and L2 are perpendicular if and only if m1 m2 = -1</p> <p>L2</p> <p>L1</p> <p>E2</p> <p>E1</p> <p>E1</p> <p>E2</p> <p>L1 L2 (a) Figure 12 (b)</p> <p>In figure 12 (a), E1 = E2. So tan E1 = tan E2. Thus, m1 = m2.</p> <p>In figure 12 (b) , let m1 = tan E1 and m2 = tan E2 be slopes of L1 and L2 respectively. It can be shown that E2 = 900 + E1. So, tan E2 = tan (900 + E1). Since tan 900 does not exist, take</p> <p>sin( 900 E1) sin 90 0 cos E1 cos 900 sin E1 ta n E 2 = = 0 cos( 90 E1) cos 90 0 cos E1 sin 90 0 sin E1 cos E1 1 = . = sin E1 tan E1</p> <p>Example 8: Prove that the points A(-2, 7), B(5, 4), C(-1, -10) and D(-8, -7) are vertices of the rectangle ABCD.A B</p> <p>D C</p> <p>We must show that the adjacent sides are perpendicular. That is, mABmBC = -1 and mADmDC = -1. We compute the following:.</p> <p>m AB !</p> <p>47 3 ! 52 7</p> <p>mDC !</p> <p>7 8</p> <p>!</p> <p>3 7</p> <p>77 7 ! mAD ! 8 2 3</p> <p>and</p> <p>mBC !</p> <p> 0 4 7 ! 3 5</p> <p>Thus, the result follows. Also, it can be shown that opposite sides are parallel. However, that is enough to show that the figure is already a rectangle (it may be any parallelogram). It must be shown that any of the two adjacent sides are perpendicular.</p> <p>,</p> <p>.</p> <p>2. Draw the triangle with the vertices A(-1, 1), B(6, -2) and C(4, 3) and show that the triangle is a right triangle.C</p> <p>A</p> <p>B</p> <p>Finding the slopes of the sides yield the following:mAB !2 3 ! 6 7m BC ! 3 2 5 ! 4 6 2</p> <p>and</p> <p>m AC !</p> <p>3 4</p> <p>!</p> <p>2 5</p> <p>Since the slopes of the sides BC and AC are negative reciprocal to each other, these sides must be perpendicular. Hence the triangle must be a right triangle.</p> <p>Angle Between Two Lines:Let J be an angle between two intersecting lines L1 and L2. Suppose E1 and E2 are the inclinations of L1 and L2 respectively. Refer to Figure 13.L2 L1 J</p> <p>E1</p> <p>E2</p> <p>Figure 13</p> <p>It can be checked that J + E1 = E2. So, J = E2 - E1 and tan J = tan (E2 - E1). Thus, tan J =tanE2 tanE1 . 1 tanE2 tanE1</p> <p>But tan E1 is the slope of L1 and tan E2 is the slope of L2. Hence, tan J =m 2 m1 , 1 m 2m 1</p> <p>where J is obtained by taking the inverse tangent of both sides. It must be noted that m1 is the slope of the initial line and m2 is the slope of the terminal side.</p> <p>Definition 1.8 If m1 and m2 are the slopes of lines L1 and L2 respectively, then the angle J from L1 to L2 is obtained from the equation tan J =m 2 m1 1 m 2m 1</p> <p>.</p> <p>Example 9: 1.The slope of one line is 3/2. Find the slope of the other line if the angle between the two lines is 1350. Solution: From Definition 1.8, either tan 1350 =m 1 3 2 1 </p> <p> m 1</p> <p>or tan 1350 =</p> <p>3 2</p> <p>3 m 1 2</p> <p>1 </p> <p>3 m 1 2</p> <p>.</p> <p>L1 L2</p> <p>L2</p> <p>L1</p> <p>1350</p> <p>1350</p> <p>2. Find all the angles of a triangle whose vertices are at (-4, 2), (3, 3), and (2, -3).L2 (-4, 2) (3, 3)</p> <p>L1 L3</p> <p>(2, -3)</p> <p>The slopes of L1, L2, and L3 are m1 = 6, m2 = 1/7 and m3 = -5/6 respectively.</p> <p>Let J1 be the angle f om L2 to L1, J2 be the angle f om L1 to L3, and J3 be the angle f om L3 to L2.</p> <p>1 1 7 ! and By Definition 1.8, tan J1 = 13 1 1 6 7 5 6 4 6 . Also, J3 = 1800 (J 1 + J2). tan J2 = ! 24 5 6 6 6Thus, J1 = tan-1 J2 = tan-1</p> <p> 1 = 2.40 13 </p> <p>,</p> <p> 41 = 59. 0, and J3 = 1800 ( 2.40 + 59. 0) = 49.90. 24 </p> <p>Example: Page 19 A television camera is located along the 40-yd line at a football game. If the camera is 20 yds back from the sideline, through what angle should it be able to pan in order to cover the entire field of play, including the end zones, which are 10 yds deep?</p> <p>Area of a Triangle Given the VerticesLet A(x1, y1), B(x2, y2), and C(x3, y3) be the vertices of a triangle. Figure below shows that the area of the triangle ABC is equal to the sum of the areas of the trapezoids ADEC and CEFB minus the area of the trapezoid ADFB. The area of a trapezoid is one-half the sum of the parallel sides times the altitude. So,C</p> <p>B A</p> <p>D</p> <p>E</p> <p>F</p> <p>AADEC =</p> <p>1 1 AD CE DE = ? 1 y 3 A x 3 x1 ) , y ( 2 2 1 1 CE BF EF = ? 3 y 2 A x 2 x 3 ) , y ( ACEFB = 2 2</p> <p>?</p> <p>A</p> <p>?</p> <p>A</p> <p>AADFB =</p> <p>1 1 AD BF DF = ? 1 y 2 A x 2 x1 ) , y ( 2 2</p> <p>?</p> <p>A</p> <p>Thus, AABC =</p> <p>1 ?y1 y3 A(x3 x1 ) + 1 ?y3 y2 A(x2 x3 ) 2 2 1 ?y1 y2 A(x1 x2 ) 2</p> <p>which i al o q al to th</p> <p>t</p> <p>mi a t</p> <p>AABC</p> <p>x x x3</p> <p>y y y3</p> <p>Eq ival</p> <p>tly w ca w it</p> <p>AABC a cho i a co</p> <p>x yt</p> <p>x y</p> <p>x3 y3</p> <p>x yo</p> <p>wh</p> <p>th poi t</p> <p>clockwi</p> <p>Example: Find the area of the triangle with vertices (2, 2), (-4, -1), and (6, -5).(2, 2)</p> <p>(-4, -1)</p> <p>( , -5)</p> <p>S</p> <p>i</p> <p>:</p> <p>2 2 1 1 4 1 1 2 5 127 are nit .</p> <p>1 2 12 20 6 8 10 2</p> <p>Assignment: (p22 fuller) 54. A television camera is 30 ft from the sideline of a basketball court 94 ft long. The camera is located 7 ft from the midcourt. Through what angle must it sweep in order to cover all action on the court? 55. A bridge is trussed as shown below. Find the slopes and inclinations of the sections AB and BC.</p> <p>The Straight Lines</p> <p>The Straight Lines Theorem 2.1 The equation of every straight line is expressible in terms of the first degree. Conversely, the graph of a first-degree equation is a straight line.</p> <p>Equations of LinesVertical Lines and Horizontal Lines</p> <p>x=a</p> <p>y=b</p> <p>y b</p> <p>ya</p> <p>Standard Forms of the Equation of a Line 1. Slope-Intercept Form Let m be the slope and b be the y-intercept of the line. Using slope given two points (0, b) and (x, y), we get yb m= .</p> <p>y (x, y)</p> <p>x0</p> <p>y (0, b)</p> <p>Thus, mx = y b. So, y = mx + b. This the slopeintercept form of a line.</p> <p>Example: Find the equation of the line with slope -1/2 and yintercept 4.Solution: Let m we get</p> <p>/ and b</p> <p>4. Substitute to the</p> <p>m</p> <p>b</p> <p>!</p> <p> 4.y is the e uation of the line.</p> <p>Simplifying</p> <p>y b=4</p> <p>x + 2y 8 = 0</p> <p>2.</p> <p>What are the slope and y-intercept of the line whose equation is 5x 3y = -9.</p> <p>Solution: Transform the given equation to slope-intercept form we have 3y = 5x + 9 and so, y = and the y-int is b = .5x 3y = -9</p> <p>5 x 3 . Thus, the slope m = 3</p> <p>m=</p> <p>5 3</p> <p>y b=3</p> <p>2. Point-Slope FormLet P(x1, y1) be a point on the line and let m be the slope of the line.The slope of the line is</p> <p>y y1 m= x x1y (x, y)</p> <p>y P (x1, y1)</p> <p>Simplifying, y y1 = m(x x1). This is the Point-Slope Form of a line</p> <p>Example: What is the equation of the line that passes through the points (-1, 4) and (3, -6)?T(-1, 4)</p> <p>s p m</p> <p>t</p> <p>s</p> <p>46 ! . 1 2p t( , ), w t s</p> <p>Tak t</p> <p>yyy (3, -6)</p> <p>!3</p> <p>( ) w c0.</p> <p>3. Intercept Form Let a and b be the x and y intercept of the line respectively.</p> <p>The slope of the line is m=y (0, b)</p> <p>b0 b ! . 0a a</p> <p>Using point-intercept form,</p> <p>b y b ! x, aay ab = -bx, bx + ay = ab. Divide both sides by ab,</p> <p>(a, 0) y</p> <p>x y ! 1. a bThis is the Intercept Form of line.</p> <p>Directed Distance of a Line to a PointTheorem 2.3: The directed distance from the slant line Ax+By+C=0 to the point P1(x1,y1) is given by the formula</p> <p>d!</p> <p>Ax1 By1 C s A B2 2</p> <p>Where the denominator is given the sign of B. The distance is positive if the point P is above the line, and negative if P is below the line.</p> <p>Example1:Find the distance from the line 5x=12y+26 to the points P1(3,-5), P2(-4,1) and P3(9,0).</p> <p>Example2:Find the distance between the parallel lines 15x+8y+68=0 and 15x+8y-51=0.</p> <p>Example3:Find the equation of the bisector of the pair of acute angles formed by the line x-2y+1=0 and x+3y-3=0.</p> <p>CIRCLES</p> <p>Definition 3.1 A circle is a set of all points in a plane equidistant from a fixed point called the center. The constant equal distance is called the radius.</p> <p>Circle with Center at the Origin:</p> <p>If the center of the circle is at the origin and if the point (x, y) is any point on the circle, then by distance formula,(x 0)2 (y 0)2 ! r , hich isx2 y2 ! r2T is is the e ation of the circle ith center at the origin. If the radi s is 1, it is called a unit circle.</p> <p>r</p> <p>y (x, y)</p> <p>Circle with Center at C(h, k) Let the point (h, k) be the center of a circle and let r be its radius. By distance formula, (x h)2 + (y k)2 = r2. This is the center-radius form of an equation of a circle. This is also called the standard form of the equation.</p> <p>r</p> <p>y P(x, y)</p> <p>y C(h, k)</p> <p>Example:1. Find an equation of the circle where the endpoints of the diameter are at (-2, 4) and (5, 2).Solution: Solving for the distance between the two points will give d=(2 5)2 (4 2)2 ! 5 , which is the diameter of the circle.</p> <p>53 . To find the center of the circle, we need to get the 2 midpoint of the diameter. Thus, the center is at (3/2, 3). If follows that the equation is (x 3/2)2 + (y 3)2 = 53/4.</p> <p>So, the radius is r =</p> <p>Simplifying, x2 + y2 3x 6y 2 = 0</p> <p>(-2, 4)</p> <p>(3/2, 3) (5, 2)</p> <p>2. Find an equation of the circle with center at (-3,-4) and tangent to the ySolution: axis. Since the center is already given, we must be able to find the radius. Note that the circle is tangent to the y-axis. It means that the center is 3 units from the left of the y-axis. Thus, the radius is 3. Hence, the equation is (x + 3)2 + (y + 4)2 = 9. Simplifying, x2 + y2 + 6x + 8y + 16 = 0</p> <p>(-3, -4)</p> <p>3. Find an equation of the circle with center at (4, -1) and tangent to the line x Solution: 2y = 2. radius of the circle is the distance of the line x 2y = 2 from Thethe point (4 -1). It follows that 1(4) 2(1) 2 4 4 ! and so r = . d= 5 5 12 (2)2 The equation of the circle is (x - 4)2 + (y + 1)2 = 16/</p> <p>which is also x2 + y2 8x + 2y + 69/5 = 0.</p> <p>(4, 1)</p> <p>General Form of an Equation of a Circle x2 + y2 + dx + ey + f = 0 where d, e, and f are constants.Example: Reduce the equation to center-radius form and draw the circle. 4x2 + 4y2 + 24x 4y + 1 = 0Fir t, ivi e tTr s s s</p> <p>i ey</p> <p>y . We et,y .</p> <p>4</p> <p>! 0.</p> <p>y</p> <p>T i i eq</p> <p>lt</p> <p>2</p> <p>1 y - 2 </p> <p>2</p> <p>! 9 w ere t e</p> <p>t 3, </p> <p>1 2</p> <p>d r di</p> <p>i 3.</p> <p>rigy</p> <p>sid4!</p> <p>d y49</p> <p>m l i g4</p> <p>s</p> <p>r s,</p> <p>e ter i</p> <p>(-3, )</p> <p>3. Find an equation of the circle passing through the points (-1,2), (1,1) and (3,2).</p>


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