Linear Equations Non Vertical Lines Ex: 1, 2, 3, 4, & 5 Parallel and Perpendicular Lines Ex: 6, 7 & 8 General Form Ex: 9 Proof of the product of the slopes of two perpendicular lines is negative one. Using Similar Triangle Using Pythagorean Theorem Equation of a non Vertical Line Let’s P0 =(0,b) be the point of intersection of a non vertical line and the yaxis ( b is usually called the y-intercept). For any two (arbitrary) points P1(x1, y1) & P2 = (x2 , y2) we define P2 the slope of P1P2 as If P0, P1 , P2 are on the same line, then mP0 P1  mP1P2 Proof: Since the blue and the red right triangles are similar as shown in the figure, so the ratios of the corresponding sides are in proportion, therefore: AP1 BP2  AP0 BP1  y1  y0 y2  y1  x1  x0 x2  x1  mP0 P1  mP1P2 y2-y1 P1 x2-x1 B y1-y0 P0 x1-x0 A b This means that the line has an unique slope and we will call it m. If P=(x,y) is an arbitrary point on the line besides P0 then : Any point (x,y) satisfying the equation y = mx+b lies on the (non vertical) line and vice-versa. Slope-intercept equation Return to Table More of non Vertical Lines Let  be a line with slope m, and P1(x1 , y1) a given point Then for any arbitrary point P(x,y) on , the slope m is given by: Multiplying both side by x - x1 , we get the equation of  y – y1 = m (x - x1 ) Point-slope equation Example1: Find m and b for the line 10x – 5y = 4 Solving for y 5y = 10x – 4  y = 2x – 4/5 Comparing with y = mx + b we have m = 2 & b = – 4/5 Example 2: Find the equation of line passing though (5,-2) & (-3,-4). Using slope formula Point-slope equation using Working on the equation Return to Table More Example 3: Find x if (x, -5/4) is a point on the line 8x-20y = 7. Since (x, -5/4) satisfies the equation 8x-20(-5/4) = 7  8x + 25 = 7  x = – 9/4 Example 4: Graph of line y = 2x – 4 We need to find two points on the line. Let’s fill the blanks in the following table x y 0 -4 2 0 2 If x = 0 ; y =2(0) – 4 = – 4 So, A = (0, -4) is on the line. If y=0 ; 0 = 2x – 4  x=2 So, B = (2 , 0) is on the line. -4 Once we plot the points, we can draw the line. Example 5: Find slope of the line as shown in the figure. Pick two points on the line, whose coordinates can read accurately, such as P1= (0,3) & P2 = (4,1). Using P1 & P2 , we get: Return to Table 1 1 Parallel & Perpendicular Lines Example 6: Given the line 3x+5y =30, find the area of OAB in the figure. From figure A = (10,0) & B = (0, 6) Since OAB is a right triangle So, Area = ½( OA )(OB) =½ (10)(6) = 30. Parallel & Perpendicular Lines Let L1 & L2 be two non vertical lines: L1 // L 2  m1 = m2 L1  L2  m1 m2 = -1 Click here to see the proof Example 7: Specify the following lines that are //, or , or neither. L1: L2: L3: L4: y = -2x + 5 4x +2y = 7 5x + 10 y = 3 x – 2y = 8     m1 = - 2 m2 =-2 m3 = -1/2 m4 = 1/2 L1 // L2 L1  L4, & L2  L4 But L1&L3; L2&L3; and L3&L4 are neither. Return to Table More Example 8: Find an equation of the line passing through (15, 4) and is a) // to the line 5y = 3x + 15 b)  to the line 5y = 3x + 15 Solving for y , we get: y = (3/5)x +3  m = 3/5 a) Two parallel lines have slope the same, so m = 3/5 The equation should have the form y = (3/5)x + b Since the line passing through (15,4) 4 =(3/5)15 + b  b = -5 So the required parallel line is y = (3/5)x – 5 b) Perpendicular line has slope –1/m, which is equal to – 1/(3/5) = – 5/3 The equation should have the form y = –(5/3)x + b Since the line passing through (15,4) 4 = –(5/3)15 + b  b = 29 So the perpendicular line is y = –(5/3)x + 29 Return to Table General Equation of a Line The linear equation Ax + By = C (with A & B are not simultaneously zero) is called the general equation of a line (in the Cartesian Plane), because: If B  0, then solving for y we get y = -Ax /B + C/B , that is, y = mx + b with m = -A/B & b= C/B. So the graph is a line. If B = 0, then A  0. So Ax = C, then x = C/A and the graph is a line parallel to the Y axis (called a vertical line ). Note that for vertical lines, their slopes are not defined. So the graph of Ax + By = C (with A & B not simultaneously zero) is a line. Equivalently, every point (x,y) not on the line Ax+By=C satisfying Ax+ By  C, is either in Ax+ By >C or in Ax+By < C. The line divides the plane into two regions and it can be proved that all points in the plane satisfying one of the inequalities are located in the same region. This idea will be crucial to solve (by graphing ) systems of linear inequalities. Return to Table Linear Equations in General or Standard From Example 9: Find an equation of the line with slope – 2/3 and y intercept 2. Write the final answer in standard form. Solution: Use slope-intercept formula: y = mx + b with m = -2/3 & b = 2. We have: y = (-2/3)x + 2 Multiply both sides by 3: 3y = -2x +6 Move – 2x to left side and we have the required equation in standard form: 2x + 3y = 6 Return to Table The product of the slopes of two perpendicular lines is negative one. 1. Prove using similar triangles. Let L1 & L2 be perpendicular lines. For simplicity, we shift the point of intersection of lines in figure A to the origin. We draw the perpendicular line at x=1 and determine the red and the blue right triangles (as shown in figure). Blue and red triangles are similar.  So the ratio of the corresponding sides are in proportion: We form the ratio of the opposite side to the adjacent side with respect to the angle  And we have   y1 1  So y1 | y2 |  1 1 | y2 | The slope of line L1 is m1  y1  y1 1 y2 m   y2 The slope of line L2 is 2 1 y1 x 1   m1 m2  1  m1m2  1 For any x: m1 = = x y2 m2 Therefore the product of the slopes is negative one. Continue with second Proof Return to Table y1  1  |y2| 1 The product of the slopes of two perpendicular lines is negative one. 2. Prove using Pythagorean Theorem. For simplicity, we shift the point of intersection of two perpendicular lines L1 & L2 to the origin. Using the Pythagorean Theorem: On red right triangle: On blue right triangle:  OA  OB 2 2  12  m12  1  m12 But AB  m1  m2 o  12  m22  1  m22  AB    OA   OB 2 On right triangle OAB: A 2 2 . Thus Therefore the product of the slopes is negative one. B Return to Table