Lecture 4: Lines in the Plane PDF

The slope m of the nonvertical line passing through the points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is 𝑦2 −𝑦1 change in 𝑦 𝑚= = 𝑥2 −𝑥1 change in 𝑥 where 𝑥1 ≠ 𝑥2. Note Example Solution The graphs of the four lines are shown in the following Figures. ...

The slope m of the nonvertical line passing through the points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is 𝑦2 −𝑦1 change in 𝑦 𝑚= = 𝑥2 −𝑥1 change in 𝑥 where 𝑥1 ≠ 𝑥2. Note Example Solution The graphs of the four lines are shown in the following Figures. Point-Slope Form of the Equation of a Line The point-slope form of the equation of the line that passes through the point (𝑥1 , 𝑦1 ) and has a slope of m is y − y1 = m ( x − x1 ). Example Find an equation of the line that passes through 1, −2 and has a slope of 3. Solution Example Find the slope and y-intercept of each line (a) (b) Solution (a) (b) Example Figure. Solution So, the given line has a slope of 𝑚 = 2/3. Because any line parallel to the given line must also have a slope of 2/3, the required line through 2, −1 has the following equation. Example Solution Example Find the equation of the line through the point of intersection of the lines with equations 3𝑥 + 4𝑦 = 8 and 6𝑥 − 10𝑦 = 7 that is perpendicular to the first of these two lines. Solution To find the point of intersection of the two lines, we multiply the first equation by −2 and add it to the second equation −6 x − 8 y = −16 6 x −10 y = 7 − 18 y = −9 1 y= 2  1 Substituting 𝑦 = in either of the original equations yields 𝑥 = 2. The point of 2 1 intersection is 2,. When we solve the first equation for y (to put it in slope- 2 intercept form), we get 3 𝑦= − 𝑥 + 2. 4 4 A line perpendicular to it has slope. 3 The equation of the required line is Example Solution Example Solution  2 c. 2 x + 3 y = 6  3 y = −2 x + 6  y = − x + 2 3 3 k 3 9 m= and = k = 2 3 2 2 Example Solution y = 3(3) − 1 = 8  9 Therefore,