Being that my first name starts with “J” The equation which will be used by me is y = -x+ 2 and the parallel line must pass through ⅔ point (9, -3). The slope of the given line is -. So, the ⅔ slope of the parallel line will also be -. We now ⅔ know the slope of the new line and we know that the line passes through (9,-3). I will therefore use the point-slope form of a linear equation to write my new equation. y – y1 = m(x – x1) This is the general form of the point-slope equation y – (-3) = -x [x – (9)] ⅔ Substituting in my known slope and ordered pair y + 3 = -x + (-)-9 ⅔ ⅔ Simplifying double negatives and distributing the slope y = -x +6 – 3 ⅔-3 is subtracted from both sides y = -x + 3 ⅔ Equation of my parallel line! This line falls as one goes from left to right across the graph of it, the y-intercept is -3 units below the origin , and the x-intercept is 4.5 unit to the left of the origin . Now we have to write the equation of the perpendicular line.
Math 221: Week 3- Parallel and PerpendicularAssigned number: y=1/2x+3; (-2, 1) – Parallely=1/2x+3; (-2, 1)-Discuss the steps to solve it.-Answer two questions●What does it mean for one line to be parallel to another?●What does it mean for one line to be perpendicular to another?Five math vocabulary to use in discussion●Origin●Ordered pair●X- or y-intercept●Slope●ReciprocalIn this week discussion, we are going toascertainin small stagesthe equations of lines parallel and perpendicular to the given line which willtoleratea given purpose. (A) The equation that is given in my assigned number 1 isy= 1/2x + 3 The parallel line that will pass through a point: (-2,1)First we have to remember that for parallel, the lines are only parallel if they have the same slope. This given equation will indicate that the slopeof the parallel line is going to ½ and with the ordered pairs(-2, 1), I will use these numbers in the slope intercept form y= mx + b, where m= ½, y =1, and x= -2. y= ½x + 3 ----> 1= ½(-2) + b