# Forthe second problem, I needed to find a line that was perp

Forthe second problem, I needed to find a line that was perpendicular anothergiven equation. The given equation was y=-2x-4 and passedthrough (1, 3). The ordered pair arex1 = 1 and y1 = 3.Thedefinition of perpendicular lines is that they must both have the reciprocal slope with opposite signsand intersect each other in a coordinate plane.Againcomparing the given equation to the general formula y= mx + b, m= -2 is the slope and b= -4 is the y-intercept. To fine the perpendicularline, based on the definition, I had to find the opposite reciprocal of -2 which 2. The new slope was 1/2 and the given point was (1, 3). I then used thepoint-slope equation y-y1 = m (x – x1).First,I substituted the new slope and ordered pair into the equation. The newequation looked like this: y – 3 = 1/2 (x -1). I then used the distributionproperty to alter the equation to look like this: y – 3= 1/2x -1/2. The laststep was to add 3 on both sides. The answer was y = ½ x + 5/2. But for the purposeof graphing, I changed 5/2 to 2 ½. The final answer was y = ½ x+ 2 ½.Mathematical steps below:y– y1 = m (x – x1)y– 3 = ½ (x – 1)y – 3 = ½ x – 1/2y = ½ x + 5/2y = ½ x + 2 ½ For this problem, one line rises while the otherfalls. Both lines then intersect each other. The y-intercept is 2 ½ above the originwhile the x-intercept is 5 units tothe left of the origin.

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