Response to classmate`s answer Part 2 Unit

RESPONSE

Responseto classmate’s answer: Part 2

Unit

Question 13

Response

Parallel line

I love the way youintroduced your work by noting the question you were tackling withthe equation being y=x+4 with the ordered pair (-7, 1). Again, youhave clearly indicated the steps that you are making such as what youare solving. This goes on well and you indicate formulas you areusing and for what purpose. You also correctly identify the as 1.Your progression is well up to where you get your equation asy-1=1(x-(-7). Here you simplify further to get y-1=1x=7. Here, Ithink there is typo. I believe you meant to write y-1=1x+7 ory-1=x+7.

Assuming that youhave this, y-1=x+7, you cannot subtract 1 from both sides as you havedone. Instead you have to add one to both sides with the intention ofleaving y alone on one side of the equation. In that case your answershould be y=x+7+1 thus y = x+8. Alternatively you can choose totransfer -1 from one side to the other side of the equals sign whereit will change from -1 to +1 to give y =x+7+1.

On theperpendicular line,

Using the sameequation y=x+4 with the ordered pair (-7,1), you applied the generalform of the pint slope equation as y– y1 =m(x – x1) which is impressive. However, the slope in this case for aperpendicular line must have a slope which is the negative reciprocalof the slope of the other line. In this case, the slope, which is thecoefficient of x is 1. The negative reciprocal of 1 should be -1.However, you have just used 1 in your equation instead of -1.

Your equation shouldthus read y-1=-1x+7+4 instead of what you have as y-1=1x+7+4.

I believe this canbe easily adjusted. Generally, you have to be keen about negative andpositive numbers around equations.