a) You made a mistake about the sign:
b) The slope of the perpendicular line is
I don't understand. I looked at how to do it and used their same method, but I didn't arrive at the same answer as the one provided in the back of the book.
The question is:
Find equations of lines that pass through the point (3,8) and are (a) parallel to and (b) perpendicular to -4x+7y=-5.
The answer in the back of the book for this chapter test question is
OK, originally I messed up with the arithmetic, ugh. But I think I know how, am I right
You multiply the denominator and get 7y =4x-44 and move all the terms on one side...
4x-7y+44 = 0.
There are two questions here, (a) and (b). Which one are you talking about?
Okay, the slope of the equation -4x+ 7y= -5, which is the same as 7y= 4x- 5 or y= (4/7)x- 5/7 is 4/7 and so a line through (3, 8) is y- 8= (4/7)(x- 3)
You are solving (a)?
Sign error: 7(y- 8)= 7y- 56, not 7y+ 56.
Why divide? The solution in the back of the book, that you apparently are trying to duplicate, has no fractions. Just subtract 7y from both sides and add 12 to both sides: -56+12= -44= 4x+ 7y so 4x+7y+ 44= 0.
But what about (b)? A line perpendicular to a line with slope 4/7 has slope -7/4 (so the product is -1). The equation of a line through (3, 8) with slope -7/4 is y- 8= (-7/4)(x- 3). Multiply on both sides by 4 to get 4(y- 8)= -7(x- 3). Multiply that out and simplify.
OK, originally I messed up with the arithmetic, ugh. But I think I know how, am I right
You multiply the denominator and get 7y =4x-44 and move all the terms on one side...
4x-7y+44 = 0.