I had a few questions I needed help with but saw someone had already posted it, so I am re-posting it because I need help with the same questions.
Here's the link: https://web2.0calc.com/questions/please-help-me-with-these-question-i-am-stuck-and
Thank you for your time.
+23245 Given two lines, one with a slope of 13/9 and the other with a slope of 1/3, find the slope of the bisector.
According to Wolfram (a very authoritative site), to find this slope, use this formula:
slope = [ m1·m2 - 1 + sqrt{ (m12 + 1)·(m22 + 1) } ] / ( m1 + m2 )
For this problem: m1 = 13/9 and m2 = 1/3.
Putting these numbers into the formula:
numerator = [ (13/9)·(1/3) - 1 + sqrt{ ( (13/9)2 + 1 ) · ( (1/3)2 + 1 ) } = [13/27 - 1 + 50/27 ] = 36/27
denominator = ( 13/9 + 1/3 ) = 16/9
slope = ( 36/27) / ( 16/9 ) = 3/4.
I checked these out using inverse tangent for 13/9, 1/3, and 3/4 and found that they work.
+23245 For the second question: Find the largest x-value for the equation: x2 + y2 = 4x + 4y.
This is the equation of a circle.
I plan to find the center and radius of the circle by completing the square on both x and y.
x2 + y2 = 4x + 4y. ---> (x2 - 4x ) + (y2 - 4y ) = 0
complete the square ---> (x2 - 4x + 4) + (y2 - 4y + 4) = 0 + 4 + 4
(x - 2)2 + (y - 2)2 = 8
This is a circle with center (2, 2) and radius sqrt(8).
So, the largest x-value is a distance of the sqrt(8) to the right of the x-value of the center: 2 + sqrt(8)
