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.

Guest May 10, 2020

#1**+1 **

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 = [ m_{1}·m_{2} - 1 + sqrt{ (m_{1}^{2} + 1)·(m_{2}^{2} + 1) } ] / ( m_{1} + m_{2} )

For this problem: m_{1} = 13/9 and m_{2} = 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.

geno3141 May 10, 2020

#2**+1 **

For the second question: Find the largest x-value for the equation: x^{2} + y^{2} = 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.

x^{2} + y^{2} = 4x + 4y. ---> (x^{2} - 4x ) + (y^{2} - 4y ) = 0

complete the square ---> (x^{2} - 4x + 4) + (y^{2} - 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)**

geno3141 May 10, 2020