Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Find an equation of the line satisfying the given conditions

Through (6,4); perpendicular to 3X + 5Y =38

Sagot :

Cam943
To answer this, we will need to know:

• The slope of the equation we are trying to get
• The point it passes through using the 

First, we will need to find the slope of this equation. To find this, we must simplify the equation [tex]3x+5y=38[/tex] into [tex]y=mx+b[/tex] form. Lets do it!

[tex]3x+5y=38[/tex]
= [tex]5y = -3x+38[/tex] (Subtract 3x from both sides)
= [tex]y= -\frac{3}{5}x+ \frac{38}{5} [/tex] (Divide both sides by 5) 

The slope of a line perpendicular would have to multiply with the equation we just changed to equal -1. In other words, it would have to equal the negative reciprocal.

The negative reciprocal of the line given is [tex] \frac{5}{3} [/tex]. 

Now that we know the slope, we have to find out the rest of the equation using the slope formula, which is:

[tex] \frac{y-y _{1} }{x- x_{1} }=m[/tex]

Substituting values, we find that:

[tex] \frac{y-4}{x-6}= \frac{5}{3} [/tex]

By simplifying this equation to slope-intercept form (By cross-multiplying then simplifying), we then get that: 

[tex]y= \frac{5}{3}x-6[/tex] , which is our final answer.

Thank you, and I wish you luck.
Lilith
[tex](6,4); 3x + 5y =38 \ subtract \ 3x \ from \ each \ side \\ \\ 5y = -3x + 8 \ divide \ each \term \ by \ 5 \\ \\ y = -\frac{3} {5}x + \frac{38}{5}\\ \\ The \ slope \ is :m _{1} = - \frac{3}{5} \\ \\ If \ m_{1} \ and \ m _{2} \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \ m _{1}*m _{2} = -1[/tex]

[tex]m _{1} \cdot m _{2} = -1 \\ \\ -\frac{3}{5} \cdot m_{2}=-1 \ \ / \cdot (-\frac{5}{3}) \\ \\ m_{2}=\frac{5}{3}[/tex]

[tex] Now \ your \ equation \ of \ line \ passing \ through \ (6,4) would \ be: \\ \\ y=m_{2}x+b \\ \\4=\frac{5}{\not3^1} \cdot \not 6^2 + b [/tex]

[tex] 4=5 \cdot 2+b\\ \\4=10+b \\ \\b=4-10\\ \\b=-6 \\ \\ y = \frac{5}{3}x -6 [/tex]

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.