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Find an equation of the line satisfying the given conditions

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

Sagot :

Cam943idn
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.
Lilithidn
[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]

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