Find an equation of the line satisfying the given conditions

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

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 $$3x+5y=38$$ into $$y=mx+b$$ form. Lets do it!

$$3x+5y=38$$
= $$5y = -3x+38$$ (Subtract 3x from both sides)
= $$y= -\frac{3}{5}x+ \frac{38}{5}$$ (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 $$\frac{5}{3}$$.

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

$$\frac{y-y _{1} }{x- x_{1} }=m$$

Substituting values, we find that:

$$\frac{y-4}{x-6}= \frac{5}{3}$$

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

$$y= \frac{5}{3}x-6$$, which is our final answer.

Thank you, and I wish you luck.

$$(6,4); 3x + 5y =38 \ subtract \ 3x \ from \ each \ side \\ \\ 5y = -3x + 8 \ divide \ each \term \ by \ 5 \\ \\ y = -\frac{3} {5}x + \frac{38}{5}\\ \\ The \ slope \ is : m _{1} = - \frac{3}{5} \\ \\ If \ m_{1} \ and \ m _{2} \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \ m _{1}*m _{2} = -1$$

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

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

$$4=5 \cdot 2+b\\ \\4=10+b \\ \\b=4-10\\ \\b=-6 \\ \\ y = \frac{5}{3}x -6$$

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