What is an equation of the line that passes through the point (5, 4) and is parallel to
the line 6x – 5y = 15?
Someone please help me

Answer & Step-by-step explanation:

When two lines are parallel, their slopes are the same. To find the slope of the given equation, convert to slope-intercept form:

[tex]y=mx+b\\\\[/tex]

[tex]6x-5y=15[/tex]

Solve for y. Subtract 6x from both sides:

[tex]6x-6x-5y=15-6x\\\\-5y=-6x+15[/tex]

Divide both sides by -5:

[tex]\frac{-5y}{-5} =\frac{-6x}{-5} +\frac{15}{-5} \\\\y=\frac{6}{5}x-3[/tex]

The slope is [tex]\frac{6}{5}[/tex].

Now you need to make an equation in point-slope form:

[tex]y-y_{1}=m(x-x_{1})[/tex]

Insert known values:

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

You can use this equation, or you can convert to slope-intercept form by solving for y:

Use the distributive property:

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

Add 4 to both sides:

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

:Done

The equation of the perpendicular line is [tex]y = \frac 65x - 2[/tex]

The equation of the line is given as:

[tex]6x -5y = 15[/tex]

Start by calculating the slope (m) of the line as follows:

[tex]6x -5y = 15[/tex]

Subtract 6x from both sides

[tex]-5y = 15 - 6x[/tex]

Divide both sides of the equations by -5

[tex]y = -3 + \frac65x[/tex]

Rewrite as:

[tex]y = \frac65x-3[/tex]

For a linear equation, y = mx + b, the slope of the line is m

So, the slope is:

[tex]m = \frac65[/tex]

The equation of the line is said to be parallel to the above equation

So, the slope of the line is:

[tex]m = \frac65[/tex]

The line is said to pass through point (5,4).

So, the equation of the line is:

[tex]y = m(x - x_1) +y_1[/tex]

This gives

[tex]y = \frac 65(x - 5) +4[/tex]

Expand

[tex]y = \frac 65x - 6 +4[/tex]

[tex]y = \frac 65x - 2[/tex]

Hence, the equation of the line is [tex]y = \frac 65x - 2[/tex]

Read more about line equations at:

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