Determine and state an equation of the line perpendicular to 5x-4y=10nand passing through point (5,12).

The equation of line perpendicular to and passing through the point (5,12) is

The line given in the problem is in Standard Form. The standard form of a linear equation is:

Where, if at all possible, A, B, and C are integers, and A is non-negative, and, A, B, and C have no common factors other than 1.

The slope of an equation in standard form is: . Substituting the values from the equation in the problem gives:

Let's call the slope of the line perpendicular to this line . The slope of a perpendicular line is:

We can now use the point-slope formula to find the equation of the line. The point-slope formula states:

Therefore, the equation of line perpendicular to and passing through the point (5,12) is

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4x + 5y = 80

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange 5x - 4y = 10 into this form

Subtract 5x from both sides

- 4y = - 5x + 10 ( divide all terms by - 4 )

y = x + ← in slope- intercept form

with slope m =

Given a line with slope m then the slope of a line perpendicular to it is

= -  = - = - , thus

y = - x + c ← is the partial equation

To find c substitute (5, 12) into the partial equation

12 = - 4 + c ⇒ c = 12 + 4 = 16

y = - x + 16 ← in slope- intercept form

Multiply through by 5

5y = - 4x + 80 ( add 4x to both sides )

4x + 5y = 80 ← in standard form

there are no short passages to look at

$12. that is my answer.