Question

Write an equation in standard form of a line perpendicular to 5x− 3y = 1, and passing through (−3, 2)

Answer 1

The equation of the line in the standard form is 3x - 5y = 19.

According to the statement

we have to find the equation of line in the standard form.

So, For this purpose, we know that the

The equation of line is an algebraic form of representing the set of points, which together form a line in a coordinate system.

From given information:

The perpendicular line is 5x− 3y = 1.

and

Passing points are (−3, 2).

So, The perpendicular line is:

5x− 3y = 1

Rearrange it then

3y = -5x +1

y = -5/3x +1/3

So,

It is in the form of y = mx + b then the slope m become

m = -5/3.

Then

A perpendicular slope is the opposite sign reciprocal. So, we change the sign and switch the numerator and denominator.

Perpendicular slope = m =3/5.

To find the equation of the new line, use the point slope equation which is

$y - y_{1} = m(x - x_{1} )$

here the the points are (−3, 2) and the m is 3/5.

then substitute the values in it

y - 2 = 3/5(x +3)

5(y-2) = 3(x +3)

5y - 10 = 3x +9

3x - 5y = 9 +10

3x - 5y = 19.

So, The equation of the line in the standard form is 3x - 5y = 19.

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