How do you find the equation, in standard form, of the line perpendicular to #2x + 3y = -5# and passing through (3, -5)?

Question

Hazel Anglin · Accepted Answer

The first information that we have is that we're looking for a line passing through #(3,-5)#. The family of lines passing through a given point #(x_0,y_0)# is #y-y_0=m(x-x_0)#, where #m# is the slope of the line. So, the lines passing through #(3,-5)# are of the form #y+5 = m_p(x-3)#, where i called the slope #m_p# for perpendicular. To find #m_p#, let's find the slope #m# of the original line before: bringing into standard form the equation #2x+3y=-5#, we get #y=\frac{-2x-5}{3} = -2/3 x - 5/3#. So, the original line has slope #-2/3#.  Now, two lines are perpendicular if their slopes #m# and #m'# are in the relation #m=-1/{m'}#. From this relation, we have #m_p=3/2#, and the solution is thus #y+5 = 3/2 (x-3)# Here's a link where you can check that the two lines are perpendicular, and that the second one passes through #(3,5)#.

Genesis Allen · Answer

undefined To find the equation of a line perpendicular to 2x + 3y = -5, we first need to determine the slope of the given line. The given line is in the form Ax + By = C, where A = 2 and B = 3. So, the slope of the given line is calculated as -A/B, which is -2/3.

Since perpendicular lines have slopes that are negative reciprocals of each other, the slope of the line perpendicular to the given line is the negative reciprocal of -2/3, which is 3/2.

Now that we have the slope of the perpendicular line, we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting (3, -5) for (x1, y1) and 3/2 for m, we get: y - (-5) = (3/2)(x - 3).

Simplify the equation: y + 5 = (3/2)(x - 3).

Now, rewrite the equation in standard form by multiplying both sides by 2 to eliminate the fraction and rearranging terms: 2y + 10 = 3(x - 3).

Expand the right side: 2y + 10 = 3x - 9.

Move all terms to one side to get the standard form: 3x - 2y = 19.

So, the equation of the line perpendicular to 2x + 3y = -5 and passing through (3, -5) in standard form is 3x - 2y = 19.