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A line passes through the origin and has a slope of -3 what is the equation of the line that is perpendicular to the first line and passes through the point (3,4)?

Respuesta :

luisejr77

Answer: [tex]y=\frac{1}{3}x+3[/tex]

Step-by-step explanation:

The equation of the line is Slope-intercept form is:

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

Where "m" is the slope and "b" the y-intercept.

The slopes of perpendicular lines are negative reciprocal.

Then, if the slope of the first line is -3, the slope of the other line must be:

[tex]m=\frac{1}{3}[/tex]

Substitute the point (3,4) into the equation and solve for b:

[tex]4=\frac{1}{3}(3)+b\\ 4-1=b\\b=3[/tex]

Then the equation of this line is:

[tex]y=\frac{1}{3}x+3[/tex]

kudzordzifrancis

ANSWER

[tex]y= \frac{1}{3} x + 3[/tex]

EXPLANATION

We want to find the equation of a line which is perpendicular to another line with slope -3 and passes through (3,4).

Our line of interest is has a slope that is the negative reciprocal of -3

[tex]m = - \frac{1}{ - 3} = \frac{1}{3} [/tex]

The equation is given by

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

We substitute the point and slope to get:

[tex]y-4= \frac{1}{3} (x-3)[/tex]

Expand

[tex]y-4= \frac{1}{3} x-1[/tex]

[tex]y= \frac{1}{3} x-1 + 4[/tex]

[tex]y= \frac{1}{3} x + 3[/tex]