46.1k views
5 votes
Find the equation of the straight line passing through the point (3,5)

which is perpendicular to the line
y = 3x + 2

User RobSiklos
by
3.6k points

2 Answers

2 votes

Final answer:

The equation of the line that is perpendicular to the line y = 3x + 2 and passes through the point (3,5) is x + 3y = 18.

Step-by-step explanation:

To find the equation of a line that is perpendicular to another line and passes through a given point, we must first determine the slope of the original line and then find the negative reciprocal of that slope for our perpendicular line. The given line has an equation y = 3x + 2, so its slope (m) is 3. The negative reciprocal of this slope is -1/3. This will be the slope of our new line.

With a slope of -1/3 and a point (3,5), we can use the point-slope form to write the equation of our perpendicular line. The point-slope form is given by (y - y1) = m(x - x1), where m is the slope and (x1, y1) is the point the line passes through. Substituting our values into this formula gives us (y - 5) = -1/3(x - 3).

To find the y-intercept of our line, we can simplify and rearrange the equation into slope-intercept form (y = mx + b). Multiplying both sides by 3 to clear the fraction and then distributing gives us 3(y - 5) = -1(x - 3), which simplifies to 3y - 15 = -x + 3. Adding x to both sides and adding 15 to both sides gives us x + 3y = 18.

User Geeehhdaa
by
3.3k points
3 votes

Answer:

Step-by-step explanation:

y = 3x + 2

Slope m = 3

slope of the line perpendicular to this line = -3

Point (3,5)

y- y1 = m(x-x1)

y - 5 = -3(x-3)

y - 5 = -3*x-3 * -3

y - 5 = -3x + 9

y = -3x + 9 +5

y = -3x + 14

User Ralkie
by
2.9k points