The line L1 has the equation y=3x. The line L2 is perpendicular to L1 and passes through the point P(0,0). Write an equation of L2
The slope must be -1/3, since it is perpendicular to a line with slope equal to 3.
The only line with that slope that goes through the origin is y = -x/3
To find an equation of a line perpendicular to line L1, we need to determine the slope of L1 and then find the negative reciprocal of that slope.
The slope of L1 is given by the coefficient of x in the equation y = 3x, which is 3.
The negative reciprocal of 3 is -1/3.
Since line L2 passes through the point P(0,0), we can use the point-slope form of a line to find the equation of L2:
y - y₁ = m(x - x₁),
where (x₁, y₁) are the coordinates of the given point, and m is the slope of the line.
Substituting (x₁, y₁) = (0,0) and m = -1/3, the equation of L2 is:
y - 0 = (-1/3)(x - 0).
Simplifying:
y = (-1/3)x.
Therefore, the equation of line L2 is y = (-1/3)x.
To find the equation of line L2, which is perpendicular to L1 and passes through point P(0,0), we need to determine the slope of L2.
Since L2 is perpendicular to L1, the slopes of L1 and L2 will be negative reciprocals of each other. The slope of L1 is 3, so the slope of L2 will be -1/3.
Now, we have the slope of L2 (-1/3) and a point on L2 (0,0). We can use the point-slope form of a line to write the equation of L2.
The point-slope form of a line is: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Substituting the values into the point-slope form, we get:
y - 0 = (-1/3)(x - 0)
Simplifying the equation, we have:
y = (-1/3)x
Therefore, the equation of L2 is y = (-1/3)x.