1 more question.... how would I make this equation perpendicular?
The question is....
The line Y=3x-1 and a line perpendicular to it intersect at R(1,2). Determine the equation of the perpendicular line.
Can someone please tell me the steps to making that line perpendicular?
THe slope of lines perpendicular is the negative reciprocal. In this case, the equation of the line perpendicular will be
y= -1/3 x + b (you have to solve for b).
To make the equation of a line perpendicular to another line, you need to understand the concept of slope. The slope of a line measures how steep it is. Two lines are perpendicular if and only if their slopes are negative reciprocals of each other.
In this case, you are given the equation of the line Y = 3x - 1, and you want to find the equation of a line that is perpendicular to it and passes through the point R(1,2).
Step 1: Find the slope of the given line
The given line is in the form y = mx + b, where m represents the slope. By comparing the equation to this form, you can see that the slope of the given line is 3.
Step 2: Find the negative reciprocal of the slope
To find the slope of the perpendicular line, take the negative reciprocal of the slope of the given line. The negative reciprocal of 3 is (-1/3).
Step 3: Use the point-slope form to determine the equation
Next, you can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. In this case, you know that the point R(1,2) is on the perpendicular line, and the slope is (-1/3). Therefore, you can substitute these values into the point-slope form:
y - 2 = (-1/3)(x - 1)
Step 4: Simplify the equation
To simplify, distribute the (-1/3) through the parentheses:
y - 2 = (-1/3)x + 1/3
Finally, rearrange the equation into slope-intercept form (y = mx + b) by adding 2 to both sides:
y = (-1/3)x + 7/3
So, the equation of the perpendicular line is y = (-1/3)x + 7/3.