Plans for a stadium are drawn on a coordinate grid. One wall lies on the line y = 3x + 2. A perpendicular wall passes through the point (6, -8. Write the equation of the line that contains the new wall
y=-1/3x-6
y=3x-26
y=1/3x-10
y=-3x-14
To find the equation of a line that is perpendicular to the line y = 3x + 2 and passes through the point (6, -8), we can first find the slope of the given line. The slope of a line in the form y = mx + b is equal to m. Therefore, for the line y = 3x + 2, the slope is 3.
A line that is perpendicular to another line has a slope that is the negative reciprocal of the slope of the given line. Therefore, the slope of the new line is -1/3.
Using the point-slope form of a linear equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute the values into the equation:
y - (-8) = (-1/3)(x - 6)
y + 8 = (-1/3)x + 2
y = (-1/3)x + 2 - 8
y = (-1/3)x - 6
Therefore, the equation of the line that contains the new wall is y = (-1/3)x - 6.
To find the equation of the line that contains the new wall, we need to know that the new wall is perpendicular to the line y = 3x + 2.
First, we can determine the slope of the line y = 3x + 2. Since this equation is in the form y = mx + b, where m represents the slope, we can see that the slope is 3.
Because the new wall is perpendicular to the given line, we know that the product of their slopes should be -1.
The slope of the new wall, let's call it m2, can be found using the formula m1 * m2 = -1. Substituting the known values, we have:
3 * m2 = -1.
Now let's solve for m2.
Dividing both sides of the equation by 3 gives us m2 = -1/3.
So, the slope of the new wall is -1/3.
Next, we need to find the equation of the line that passes through the point (6, -8) and has a slope of -1/3.
Using 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, we can substitute the values:
y - (-8) = -1/3(x - 6).
Simplifying the equation gives us:
y + 8 = -1/3x + 2.
To isolate y, we subtract 8 from both sides:
y = -1/3x + 2 - 8.
Combining like terms gives us:
y = -1/3x - 6.
So, the equation of the line that contains the new wall is y = -1/3x - 6.
To find the equation of the line perpendicular to y = 3x + 2 and passing through the point (6, -8), we need to follow these steps:
Step 1: Determine the slope of the given line
The equation y = 3x + 2 is in slope-intercept form (y = mx + b), where m represents the slope. In this case, the slope of the given line is 3.
Step 2: Find the negative reciprocal of the slope
To find the slope of a line perpendicular to the given line, we need to take the negative reciprocal of the slope. The negative reciprocal of 3 is -1/3.
Step 3: Use the slope and the given point to write the equation.
Now we have the slope (-1/3) and the point (6, -8), we can use the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the given point. Substituting the values, we get:
y - (-8) = -1/3(x - 6)
y + 8 = -1/3(x - 6)
Step 4: Simplify the equation
To simplify, we distribute -1/3 to (x - 6):
y + 8 = -1/3x + 2
Next, move 8 to the other side of the equation:
y = -1/3x + 2 - 8
y = -1/3x - 6
Therefore, the equation of the line that contains the new wall is y = -1/3x - 6.