Hi. I am having some issues with this one. Give the equation of the line that is perpendicular to the line y=x+6 and passes through point (0,-4). Thank you so very much!!!
The slope of y = x+6 is 1
so the slope of the perpendicular is -1
new equation:
y = -x + b , but (0,-4) lies on it and it is the y-intercept, so new equation....
y = -x - 4
ok I got it! thank you for your help
To find the equation of a line that is perpendicular to another line, we need to determine the slope of the given line and then find the negative reciprocal of that slope.
The given line is y = x + 6, which is in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope, m, is 1 (since the coefficient of x is 1).
To find the negative reciprocal of the slope, we take the reciprocal of the slope (1/1 = 1) and then multiply it by -1: -1 * 1 = -1.
So, the perpendicular line will have a slope of -1.
Now, we can use the point-slope form of a linear equation to find the equation of the line that passes through the point (0, -4) with a slope of -1.
The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Substituting the values, we have y - (-4) = -1(x - 0).
Simplifying this equation, we get y + 4 = -x.
To put it in slope-intercept form, we rearrange the equation by subtracting x from both sides: x + y + 4 = 0.
Thus, the equation of the line that is perpendicular to y = x + 6 and passes through (0, -4) is x + y + 4 = 0.