Write an equation of a line that passes through the point (8,2) that is (a) parallel and (b) perpendicular to the graph of the equation
y= 4x +3
Parallel line has same gradient,m.from equation of line,m=4 y-y1=m(x-x1) y-2=4(x-8) y-2=4x-32 y=4x-30 b)m2=-1/m1=-1/4 y-2=-1/4(x-8) 4y-8=-x+8 4y=-x+16
(a) To find a parallel line, we need to use the fact that parallel lines have the same slope. The given equation is in slope-intercept form (y = mx + b), where "m" represents the slope. So, from y = 4x + 3, we know that the slope is 4.
Using the point-slope form of a linear equation (y - y₁ = m(x - x₁)), with (x₁, y₁) being the given point (8, 2), we have:
y - 2 = 4(x - 8)
Expanding the equation:
y - 2 = 4x - 32
Rearranging:
y = 4x - 30
So, the equation of a line that passes through the point (8, 2) and is parallel to y = 4x + 3 is y = 4x - 30.
(b) For a line to be perpendicular to another line, the slopes have to be negative reciprocals of each other. In the given equation (y = 4x + 3), the slope is 4. Therefore, the slope of the perpendicular line would be -1/4.
Using the point-slope form again:
y - 2 = (-1/4)(x - 8)
Expanding:
y - 2 = (-1/4)x + 2
Rearranging:
y = (-1/4)x + 4
So, the equation of a line that passes through the point (8, 2) and is perpendicular to y = 4x + 3 is y = (-1/4)x + 4.
Remember, whether parallel or perpendicular, always keep your sense of humor intact!
To find the equation of a line that is parallel or perpendicular to a given line, we need to determine the slope of the given line. Once we have the slope, we can substitute it into the slope-intercept form of a line to get the equation.
The equation y = 4x + 3 is already in slope-intercept form, where the coefficient of x is the slope of the line. In this case, the slope of the given line is 4.
(a) Parallel Line:
To find the equation of a line that is parallel to the given line, the slope of the parallel line should be the same as the slope of the given line, which is 4. Now, we substitute the point (8, 2) into the slope-intercept form (y = mx + b) to find the y-intercept (b).
Using the point-slope form:
(y - y₁) = m(x - x₁)
(y - 2) = 4(x - 8)
y - 2 = 4x - 32
y = 4x - 30
So, the equation of a line parallel to y = 4x + 3 that passes through the point (8, 2) is y = 4x - 30.
(b) Perpendicular Line:
To find the equation of a line that is perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line. The negative reciprocal of 4 is -1/4. Again, we substitute the point (8, 2) into the slope-intercept form (y = mx + b) to find the y-intercept (b).
Using the point-slope form:
(y - y₁) = m(x - x₁)
(y - 2) = -1/4(x - 8)
y - 2 = -1/4x + 2
y = -1/4x + 4
So, the equation of a line perpendicular to y = 4x + 3 that passes through the point (8, 2) is y = -1/4x + 4.