\(m=\frac{1}{2}\)For the given equation 4x - 8y = 8,
a. The slope-intercept form for the given equation is y = \(\frac{1}{2} x-1\)
b. The slope of the line is \(m=\frac{1}{2}\) and the y-intercept of the line is c = -1
c. The equation of the line that is perpendicular to the given line 4x - 8y = 8 and passes through the point (1,2) is (y-2)= -2(x-1) or y= -2x+4.
How to represent a line slope-intercept form?
The standard slope-intercept form of a line is y=mx+c. Where m is the slope and c is the y-intercept.
For a line given in the form of ax+by+c=0,
The slope m is calculated as -a/b and the y-intercept is calculated as -c/b.
How to find a perpendicular line for a given line passing through a point?
The given line has slope m=-a/b and y-intercept c. So, for the perpendicular line, the slope is calculated as -1/m (the negative reciprocal of the slope of the given line)
The equation of the line passing through a point (x1, y1) and with a slope -1/m is
(y-y1) = -1/m (x-x1) (point-slope form)
Calculation:
Given line is 4x - 8y = 8
we can write it in the form of ax+by+c=0
⇒ 4x - 8y -8 =0
⇒ 4 (x-2y-2) =0
⇒ x-2y-2=0
On comparing with ax+by+c=0, we have a=1, b=-2 and c=-2
a). Transforming the equation into the slope-intercept form:
For the line Slope m = -a/b
⇒ \(m=-\frac{1}{-2} =\frac{1}{2}\)
The y-intercept c = -c/b
⇒ \(c=-\frac{-2}{-2}=-1\)
Then, the slope-intercept form of the given line is \(y=\frac{1}{2}x-1\).
b). The slope of the given line is \(m=\frac{1}{2}\) and the y-intercept of the line is c = -1
c). Finding the perpendicular line:
The given line is x-2y-2=0
It has a slope \(m=\frac{1}{2}\) and the y-intercept is c = -1
Then, the slope for the perpendicular line is calculated as follows:
The slope of the perpendicular line m = \(-\frac{1}{m} = -2\)
Then the line that is perpendicular to the given line x-2y-2=0 and passes through the point (1,2) is
(y-y1) = m(x-x1) (point-slope form)
(y-2) = -2(x-1)
or
y-2 = -2x+2
⇒ y = -2x+4
Therefore, the perpendicular line is (y-2) = -2(x-1) or y = -2x+4.
Learn more about the slope-intercept form of a line here:
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