Find an equation of the line that is parallel to `y=3/4 x -1` and passes through the point (4,0). Find an equation of the line that is perpendicular to `y=x+4` and through the point (-7, 1).
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An equation paralle to `y= 3/4x-1` and passes through the point (4,0). Since the equation is written in slope intercept form the line has a slope of `3/4` .
In order to be parallel the line must havet he same slope, `3/4.`
Use point slope form `y - y_(1)= m(x-x_(1))` , to write the equation of the line.
`y - 0 = 3/4(x-4) rArr y = 3/4x - 3`
A line perpendicular to `y = x+4` , must have a slope of `-1` as this is the opposite reciprocal of the line's slope of 1.
Again use point...
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y = 3/4 x - 1
Any straight line parallel to the given line will be of the form
y = 3/4 x + k
When this line passes through the point ( 4 , 0 ) we have
0 = 3/4 * 4 + k
Hence k + 3 = 0 .
k = - 3 .
Hence y= 3/4 x - 3 is the required equation.
__________________________________________
y = x + 4
That is x - y + 4 =0 is the given line .
Any line perpendicular to this line will be of the form
x + y +k =0 .
Since this passes through the point ( - 7 , 1 )
1 - 7 + k = 0 . ...... That is k = 6
Hence the required equation is x + y + 6 = 0 ..
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