Respuesta :
Equation of a line given [tex] 7x-2y = 1 [/tex]
We have to find an equation of a line which is perpendicular to the given line.
If the general equation of a line is [tex] y=mx+c [/tex], m is the slope of the line there. And the slope of the perpendicular line will be the negative reciprocal of m which is [tex] (-\frac{1}{m}) [/tex].
So first here we have to make the given equation as [tex] y=mx+c [/tex]
[tex] 7x-2y = 1 [/tex]
First we have to move 7x to the right side by subtracting it from both sides.
[tex] -7x+7x-2y =-7x+1 [/tex]
[tex] -2y = -7x+1 [/tex]
Now to get y we have to move -2, by dividing it on both sides.
[tex] \frac{(-2y)}{(-2)} = \frac{(-7x+1)}{(-2)} [/tex]
[tex] y = \frac{(-7x)}{(-2)} + \frac{1}{(-2)} [/tex]
[tex] y= (\frac{7}{2} )x - \frac{1}{2} [/tex]
So here the slope [tex] m= \frac{7}{2} [/tex]
Now the slope for the perpendicular equation is
[tex] -\frac{1}{m}= -\frac{1}{(7/2)} [/tex]
[tex] -\frac{1}{m} = -\frac{2}{7} [/tex]
So slope of the perpendicular line is [tex] -\frac{2}{7} [/tex]
We can write the perpendicular equation as
[tex] y=(-\frac{2}{7} )x+c [/tex]
Now this equation is passing through the point (-4,5)
We have to plug in x = -4 and y = 5 in the line to get c.
[tex] 5= (-\frac{2}{7})(-4) + c [/tex]
[tex] 5= \frac{8}{7} +c [/tex]
[tex] 5-\frac{8}{7} = c [/tex]
[tex] \frac{35}{7} -\frac{8}{7} =c [/tex]
[tex] \frac{(35-8)}{7} = c [/tex]
[tex] \frac{27}{7} =c [/tex]
So we have got the value of c. Now we can write the perpendicular equation as,
[tex] y= (-\frac{2}{7})x+\frac{27}{7} [/tex]
[tex] y = \frac{(-2x+27)}{7} [/tex]
[tex] 7y = -2x+7 [/tex]
[tex] 7y+2x = -2x+2x+7 [/tex]
[tex] 2x+7y = 7 [/tex]
So we have got the required perpendicular line.
The equation of the perpendicular line is [tex] 2x+7y = 7 [/tex]
