MATH SOLVE

4 months ago

Q:
# Find an equation of the line perpendicular to the graph of 14x-7y=1 that passes through the point at (-2,4)A. y= 2x+3B. y=(1/2)x+3C. y= -(1/2)-3D. y= -(1/2)+3please help

Accepted Solution

A:

To solve first this needs to be in the general form of

[tex]y = a {x}^{2} + b[/tex]

so we do the following

[tex]14x - 7y = 1 \\ - 7y = - 14x + 1\\ y = 2x - \frac{1}{7} [/tex]

perpendicular lines have slopes that are the negative inverse of one another as such the slope of the line that is perpedicular to this one will have

[tex]slope = \frac{ - 1}{2} [/tex]

to determine if it passes through the desire point we use

[tex]y = - \frac{1}{2}x + b \\ 4 = - \frac{1}{2} \times( - 2) + b \\ 4 = 1 + b \\ 3 = b[/tex]

so we have the final following equation as the answer

[tex]y = - \frac{1}{2}x + 3[/tex]

[tex]y = a {x}^{2} + b[/tex]

so we do the following

[tex]14x - 7y = 1 \\ - 7y = - 14x + 1\\ y = 2x - \frac{1}{7} [/tex]

perpendicular lines have slopes that are the negative inverse of one another as such the slope of the line that is perpedicular to this one will have

[tex]slope = \frac{ - 1}{2} [/tex]

to determine if it passes through the desire point we use

[tex]y = - \frac{1}{2}x + b \\ 4 = - \frac{1}{2} \times( - 2) + b \\ 4 = 1 + b \\ 3 = b[/tex]

so we have the final following equation as the answer

[tex]y = - \frac{1}{2}x + 3[/tex]