find the equation of the line which passes through the point (7,2) and is perpendicular to y=5x-2

Answer:

[tex]\displaystyle y=-\frac{1}{5}x+\frac{17}{5}[/tex]

Step-by-step explanation:

Equation of a line

A line can be represented by an equation of the form

[tex]y=mx+b[/tex]

Where x is the independent variable, m is the slope of the line, b is the y-intercept and y is the dependent variable.

We need to find the equation of the line passing through the point (7,2) and is perpendicular to the line y=5x-2.

Two lines with slopes m1 and m2 are perpendicular if:

[tex]m_1.m_2=-1[/tex]

The given line has a slope m1=5, thus the slope of our required line is:

[tex]\displaystyle m_2=-\frac{1}{m_1}=-\frac{1}{5}[/tex]

The equation of the line now can be expressed as:

[tex]\displaystyle y=-\frac{1}{5}x+b[/tex]

We need to find the value of b, which can be done by using the point (7,2):

[tex]\displaystyle 2=-\frac{1}{5}*7+b[/tex]

Operating:

[tex]\displaystyle 2=-\frac{7}{5}+b[/tex]

Multiplying by 5:

[tex]10=-7+5b[/tex]

Operating:

[tex]10+7=5b[/tex]

Solving for b:

[tex]\displaystyle b=\frac{17}{5}[/tex]

The equation of the line is:

[tex]\boxed{\displaystyle y=-\frac{1}{5}x+\frac{17}{5}}[/tex]

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