Question

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

Answer 1

Answer:

$\displaystyle y=-\frac{1}{5}x+\frac{17}{5}$

Step-by-step explanation:

Equation of a line

A line can be represented by an equation of the form

$y=mx+b$

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:

$m_1.m_2=-1$

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

$\displaystyle m_2=-\frac{1}{m_1}=-\frac{1}{5}$

The equation of the line now can be expressed as:

$\displaystyle y=-\frac{1}{5}x+b$

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

$\displaystyle 2=-\frac{1}{5}*7+b$

Operating:

$\displaystyle 2=-\frac{7}{5}+b$

Multiplying by 5:

$10=-7+5b$

Operating:

$10+7=5b$

Solving for b:

$\displaystyle b=\frac{17}{5}$

The equation of the line is:

$\boxed{\displaystyle y=-\frac{1}{5}x+\frac{17}{5}}$