Question

Write the equation of a line that is perpendicular to y=7/5x+6 and passes through the point (2 -6)

Answer 1

The equation of a line that is perpendicular to y=7/5x+6 and passes through the point (2 -6) is:

$y=-\frac{5}{7}x-\frac{32}{7}$

Step-by-step explanation:

Given equation of line:

y=7/5x+6

The equation is in slope-intercept form. In this case, the co-efficient of x is the slope of the given line

So the slope will be: 7/5

As we know that the product of slopes of perpendicular lines is -1

$\frac{7}{5}*m=-1\\m=\frac{5}{7} * (-1)\\m=-\frac{5}{7}$

The general form is:

$y=mx+b$

Putting the value of slope

$y=-\frac{5}{7}x+b$

To find the value of b, putting the point (2,-6) in the equation

$-6=-\frac{5}{7}(2)+b\\-6=-\frac{10}{7}+b\\b=-6+\frac{10}{7}\\b=\frac{-42+10}{7}\\b=-\frac{32}{7}$

Putting the values of b and m

$y=-\frac{5}{7}x-\frac{32}{7}$

The equation of a line that is perpendicular to y=7/5x+6 and passes through the point (2 -6) is:

$y=-\frac{5}{7}x-\frac{32}{7}$

Keywords: Equation of line, Slope

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