Question

Someone please help, its for a hw assignment I have due today. What is the equation of the line through (3, 7) that is perpendicular to the line through points (-1, -2) and (5, 3)?

Answer 1

The required equation is:

$y=-\frac{6}{5}x+\frac{53}{5}$

Step-by-step explanation:

Let l_1 be the line through  (-1, -2) and (5, 3)

and l_2 be the line we require which passes through (3,7)

We have to find the lope of l_1 first

$m_1=\frac{y_2-y_1}{x_2-x_1}\\=\frac{3-(-2)}{5-(-1)}\\=\frac{3+2}{5+1}\\=\frac{5}{6}$

we have to find the equation of line perpendicular to l1

The product of slopes of two perpendicular lines is -1

Let m_2 be the slope of l_2

Then

$\frac{5}{6}*m_2=-1\\m_2=-\frac{6}{5}$

The general slope-intercept form is:

y=mx+b

Putting the value of slope

$y=-\frac{6}{5}x+b$

To find the value of b, we will put (3,7) in the equation

$7=-\frac{6}{5}(3)+b\\7=-\frac{18}{5}+b\\b=7+\frac{18}{5}\\b=\frac{35+18}{5}\\b=\frac{53}{5}$

Putting the values of b and m in standard slope intercept form:

$y=-\frac{6}{5}x+\frac{53}{5}$

Hence,

The required equation is:

$y=-\frac{6}{5}x+\frac{53}{5}$

Keywords: Equation of line

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