Question

Equation of the line perpendicular to 2x+y=3 & passes through points (2,-5)

Answer 1

the equation of line that is perpendicular to 2x+y=3 & passes through points (2,-5) is: $y = \frac{1}{2}x-6$

Step-by-step explanation:

Given equation of line is:

$2x+y = 3\\y = -2x+3$

Let m1 be the slope of given line and

m2 be the slope of line perpendicular to given line

The slope of first line will be (The coefficient of x)

m1 = -2

We know that the product of slopes of two perpendicular lines is -1

So,

$m_1.m_2 = -1\\-2 . m_2 = -1\\m_2 = \frac{-1}{-2}\\m_2 = \frac{1}{2}$

slope-intercept form is:

$y = m_2x+b$

Putting the value of slope we get

$y = \frac{1}{2}x+b$

Putting the point in the equation

$-5 = \frac{1}{2}(2} + b\\-5 = 1 + b\\b = -5-1\\b = -6$

Putting the value of b

$y = \frac{1}{2}x-6$

Hence,

the equation of line that is perpendicular to 2x+y=3 & passes through points (2,-5) is: $y = \frac{1}{2}x-6$

Keywords: slope, Equation of line

Learn more about equation of line at:

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