Write the equation of the line which passes through (4,-2) and is perpendicular to the line with equation 2y-x=1

y = -2x + 6

Step-by-step explanation:

Using

y - y_1 = m(x - x_1)

First, find the slope

2y - x = 1

Following this eqn

y = mx + C

2y = 1 + x

To get y, divide through by 2

2y/ 2 = (1 +x) / 2

y = 1+x/2

y = 1/2 + x /2

y = 1/2x + 1/2

Note, if two lines are perpendicular, their slope will be negative reciprocal

Slope = m = -2/1

Using this y - y_1 = m(x - x_1)

With point( 4 , -2)

x_1 = 4

y_1 = -2

y - (-2) = -2/1( x - 4)

y + 2 = -2/1(x - 4)

y + 2 = -2x + 8

y = -2x + 8 - 2

y = -2x + 6

The eqn of the line is

y = -2x + 6

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

i dont know men

step-by-step explanation: