(04.02 MC)
The equation of line QR is x + 2y = 2. What is the equation of a line perpendicular to line QR in slope-intercept form that contains point (5, 6)?
Oy=-3x+
17
2
Oy = 2x - 4
Oy = - 11x + 1 / 2
NIN
Oy = 2x + 16

Respuesta :

Answer:

y = 2x - 4

Step-by-step explanation:

Rearrange the given equation to slope-intercept form.

x + 2y = 2

2y = -x + 2

y = -1/2x + 1

The slope of the given line is -1/2.  The slope of a perpendicular line will be the negative inverse, meaning that the slope will be 2.

-1/2 = 2

Find the equation of the new line with the point-slope form using the new slope and given point.

y - y₁ = m(x - x₁)

y - 6 = 2(x - 5)

y - 6 = 2x - 10

y = 2x - 4

The equation will be y = 2x - 4.

Answer:

y = 2x - 4

Step-by-step explanation:

First find the slope of line QR. Write down in slope intercept form

Equation of QR:

x + 2y = 2

     2y = - x + 2

       y = [tex]\frac{-1}{2}x + \frac{2}{2}[/tex]

       y = [tex]\frac{-1}{2}x+1[/tex]

Slope of line QR = m1 = -1/2

Slope of line perpendicular to QR = -1/m1

                                                        = [tex]\frac{-1}{\frac{-1}{2}}[/tex] = [tex]-1*\frac{-2}{1}[/tex]

                                                     [tex]m_{2} = 2[/tex]

m2 = 2 & (5 , 6)

Equation of line perpendicular to QR:  

 y - y1 = m (x - x1)

y - 6 = 2 (x - 5)    

  y - 6 = 2x - 5 *2

 y - 6  = 2x - 10

Add 6 to both sides

 y = 2x - 10 + 6

Add like terms

y = 2x - 4