Write the slope-intercept form of the equation of the line that is perpendicular to and passes through Point X. Show all work for full credit.

Write the slopeintercept form of the equation of the line that is perpendicular to and passes through Point X Show all work for full credit class=

Respuesta :

Answer:

[tex]y =\dfrac{12}{5}x + 22[/tex]

Step-by-step explanation:

Slope-intercept form of line:

First find the slope of the line AB. ie, m

Slope of the perpendicular line = -1/m

(2 , 3)  ⇒ x₁ = 2 & y₁ = 3  

(-10, 8)  ⇒ x₂ = -10 & y₂ = 8

[tex]\boxed{Slope=\dfrac{y_2-y_1}{x_2-x_1}}[/tex]

          [tex]= \dfrac{8-3}{-10-2}\\\\=\dfrac{5}{-12}\\\\=\dfrac{-5}{12}[/tex]

[tex]\sf slope \ of \ the \ perpendicular \ line \ m_1 = \dfrac{-1}{m}= -1 \ \div \dfrac{-5}{12}[/tex]

                                                   [tex]\sf = -1 * \dfrac{12}{-5}=\dfrac{12}{5}[/tex]

Equation of the required line: y = mx + b

       [tex]y =\dfrac{12}{5}x+b[/tex]

The line passes through (-5 , 10). Substitute in the above equaiton,

         [tex]10 =\dfrac{12}{5}*(-5) + b\\\\ 10 = (-12) + b\\\\[/tex]

 10 + 12 = b

        b   = 22

Equation of the line:

  [tex]y =\dfrac{12}{5}x + 22[/tex]