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]
