Answer:
[tex]9x+3y+13=0[/tex]
Step-by-step explanation:
The point C divides the line segment AB joining the points A(2, 3) and B(-4, 1) in the ratio 2 : 1.
To determine the equation of the line passing through point C and perpendicular to AB, we need to find the coordinates of point C and the slope of the perpendicular line.
Coordinates of Point C
The section formula is a mathematical expression that determines the coordinates of a point dividing a line segment into two parts in a given ratio.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Section Formula}}\\\\P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of the line segment.}\\\phantom{ww} \bullet\;\textsf{Point $P$ divides the segment in the ratio $m : n$.}\end{array}}[/tex]
In this case:
- P = C
- (x₁, y₁) = Point A = (2, 3)
- (x₂, y₂) = Point B = (-4, 1)
- m : n = 2 : 1
Substitute the values into the section formula:
[tex]C(x,y)=\left(\dfrac{(2)(-4)+(1)(2)}{2+1},\dfrac{(2)(1)+(1)(3)}{2+1}\right)[/tex]
[tex]C(x,y)=\left(\dfrac{-8+2}{3},\dfrac{2+3}{3}\right)[/tex]
[tex]C(x,y)=\left(\dfrac{-6}{3},\dfrac{5}{3}\right)[/tex]
[tex]C(x,y)=\left(-2,\dfrac{5}{3}\right)[/tex]
Therefore, the coordinates of point C are:
[tex]C\left(-2,\dfrac{5}{3}\right)[/tex]
Slope of the Perpendicular Line
The slope of line segment AB can be found by substituting the endpoints of the line segment into the slope formula.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Slope formula}}\\\\\large\text{$m=\dfrac{y_2-y_1}{x_2-x_1}$}\\\\\textsf{where:}\\\phantom{w}\bullet\;\;m\; \textsf{is the slope.}\\\phantom{w}\bullet\;\;(x_1,y_1)\;\textsf{and}\;(x_2,y_2)\;\textsf{are two points on the line.}\end{array}}[/tex]
In this case:
- (x₁, y₁) = Point A = (2, 3)
- (x₂, y₂) = Point B = (-4, 1)
Therefore:
[tex]m_{AB}=\dfrac{1-3}{-4-2}=\dfrac{-2}{-6}=\dfrac{1}{3}[/tex]
So, the slope of line segment AB is:
[tex]m_{AB}=\dfrac{1}{3}[/tex]
The slope of a line perpendicular to AB is the negative reciprocal of the slope of AB. So, the slope of the line perpendicular to AB is:
[tex]m=\dfrac{-1}{\frac{1}{3}}=-3[/tex]
Equation of the Line
Now that we have the slope and the coordinates of point C, we can substitute these values into the point-slope form of a linear equation:
[tex]\begin{aligned}y-y_1&=m(x-x_1)\\\\y-\dfrac{5}{3}&=-3(x-(-2))\\\\3y-5&=-9(x+2)\\\\3y-5&=-9x-18\\\\3y-5+9x+18&=-9x-18+9x+18\\\\9x+3y+13&=0\end{aligned}[/tex]
Therefore, the equation of the line that passes through point C and is perpendicular to AB is:
[tex]\Large\boxed{\boxed{9x+3y+13=0}}[/tex]
