Is this Calculus?

[BigGlenntheHeavy]

$\displaystyle Determine \ the \ two \ points \ of \ trisection, \ P_1(x_1,y_1) \ and \ P_2(x_2,y_2), \ of \ the \ line \ segment \ joining$

$\displaystyle A(3,-1) \ and \ B(9,7).$

 

[pka]

$\displaystyle Determine \ the \ two \ points \ of \ trisection, \ P_1(x_1,y_1) \ and \ P_2(x_2,y_2), \ of \ the \ line \ segment \ joining$
$\displaystyle A(3,-1) \ and \ B(9,7).$

It is more like vector geometry.
The vector $\displaystyle \overrightarrow {AB} = B - A = \left\langle {6,8} \right\rangle$.
The line that contains $\displaystyle A~\&~B$ is $\displaystyle P(t) = A + t\overrightarrow {AB} = \left\langle {3 + 6t, - 1 + 8t} \right\rangle$.
Note that $\displaystyle P(0)=A~\&~P(1)=B$.
So the trisection points are $\displaystyle P\left( {\frac{1}{3}} \right)\;\& \,P\left( {\frac{2}{3}} \right)$.

 

[soroban]

Hello, BigGlenn!

$\displaystyle \text{Determine the two points of trisection, }P(x_1,y_1)\text{ and }Q(x_2,y_2)$

$\displaystyle \text{of the line segment joining: }\:A(3,-1)\text{ and }(9,7)$

                                          (9,7) B
                                            o -
                                        *   | :
                                Q   *       | :
                                o           | :
                            *               | 8
                    P   *                   | :
                    o                       | :
                *                           | :
        A   *                               | :
.(3,-1) o - - - - - - - - - - - - - - - - - * -
.       : -   -   -   -   6   -   -   -   - :

$\displaystyle P\text{ is }\tfrac{1}{3}\text{ of the distance from }A\text{ to }B.$

$\displaystyle \text{Its }x\text{-coordinate is }\tfrac{1}{3}\text{ of the horizontal distance from }A\text{ to }B.$
. . $\displaystyle \tfrac{1}{3}(9-3) \:=\:2$
$\displaystyle \text{Hence, its }x\text{-coordinate is: }\:x \:=\:3 + 2 \:=\:5$

$\displaystyle \text{Its }y\text{-coordinate is }\tfrac{1}{3}\text{ of the vertical distance from }A\text{ to }B.$
. . $\displaystyle \tfrac{1}{3}[7-(\text{-}1)] \:=\:\tfrac{8}{3}$
$\displaystyle \text{Hence, its }y\text{-coordinate is: }\:y \:=\:-1 + \tfrac{8}{3} \:=\:\tfrac{5}{3}$

$\displaystyle \text{Therefore: }P\left(5,\:\tfrac{5}{3}\right)$


$\displaystyle \text{In a similar fashion, point }Q\text{ can be determined.}$

 

[BigGlenntheHeavy]

$\displaystyle To \ find \ P_1(x_1,y_1), we \ have \ r=1:2, \ so \ x_1 \ = \ \frac{3+\frac{1(9)}{2}}{1+\frac{1}{2}} \ =5$

$\displaystyle and \ y_1 \ = \ \frac{-1+\frac{1(7)}{2}}{1+\frac{1}{2}} \ = \ \frac{5}{3}. \ Hence \ P_1(x_1,y_1) \ = \ (5,\frac{5}{3}).$

\(\displaystyle To \ find \ P_2(x_2,y_2), \we \ have \ r=2:1, \ so \ x_2 \ = \ \frac{3+2(9)}{1+2} \ = \ 7 \ and \ y_2 \ = \ \frac{-1+2(7)}{1+2} \ = \ \frac{13}{3}.\)

$\displaystyle Hence, \ P_2(x_2,y_2) \ = \ (7,\frac{13}{3}).$