points a,b, and c make the triangle ABC and are at the coordinates A(-2,9), B(-33,13) and c(-21,25) point D is the midpoint of BC and AD is a median of ABC, the equation of the median can be given by ax+by=c where A B and C

what is most simple way to answer this

[tex]\textsf{Midpoint}=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)\quad \textsf{where}\:(x_1,y_1)\:\textsf{and}\:(x_2,y_2)\:\textsf{are the endpoints}}\right)[/tex]

Define the endpoints:

[tex]\text{Let }(x_1,y_1)=\sf B=(-33,13)[/tex]

[tex]\text{Let }(x_2,y_2)=\sf C=(-21,25)[/tex]

Substitute the defined endpoints into the formula:

[tex]\textsf{Midpoint of BC}=\left(\dfrac{-21-33}{2},\dfrac{25+13}{2}\right)=(-27,19)[/tex]

Therefore, D = (-27, 19).

Step 2

Find the slope of the median (line AD) using the Slope formula.

Define the points:

[tex]\textsf{let}\:(x_1,y_1)=\sf A=(-2,9)[/tex]

[tex]\textsf{let}\:(x_2,y_2)=\sf D=(-27,19)[/tex]

Substitute the defined points into the Slope formula:

[tex]\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{19-9}{-27-(-2)}=-\dfrac{2}{5}[/tex]

Therefore, the slope of the median is -²/₅.

Step 3

Substitute the found slope and one of the points into the Point-slope formula to create an equation for the median.

[tex]\implies y-y_1=m(x-x_1)[/tex]

[tex]\implies y-9=-\dfrac{2}{5}(x-(-2))[/tex]

Simplify and rearrange the equation so it is in standard form Ax+By=C:

[tex]\implies 5(y-9)=-2(x+2)[/tex]

[tex]\implies 5y-45=-2x-4[/tex]

[tex]\implies 2x+5y-45=-4[/tex]

[tex]\implies 2x+5y=41[/tex]

Conclusion

Therefore, the equation of the median is:

2x + 5y = 41