Question

GEOMETRY 10TH GRADE NEED HELP ASAP

Answer 1

Answer:

Step-by-step explanation:

AB:BC::5:3

let the coordinates be (x,y)

$x=\frac{nx1+mx2}{m+n} \\=\frac{3(-6)+5(26)}{5+3} \\=\frac{-18+130}{8} \\=\frac{112}{8} \\=14\\y=\frac{ny1+my2}{m+n} \\=\frac{3(3)+5(3)}{5+3} \\=\frac{9+15}{8} \\=\frac{24}{8} \\=3$

coordinates of B are (14,3)

Answer 2

$\boxed{(6, -3)}$

Step-by-step explanation:

Given the ratio 5:3

and Points A and B where A is located at (-6, 3), and B is located at (26, -13).

Let point A be $(x_{1}, y_{1})$, and let point B be $(x_{2}, y_{2})$.

(-6, 3) → $(x_{1}, y_{1})$.

(26, -13) → $(x_{2}, y_{2})$.

Let 5 be n, and 3 be m.

5:3 → n:m

$(\frac{nx_{1} + mx_{2}}{n+m}, \frac{ny_{1} + my_{2}}{n + m})$.

To solve, just substitute these variables into the expressions of these coordinates to get the answer.

$(\frac{nx_{1} + mx_{2}}{n+m}, \frac{ny_{1} + my_{2}}{n + m})$ →

$(\frac{(5)x_{1} + (3)x_{2}}{(5)+(3)}, \frac{(5)y_{1} + (3)y_{2}}{(5) + (3)})$ →

$(\frac{(5)(-6)+ (3)(26)}{(5)+(3)}, \frac{(5)(3)+ (3)(-13)}{(5) + (3)})$ →

$(\frac{-30 + 78}{8}, \frac{15+ -39}{8})$ →

$(\frac{78 – 30}{8}, \frac{15 – 39}{8})$ →

$(\frac{48}{8}, \frac{-24}{8})$

→

$(\frac{6}{1}, \frac{-3}{1})$

→

$(6, -3)$.

Thus the coordinates of B are:

$\boxed{(6, -3)}$