Question

The endpoints of line RS are R(1, -3) and S(4,2). Find RS

Answer 1

Answer:

Part 1) $RS=\sqrt{34}\ units$

Part 2) $CD=\sqrt{125}\ units$

Part 3) The coordinates of endpoint C are (13,5)

Step-by-step explanation:

Part 1) The endpoints of line RS are R(1, -3) and S(4,2). Find RS

we now that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$RS=\sqrt{(2+3)^{2}+(4-1)^{2}}$

$RS=\sqrt{(5)^{2}+(3)^{2}}$

$RS=\sqrt{34}\ units$

Part 2) The endpoints of line CD are C(-8,-1) and D(2,4). Find CD

we now that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$CD=\sqrt{(4+1)^{2}+(2+8)^{2}}$

$CD=\sqrt{(5)^{2}+(10)^{2}}$

$CD=\sqrt{125}\ units$

Part 3) The midpoint of line AC is M(5,6). One endpoint is A(-3,7). Find the coordinates of endpoint C

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

Let

(x2,y2)-------> the coordinates of point C

(x1,y1) -------> the coordinates of point A

substitute the given values

$(5,6)=(\frac{-3+x2}{2},\frac{7+y2}{2})$

write the equations

$\frac{-3+x2}{2}=5\\ \\-3+x2=10\\ \\ x2=10+3\\ \\x2=13$

$\frac{7+y2}{2}=6\\\\ 7+y2=12\\ \\ y2=12-7\\ \\y2=5$

therefore

The coordinates of endpoint C are (13,5)