Question

three collinear points on the coordinate plane are r(x,y), s(x+8h, y+8k), and p (x+6h, y+6k)

Answer 1

Answer:

A.$\frac{RP}{SP}=3$

B.$\frac{RP}{RS}=\frac{3}{4}$

Step-by-step explanation:

The complete question is

Three collinear points on the coordinate plane are R(x, y), S(x+8h, y+8k), and P(x+6h, y+6k).

Part A: Determine the value of RP/SP

Part B: Determine the value of RP/RS

we know that

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

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

we have

$R(x,y),S(x+8h,y+8k) and P(x+6h,y+6k)$

Part A.We have to find the value of $\frac{RP}{SP}$

step 1

Find the distance RP

$R(x,y),P(x+6h,y+6k)$

substitute the values in the formula

$RP=\sqrt{(x+6h-x)^2+(y+6k-y)^2}$

$RP=\sqrt{36h^2+36 k^2}$

$RP=6\sqrt{h^2+k^2}$

step 2

Find the distance SP

$S(x+8h,y+8k),P(x+6h,y+6k)$

substitute the values in the formula

$SP=\sqrt{(x+6h-x-8h)^2+(y+6k-y-8k)^2}$

$SP=\sqrt{4h^2+4k^2}$

$SP=\sqrt{4(h^2+k^2)}$

$SP=2\sqrt{h^2+k^2}$

step 3

Find the ratio  RP/SP

$\frac{RP}{SP}=\frac{6\sqrt{h^2+k^2}}{2\sqrt{h^2+k^2}}$

$\frac{RP}{SP}=3$

Part B. We have to determine the value of $\frac{RP}{RS}$

step 1

Find the distance RS

$R(x,y),S(x+8h,y+8k)$

$RS=\sqrt{(x+8h-x)^2+(y+8k-y)^2}$

$RS=\sqrt{64h^2+64k^2}$

$RS=\sqrt{64(h^2+k^2)}$

$RS=8\sqrt{h^2+k^2}$

step 2

Find the ratio RP/RS

$\frac{RP}{RS}=\frac{6\sqrt{h^2+k^2}}{8\sqrt{h^2+k^2}}$

$\frac{RP}{RS}=\frac{3}{4}$