Question

d is located at (-2,6). What is the coordinate of E if the length of DE is 10 units?

Answer 1

Answer:

Point E could be any point on the circumference of the circle $(x+2)^2+(y-6)^2=100$

Step-by-step explanation:

Let

(x,y) ----> the coordinates of point E

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

$points\ D(-2,6),E(x,y)$

$d_E_D=10\ units$

substitute in the formula

$10=\sqrt{(y-6)^{2}+(x+2)^{2}}$

squared both sides

$100=(y-6)^2+(x+2)^2$

Rewrite

$(x+2)^2+(y-6)^2=100$

This is the equation of a circle with center at (-2,6) and radius equal to 10 units

therefore

Point E could be any point on the circumference of the circle $(x+2)^2+(y-6)^2=100$