Question

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point (6, -7) are Simplify your answer. Type an ordered pair. Use a comma to separate answers as needed.)

Answer 1

Answer:
`(6+7√(3),0)\text{ and }(6-7√(3),0)` are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

`√((x-6)^2+(0-(-7))^2)=14\ \ \ [\ \because distance=√((x_2-x_1)^2+(y_2-y_1)^2)]`

Taking square on both the sides , we get

`(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0`

Using quadratic formula :
`x=(-b\pm√(b^2-4ac))/(2a)`

`x=(12\pm√((-12)^2-4(1)(-111)))/(2)\\\\\Rightarrow\ x=(12\pm√(144+444))/(2)\\\\\Rightarrow\ x=(12\pm√(588))/(2)\\\\\Rightarrow\ x=(12\pm√(2^2*7^2*3))/(2)\\\\\Rightarrow\ x=(12\pm14√(3))/(2)\\\\\Rightarrow\ x=6\pm7√(3)`

so,
`(6+7√(3),0)\text{ and }(6-7√(3),0)` are the required points.

since
`√(3)=1.732`

so,
`(6+7(1.732),0)\text{ and }(6-7(1.732),0)` are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.