Question

NO LINKS!! URGENT HELP PLEASE!!!

Find all points on the x-axis that are a distance 5 from P(-6, 3)

(x, y) = _________ (smaller x-value)
(x,y)= __________ (larger x-value)

Answer 1

(x, y) = (-10, 0) and (x, y) = (-2, 0)

Explanation:

To find the points on the x-axis that are a distance 5 from P(-6,3), we can use the distance formula:

d = √((x2 - x1)^2 + (y2 - y1)^2)

Since we want to find the points that are a distance of 5 from P(-6,3) on the x-axis, we know that y = 0 for these points. So we can simplify the distance formula to:

d = √((x - (-6))^2 + (0 - 3)^2)

d = √((x + 6)^2 + 9)

We want to find the values of x that make d = 5. So we can set up the equation:

5 = √((x + 6)^2 + 9)

Squaring both sides, we get:

25 = (x + 6)^2 + 9

Subtracting 9 from both sides, we get:

16 = (x + 6)^2

Taking the square root of both sides (remembering to include both the positive and negative square root), we get:

x + 6 = ±4

Subtracting 6 from both sides, we get:

x = -10 or x = -2

So the two points on the x-axis that are a distance of 5 from P(-6,3) are (-10, 0) and (-2, 0).

Therefore, the two answers are:

(x, y) = (-10, 0) and (x, y) = (-2, 0)

Answer 2

(x, y) = (-10, 0) (smaller x-value)

(x, y) = (-2, 0) (larger x-value)

Explanation:

To find the all the points on the x-axis that are a distance of 5 from P(-6, 3), we can use the distance formula.

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance Formula}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where:\\ \phantom{ww}$\bullet$ $d$ is the distance between two points. \\\phantom{ww}$\bullet$ $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

As the distance is 5, d = 5.

Let (x₁, y₁) = (-6, 3).

As the points to be found are on the x-axis, their y-coordinates are zero.

Therefore, let (x₂, y₂) = (a, 0).

Substitute these values into the distance formula:

`\implies √((a-(-6))^2+(0-3)^2)=5`

Simplify and solve for a:

`\implies \left(√((a+6)^2+(-3)^2)\right)^2=5^2`

`\implies (a+6)^2+9=25`

`\implies (a+6)^2+9-9=25-9`

`\implies (a+6)^2=16`

`\implies √((a+6)^2)=√(16)`

`\implies a+6=\pm 4`

`\implies a+6-6=-6\pm 4`

`\implies a=-10, -2`

Therefore, the points on the x-axis that are a distance of 5 units from P(-6, 3) are:

• (x, y) = (-10, 0) (smaller x-value)
• (x, y) = (-2, 0) (larger x-value)