Question

Find all of the points on the line y=8-x which are 2 units from (1,6).

Answer 1

The points on the line y = 8 - x that are 2 units from the point (1,6) are (2, 6) and (4, 4).

Step-by-step explanation:

To find the points on the line y = 8 - x that are 2 units from the given point (1,6), we can use the distance formula. The distance formula between two points
`\((x_1, y_1)\)` and
`\((x_2, y_2)\)` is given by
`\(d = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}\).`

In this case, we want the distance to be 2 units, and the given point is (1,6). Substituting these values into the formula, we get:

`\[2 = \sqrt{{(x - 1)}^2 + {(y - 6)}^2}\]`

Squaring both sides to eliminate the square root, we have:

`\[4 = (x - 1)^2 + (y - 6)^2\]`

Now, we substitute the equation of the line \(y = 8 - x\) into the above equation:

`\[4 = (x - 1)^2 + (8 - x - 6)^2\]`

Solving for x, we find two solutions: x = 2 and x = 4. Substituting these values back into the equation of the line, we get the corresponding y-values:

For x = 2,

y = 8 - 2 = 6, so the point is (2, 6).

For x = 4,

y = 8 - 4 = 4, so the point is (4, 4).

Therefore, the points on the line y = 8 - x that are 2 units from (1,6) are (2, 6) and (4, 4).