+1459 Find all points $(x,y)$ that are $5$ units away from the point $(2,7)$ and that lie on the line $y = 5x - 28.$
+1768 We can use the distance formula to find the distance between two points, and the equation of the line to find the points that lie on the line.
The distance formula tells us that the distance between two points (x1,y1) and (x2,y2) is:
sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
The equation of the line tells us that for any point (x,y) on the line, the following equation must hold:
y = 5x - 28
We can use these two equations to find all points (x,y) that are 5 units away from the point (2,7) and that lie on the line y=5x−28.
First, we can find the distance between the point (2,7) and any point (x,y) on the line. Using the distance formula, we get:
sqrt((x - 2)^2 + (y - 7)^2)
We can then substitute the equation of the line into this expression to get:
sqrt((x - 2)^2 + (5x - 35)^2)
We want this distance to be equal to 5, so we can set up the equation:
sqrt((x - 2)^2 + (5x - 35)^2) = 5
Squaring both sides of this equation, we get:
(x - 2)^2 + (5x - 35)^2 = 25
Expanding this equation, we get:
x^2 - 4x + 4 + 25x^2 - 70x + 1225 = 25
Combining like terms, we get:
26x^2 - 74x + 1249 = 0
Factoring this equation, we get:
(13x - 49)(2x - 25) = 0
Therefore, the possible values of x are 1349 and 225.
We can substitute these values of x back into the equation of the line to find the corresponding values of y. For x=1349, we get:
y = 5 \cdot \frac{49}{13} - 28 = 7
For x=225, we get:
y = 5 \cdot \frac{25}{2} - 28 = 7
Therefore, the two points that are 5 units away from the point (2,7) and that lie on the line y=5x−28 are (1349,7) and (225,7).
