Find all points on the x-axis 5 units away from the point (4, 3).
on the x axis means that the y value is 0
distance formula between 2 points (x1,y1) and (x2,y2)
D=\(\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}\)
D=5 and one point is (4,3) and the other is (x,0)
sub
5=\(\sqrt{(4-x)^{2}+(3-0)^{2}}\)
5=\(\sqrt{(4-x)^{2}+(3)^{2}}\)
square both sides
25=\((4-x)^{2}+(3)^{2}\)
25=\((4-x)^{2}+9\)
minus 9 both sides
16=\((4-x)^{2}\)
square root both sides don't forget positive and negative roots
+/-4=4-x
minus 4 from both sides
-4+/-4=-x
times -1 both sides
4+/-4=x
4+4=x or 4-4=x
8 or 0=x
the possible points are
(0,0) or (8,0)
The points 5 units away from 4 is -1 to the left and 9 to the right. (-1, 9)
The given point:
\((x, y) = (4,3)\)
To find:
The points on x-axis 5 units away from the given points is calculated as:
\(x = +/- \ \ \ 5\\\\x_1 = -5 + 4= -1\\\\x_2 = +5 + 4 = 9\)
Thus, the points 5 units away from 4 is -1 to the left and 9 to the right
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