the segment joining p1(2,3) and p2 (6,7) is produced beyond p1 to a point P(x,y) so the P(x,y) is five times as far from p2 as from p1. find the coordinates of p(x,y)
(x, y)
6-2 = 4 which is present x difference
4*4 = 16
2 - 16 = -14 new x
then y
7 - 3 = 4
lol , same difference
4*4 = 16
3 - 16 = - 13
so (-14,-13)
To find the coordinates of point P(x, y), which is five times as far from P2(6, 7) as it is from P1(2, 3), we can use the concept of midpoint and section formula.
Step 1: Find the midpoint M(xm, ym) of P1 and P2 using the midpoint formula:
xm = (x1 + x2) / 2
= (2 + 6) / 2
= 8 / 2
= 4
ym = (y1 + y2) / 2
= (3 + 7) / 2
= 10 / 2
= 5
The midpoint M is (4, 5).
Step 2: Find the coordinates of point P(x, y) using the section formula
Let the distance from P1 to P(x, y) be d. Then the distance from P2 to P(x, y) is 5d.
Using the section formula, we have:
x = (5x2 + x1) / 6
= (5 * 6 + 2) / 6
= (30 + 2) / 6
= 32 / 6
= 16 / 3
= 5.33 (rounded to 2 decimal places)
y = (5y2 + y1) / 6
= (5 * 7 + 3) / 6
= (35 + 3) / 6
= 38 / 6
= 6.33 (rounded to 2 decimal places)
The coordinates of point P are approximately (5.33, 6.33).