find the coordinates of point p which divides segment ab in the ratio 1:2 a(2,6) b(-4,1)
Answers
A (- 2 ; 5) B (4 ; 9)
Distance AB
xAB = xB - xA = 4 + 2 = 6
yAB = yB - yA = 9 - 5 = 4
AB² = xAB² + yAB² = (6²) + (4²) = 36 + 16 = 52
AB = √52
AB = 2√13
You want to get (2/3) of this distance:
AP = (2/3) * AB
AP = (2/3) * 2√13
AP = (4√13)/3
AP² = (16 * 13)/9
AP² = 208/9
Distance AP
xAP = xP - xA = xP + 2
yAP = yP - yA = yP - 5
AP² = xAP² + yAP²
AP² = (xP + 2)² + (yP - 5)²
(xP + 2)² + (yP - 5)² = 208/9 ← memorize this result as (1)
The point P belongs to the line AB. Let's calculate the equation of the line AB.
The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept
m = (yB - yA) / (xB - xA) = (9 - 5) / ( 4 + 2) = 4/6 = 2/3
The equation of the line Ab becomes: y = (2/3)x + b
You know that the line AB passes through A (- 2 ; 5), so the coordinates of this point A must verify the equation of the line AB.
y = (2/3)x + b
b = y - (2/3)x → you substitute x and y by the coordinates of the point A (- 2 ; 5)
b = 5 - [(2/3) * - 2] = 5 + (4/3) = (15/3) + (4/3) = 19/3
The equation of the line Ab is: y = (2/3)x + (19/3) ← memorize this result as (2)
Recall (1)
(xP + 2)² + (yP - 5)² = 208/9 → reccal (2): y = (2/3)x + (19/3)
(xP + 2)² + [(2/3)xP + (19/3) - 5]² = 208/9
xP² + 4xP + 4 + [(2/3)xP + (19/3) - (15/3)]² = 208/9
xP² + 4xP + 4 + [(2/3)xP + (4/3)]² = 208/9
xP² + 4xP + 4 + (4/9)xP² + (16/9)xP + (16/9) = 208/9
9xP² + 36xP + 36 + 4xP² + 16xP + 16 = 208
13xP² + 52xP + 52 = 208
13xP² + 52xP - 156 = 0
xP² + 4xP - 12 = 0
xP² + 4xP + (4 - 4) - 12 = 0
xP² + 4xP + 4 - 16 = 0
(xP + 2)² - 4² = 0
(xP + 2 + 4)(xP + 2 - 4) = 0
(xP + 6)(xP - 2) = 0
First case: (xP + 6) = 0 → xP = - 6
Second case: (xP - 2) = 0 → xP = 2
You know that the point P is located inside [AB], so its abscissa is located inside [xA ; xB]
→ xP Є [- 2 ; 4]
So you keep only the solution: xP = 2
Recall (2)
y = (2/3)x + (19/3) → you know that: x = 2
y = (4/3) + (19/3)
y = 23/3
The coordinates of the point P are (2 ; 23/3)