Advertisements
Advertisements
Question
A point P moves so that P and the points (2, 2) and (1, 5) are always collinear. Find the locus of P.
Solution
Let P (x1, y1) be any point on the locus and A (2,2) B (1,5) are the fixed points.
Given that the points P, A, B are collinear.
⇒ Area of Δ PAB = 0
⇒ `1/2 [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)] = 0`
⇒ `1/2 [x_1 (2 - 5) + 2 (5 - y_1) + 1 (y_1 - 2)] = 0`
⇒ `1/2 [x_1 (-3) + 2 (5 - y_1) + 1 (y_1 - 2)] = 0`
⇒ `1/2 [- 3x_1 + 10 - 2y_1 + y_1 - 2] = 0`
⇒ `1/2 [- 3x_1 - y_1 + 8] = 0`
⇒ - 3x1 - y1 + 8 = 0
⇒ 3x1 + y1 - 8 = 0
∴ Locus of (x1, y1) is 3x + y - 8 = 0
APPEARS IN
RELATED QUESTIONS
Find the locus of a point which is equidistant from (1, 3) and x axis.
A point moves so that it is always at a distance of 4 units from the point (3, –2)
If the distance of a point from the points (2, 1) and (1, 2) are in the ratio 2 :1, then find the locus of the point.
If A(-1, 1) and B(2, 3) are two fixed points, then find the locus of a point P so that the area of triangle APB = 8 sq.units.
The locus of the point P which moves such that P is at equidistance from their coordinate axes is:
The locus of the point P which moves such that P is always at equidistance from the line x + 2y + 7 = 0:
Length of the latus rectum of the parabola y2 = -25x:
Let O(0, 0) and A(0, 1) be two fixed points. Then the locus of a point P such that the perimeter of ΔAOP is 4, is ______.
Locus of centroid of the triangle whose vertices are (a cos t, a sin t),(b sin t, – b cos t) and (1, 0), where t is a parameter is ______.
If a variable line drawn through the intersection of the lines `x/3 + y/4` = 1 and `x/4 + y/3` = 1, meets the coordinate axes at A and B, (A ≠ B), then the locus of the midpoint of AB is ______.
