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A point both of whose coordinates are positive lies in
Find the coordinates of the point equidistant from the points A(1, 2), B (3, –4) and C(5, –6).
Let the point be P(x,y)
The point which lies on yaxis at a distance of 6 units in the positive direction of yaxis is
A point both of whose coordinates are negative lies in
The point which lies on xaxis at a distance of 4 units in the negative direction of xaxis is
The point which lies on xaxis at a distance of 3 units in the positive direction of xaxis is
The points A(2, 3), B(2, 4) and C(5, 4) are the vertices of the square ABCD, the n the coordinates of the vertex D are
Find the coordinates of the point equidistant from the points A(1, 2), B(3, 4) and C(5, 6).
The given three points are A(1,2) B(3,4) and C(5,6).
Let P (x, y) be the point equidistant from these three points.
So, PA = PB = PC
⇒ x^{2 }+ 1– 2x + y^{2} + 4 – 4y = x^{2} + 9 6x + y^{2} + 16 + 8y = x 2 + 25– 10x + y^{2} + 36 + 12y
⇒ – 2x– 4y + 5 = 6x + 8y +25= – 10x + 12y+61
– 2x– 4y + 5 = 6x + 8y +25
⇒ – 2x– 4y + 5 +6x  8y 25=0
⇒ 4x– 12y 20=0
⇒ x– 3y  5 =0....(i)
 2x– 4y + 5 = – 10x + 12y+61
⇒ 2x– 4y + 5 +10x  12y61=0
⇒8x– 16y 56=0
⇒x– 2y 7=0....(ii)
Solving (i) and (ii)
x = 11, y = 2
Thus, the required point is (11, 2)
If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then ΔOAB is
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