The point A(3, 4) lies in
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 (7, 0) lies
The point which lies on yaxis at a distance of 6 units in the positive direction of yaxis is
The point B(3, 4) lies in
A point both of whose coordinates are negative lies in
A point of the form (0, b) lies on
The point (0, 4) lies
The point which lies on xaxis at a distance of 4 units in the negative direction of xaxis is
The point C(5, 2) lies in
Abscissa of a point is positive in
The coordinates of the origin are
The point (0, 9) lies
The point which lies on xaxis at a distance of 3 units in the positive direction of xaxis is
The point D(3, 6) lies in
Abscissa of a point is negative in
The equation of xaxis is
Abscissa of all points on the 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)
Ordinate of a point is positive in
The equation of yaxis is
Abscissa of all points on the yaxis is
If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then ΔOAB is
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