The point A(3, 4) lies in
A point both of whose co-ordinates 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 y-axis at a distance of 6 units in the positive direction of y-axis is
The point B(-3, 4) lies in
A point both of whose co-ordinates are negative lies in
A point of the form (0, b) lies on
The point (0, -4) lies
The point which lies on x-axis at a distance of 4 units in the negative direction of x-axis is
The point C(-5, -2) lies in
Abscissa of a point is positive in
The co-ordinates of the origin are
The point (0, 9) lies
The point which lies on x-axis at a distance of 3 units in the positive direction of x-axis is
The point D(3, -6) lies in
Abscissa of a point is negative in
The equation of x-axis is
Abscissa of all points on the x-axis is
The points A(-2, 3), B(-2, -4) and C(5, -4) are the vertices of the square ABCD, the n the co-ordinates of the vertex D are
Find the co-ordinates 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
⇒ x2 + 1– 2x + y2 + 4 – 4y = x2 + 9 -6x + y2 + 16 + 8y = x 2 + 25– 10x + y2 + 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 - 12y-61=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 y-axis is
Abscissa of all points on the y-axis 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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