The figure obtained is a right angled triangle.

The figure obtained is a square.

(i) The point B lies in II quadrant.
(ii) The point E lies in III quadrant.
(iii) The point S lies in IV quadrant.
(iv) The point T lies in I quadrant.

The point which lies on x-axis is C.
The point which lies on y-axis is B.

\(\triangle PQR\) is an equilateral triangle
PQ = PR = QR
\(\Rightarrow \) PQ = QR
\(\Rightarrow \) PQ = 4
\(\Rightarrow \) OQ = 2
In right triangle POQ
OP² + OQ² = PQ²
\(\Rightarrow \) OP + (2)² = (4)² \(\Rightarrow \) OP = \(2\sqrt { 3 } \)
\(P\rightarrow \left( 0,2\sqrt { 3 } \right) \)

\(\triangle ABC\) is an equilateral triangle
AC = BC = AB
\(\Rightarrow \) BC = AB
\(\Rightarrow \) BC = 2a
In right triangle COB,
OB² + OC² = BC²  [By pathagoras Theorem]
\(\Rightarrow \) a + OC = (2a)² \(\Rightarrow \) OC² = 3a²
\(\Rightarrow \) OC = \(\sqrt { 3 } a\)  \(\therefore \ C\rightarrow \left( 0,-\sqrt { 3 } a \right) \)
Similarly, \(D\rightarrow \left( 0,-\sqrt { 3 } a \right) \)

(i) Coordinates of point P are (0,4)
(ii) Abscissae of point Q is 5.
(iii) The point identified by the coordinates (-4,4) is S.
(iv) The point identified by the coordinates (-3,-6) is R.

(i) The coordinates of the points B and F are (-5,-4) and (6,0) respectively.
(ii) The abscissae of points D and H are 1 and 0 respectively.
(iv) The ordinates of the points A and C are 1 and 0 respectively.
(iv) The perpendicular distance of the point G from the x-axis is 4 units.

(a) The points whose ordinates are O and Q and N.
(b) The points whose abscissae are 0 are T and P.
(c) The points whose abscissae are -3 are Q and R.
(d) The points whose ordinates are 4 are S and L.

(i) I, III
(ii) I, II, III, IV
(iii) II, IV
(iv) I, III