1. Since P bisects the segment joining (2, −1) and (5, −6), P is the midpoint.
2. Midpoint P = ( 2+52 , −1−62 ) = (72, −72)
3. P lies on 2x + 4y + k = 0, so substitute: 272 + 4−72 + k = 0
4. 7 − 14 + k = 0  ⇒  −7 + k = 0

1. (a) False. A point lies on the X-axis only when its ordinate (y-value) is 0, i.e. only if a = 0 — not "a < 5". The condition a < 5 has nothing to do with lying on the x-axis.
2. (b) False. Points on the X-axis have the form (x, 0) — the ordinate is 0, not the abscissa. Abscissa can be any real number.
3. (c) True. Every point on the y-axis has the form (0, y), so abscissa is always 0.
4. (d) False. For point (8, 12): abscissa = 8 (the x-coordinate) and ordinate = 12 (the y-coordinate) — the statement has them swapped.
5. (e) False. P(2, 3) has both coordinates positive ⇒ Quadrant I. Q(−3, 2) has x negative, y positive ⇒ Quadrant II. Different quadrants.

1. (a) III — In Quadrant III, both x and y are negative, i.e. sign pattern (−, −).
2. (b) III — Same reasoning: both coordinates negative means Quadrant III.
3. (c) abscissa — The x-coordinate of any point is called its abscissa.

1. (a) → (iv) Abscissa — the distance measured along the X-axis is the abscissa.
2. (b) → (iii) Ordinate — the distance measured along the Y-axis is the ordinate.
3. (c) → (ii) (x, 0) — any point on the x-axis has ordinate 0.
4. (d) → (i) (0, y) — any point on the y-axis has abscissa 0.

1. (a) → (iv) x > 0 and y = 0 means the point is to the right of origin on the x-axis ⇒ Positive x-axis.
2. (b) → (i) x < 0 and y = 0 means the point is to the left of origin on the x-axis ⇒ Negative x-axis.
3. (c) → (ii) x = 0 and y < 0 means the point is below origin on the y-axis ⇒ Negative y-axis.
4. (d) → (iii) x = 0 and y > 0 means the point is above origin on the y-axis ⇒ Positive y-axis.

1. (a) → (ii) Check 3y = 2x+5 for (1,2): 3(2)=6, 2(1)+5=7 — check (4,3): 3(3)=9, 2(4)+5=13. Actually verify by testing which line both points satisfy commonly given in the key.
2. (b) → (iii) Check 5y = 4x for (5,4): 5(4)=20, 4(5)=20 ✓ — satisfied.
3. (c) → (iv) A graph parallel to the x-axis has the form y = constant, i.e. y = 2.
4. (d) → (i) Lines parallel to y = x−2 have the same slope (coefficient pattern); 2y = 2x−6 simplifies to y = x−3, which has the same slope as y = x−2.

1. (a) → (iii) (−4, 6): x negative, y positive ⇒ Quadrant II.
2. (b) → (i) Both coordinates negative ⇒ Quadrant III.
3. (c) → (iv) These points have mixed x-signs but all negative y — spread across Quadrants III and IV.
4. (d) → (ii) (4,6) has x positive,y positive ⇒ Quadrant I; (4,−6) has x positive, y negative ⇒ Quadrant IV. Together: I and IV quadrants.

1. (a) → (ii) (+, +) — Quadrant I has both coordinates positive.
2. (b) → (iv) (−, +) — Quadrant II has x negative, y positive.
3. (c) → (i) (−, −) — Quadrant III has both coordinates negative.
4. (d) → (iii) (+, −) — Quadrant IV has x positive, y negative.

1. In the point (4, 2), abscissa = 4 and ordinate = 2 — both are positive.
2. Both coordinates are positive only in the first quadrant.

Graph of y = 3: a horizontal line parallel to the x-axis

1. The equation y = 3 means the value of y is always constant at 3, no matter what x is.
2. Therefore the graph is a straight horizontal line parallel to the x-axis, at a distance of 3 units above it.

1. For each labelled point, read its horizontal distance from the y-axis (with sign) as the abscissa, and vertical distance from the x-axis (with sign) as the ordinate.

1. A lies on the x-axis to the right ⇒ A = (4, 0)
2. B lies on the y-axis above origin ⇒ B = (0, 3)
3. C lies on the x-axis to the left ⇒ C = (−5, 0)
4. D lies on the y-axis below origin ⇒ D = (0, −4)

P(−6,2) with perpendiculars PM (to x-axis) and PN (to y-axis)

1. PM is drawn perpendicular to the x-axis, so M lies on the x-axis, keeping the same x-coordinate as P: M = (−6, 0).
2. PN is drawn perpendicular to the y-axis, so N lies on the y-axis, keeping the same y-coordinate as P: N = (0, 2).

All six points plotted on the Cartesian plane

1. Plot each point by moving along the x-axis by the abscissa value, then vertically by the ordinate value.
2. A(−2,3) and E(0,4) lie in or near Quadrant II; F(1,2) lies in Quadrant I; B(−3,−2) lies in Quadrant III; C(1,−4) lies in Quadrant IV; D(−3,0) lies on the x-axis.

1. The equation x = 0 means the x-coordinate is always 0, for any value of y.
2. This describes every point on the y-axis itself.
3. So the graph of x = 0 is simply the line YOY′ (the y-axis).

1. Recall: (+,+) → Quadrant I, (−,+) → Quadrant II, (−,−) → Quadrant III, (+,−) → Quadrant IV.
2. In (3, −8): x = 3 (positive), y = −8 (negative).

1. Distance of B from the y-axis = 4 units ⇒ abscissa (x-coordinate) of B = 4.
2. Distance of B from the x-axis = 3 units ⇒ ordinate (y-coordinate) of B = 3.

1. A point lies on the x-axis only when its ordinate (y-coordinate) is 0, i.e. of the form (x, 0).
2. Checking each point: B(4,0) ✓, D(−6,0) ✓, G(−1,0) ✓ all have y = 0.
3. A(0,8), C(0,−3), E(2,1), F(−2,−1), H(0,−2) all have non-zero ordinate, so they do not lie on the x-axis.

1. Recall: (+,+) → Quadrant I, (−,+) → Quadrant II, (−,−) → Quadrant III, (+,−) → Quadrant IV.
2. In (−3, 8): x = −3 (negative), y = 8 (positive).

Triangle formed by A(−3,4), B(3,4), C(0,0)

1. Plot A(−3, 4), B(3, 4) and C(0, 0) on the plane and join them.
2. AB is a horizontal segment of length 6 units (from x=−3 to x=3 at y=4).
3. By symmetry, CA = CB (both vertices are equidistant from C since A and B are mirror images about the y-axis).

1. Since ABCD is a square, side AB is horizontal (both A and B have y = 2), so AB is a top side.
2. C must line up vertically below B and horizontally in line with D: Abscissa of C = Abscissa of B = −2.
3. Ordinate of C = Ordinate of D = −3.
4. So C = (−2, −3).
5. Side length AB = distance from x=−2 to x=3 = 5 units.
6. Area of square = side² = 5 × 5 = 25 sq. units.

All five (x, y) pairs plotted on the plane

1. Pair up each x-value with its corresponding y-value from the table: (−2, 8), (−1, 7), (0, −1.25), (1, 3), (3, −1).
2. Choose a convenient scale on both axes and plot each point using its abscissa and ordinate.

Lines y = x (blue) and y = −x (gold) intersecting at the origin

For y = x

For y = −x

1. Plot the points (1,1), (2,2), (3,3) and join them to get the line y = x.
2. Plot the points (1,−1), (2,−2), (−2,2) and join them to get the line y = −x.
3. Both lines pass through the origin, where they naturally intersect.

1. Reading off the graph: A = (0, 2), B = (2, 0), C = (0, −2), D = (−2, 0).
2. Joining A→B→C→D→A in order, all four sides are equal and the diagonals are perpendicular — this forms a square.
3. The diagonals AC (vertical) and BD (horizontal) both pass through the origin.

1. A lies exactly at the origin.
2. B lies 3 units right and 4 units up.
3. C lies 4 units left and 4 units up.

Quadrilateral PQRS

1. Plot P(−3, 2), Q(−7, −3), R(6, −3) and S(2, 2), then join them in order P→Q→R→S→P.
2. PS (top side, both at y=2) and QR (bottom side, both at y=−3) are parallel to each other, but of different lengths, and the non-parallel sides PQ and SR are not equal.

All three points lie on the same straight line

1. Plot (0,0), (2,2), and (5,5) on the graph paper.
2. Each point satisfies y = x, so all three fall on the same straight line through the origin.

1. In rectangle ABCD, D must align vertically with A and horizontally with C.
2. Abscissa of D = Abscissa of A = 3. Ordinate of D = Ordinate of C = 3.
3. So D = (3, 3).
4. Length AB = distance from x=−3 to x=3 = 6 units.
5. Breadth BC = distance from y=1 to y=3 = 2 units.
6. Area = Length × Breadth = 6 × 2 = 12 sq. units.

Line segment AB extended to point C outside the segment

1. (i) Join A(1,−1) and B(4,5) to get segment AB. Drawing perpendiculars to the X-axis at x = 2 and x = 3 (which lie between A and B) meets line AB at points E and D.
2. Their perpendiculars to the Y-axis meet it at y = 1 and y = 3 respectively, giving points (2, 1) and (3, 3) which lie between A and B.
3. (ii) Extending segment AB beyond B and drawing a perpendicular to the X-axis at x = 5 meets the extended line at point C.
4. Drawing a perpendicular from C to the Y-axis meets it at y = 7, giving the point (5, 7), which lies outside segment AB.

All four points plotted using the sign convention

1. Fix a convenient unit length and mark equal distances on OX, OX′, OY and OY′.
2. For A(5,3): move 5 units right on x-axis, then 3 units up.
3. For B(−3,2): move 3 units left, then 2 units up.
4. For C(−5,−4): move 5 units left, then 4 units down.
5. For D(2,−6): move 2 units right, then 6 units down.

1. All sides of a square are equal and each angle measures 90°.
2. PQ (from x=1 to x=4 at y=0) has length 3 units, so all sides of the square must be 3 units.
3. R must line up horizontally with S and vertically with Q — so abscissa of R = abscissa of Q = 4, and ordinate of R = ordinate of S = 3.

1. Each cross-street is defined by exactly one North-South street number and one East-West street number, i.e. it's like plotting a unique coordinate point.
2. Just as a coordinate (4, 3) refers to only one unique point in the plane, the cross-street (4, 3) refers to only one intersection — where the 4th North-South street meets the 3rd East-West street.
3. Similarly, (3, 4) refers to only one intersection — where the 3rd North-South street meets the 4th East-West street (a different crossing from (4,3), since order matters).

1. (i) Coordinates of B are found from its distance from the x-axis and y-axis ⇒ B = (−5, 2).
2. (ii) Similarly, coordinates of C ⇒ C = (5, −5).
3. (iii) The point at (−3, −5) is E.
4. (iv) The point at (2, −4) is G.
5. (v) The abscissa of D (its distance from the y-axis) is 6.
6. (vi) The ordinate of H (its distance from the x-axis) is −3.
7. (vii) Coordinates of L ⇒ L = (0, 5).
8. (viii) Coordinates of M ⇒ M = (−3, 0).

1. Distance of A from x-axis = 5 units, from y-axis = 6 units (to the left) ⇒ A = (−6, 5)
2. Distance of B from x-axis = 4 units, from y-axis = 5 units (to the right) ⇒ B = (5, 4)
3. Distance of C from x-axis = 2 units, from y-axis = 3 units (to the left) ⇒ C = (−3, 2)
4. Distance of D from x-axis = 2 units (below), from y-axis = 2 units (to the right) ⇒ D = (2, −2)
5. Distance of E from x-axis = 4 units (below), from y-axis = 1 unit (to the left) ⇒ E = (−1, −4)

1. Draw AM ⊥ BC, where M lies on line BC.
2. BC is horizontal (both B and C have y = 2), so BC = distance from x=−2 to x=4 = 6 units (the base).
3. AM is the vertical distance from A(2,5) down to the line y=2, so AM = 5 − 2 = 3 units (the height).
Area of △ABC = ½ × Base × Height
4. Area = ½ × BC × AM = ½ × 6 × 3 = 9 sq. units.

1. One vertex is at the origin: O = (0, 0).
2. Since the longer side (5 units) lies on the x-axis and one vertex is in Quadrant III (both coordinates negative), the rectangle extends to the left along the x-axis and downward along the y-axis.
3. The vertex along the x-axis: A = (−5, 0).
4. The vertex diagonally opposite O (in Quadrant III): B = (−5, −3).
5. The vertex along the y-axis: C = (0, −3).

1. D must align vertically with A and horizontally with C.
2. Abscissa of D = Abscissa of A = 2. Ordinate of D = Ordinate of C = 5.
3. So D = (2, 5).
4. AB = distance from x=−3 to x=2 = 5 units.
5. BC = distance from y=2 to y=5 = 3 units.
6. Area = AB × BC = 5 × 3 = 15 sq. units.

All twelve points plotted on the coordinate plane

1. For each point, move along the x-axis by the abscissa value (right if positive, left if negative), then vertically by the ordinate value (up if positive, down if negative).
2. Note: H, I, J, K and O all lie exactly on an axis since one of their coordinates is 0.

1. Draw AL ⊥ X′OX, meeting BC (extended if needed) at M.
2. BC is horizontal (B and C both at y=2): BC = distance from x=−2 to x=3 = 5 units (base).
3. AM = vertical distance from A(1,4) to the line y=2, so AM = 4 − 2 = 2 units (height).
ar(△ABC) = ½ × BC × AM
4. Area = ½ × 5 × 2 = 5 sq. units.

1. (i) There are 12 rows of tiles, so the top row's ordinate (y-value) is 12.
2. (ii) Following the cyclic colour pattern from tile (1,1) = Orange, tile (7,7) also lands back on Orange in the 4-colour cycle.
3. (iii) Tracing the pattern to position (2, 5) gives the colour Yellow.
4. OR: Rectangle OABX (from the figure) spans 8 columns × 6 rows = 48 tiles, each of area 1 foot², so total area = 48 foot².

1. (i) In (−1, 1): x is negative, y is positive ⇒ lies in Quadrant II.
2. (ii) The ordinate (y-coordinate) of (1, 0) is 0.
3. (iii) The abscissa (x-coordinate) of (0, 2) is 0.
4. OR: A point lies on the Y-axis when its abscissa is 0. Checking the list, only (0, 2) has x = 0, so it's the point on the Y-axis.

1. (i) Taking A as the origin, reading P's position on the grid: P = (3, 4).
2. (ii) D is the bottom-right corner of the rectangular plot: D = (10, 0) (taking A as origin).
3. (iii) If D is taken as the new origin instead of A, every x-coordinate shifts by subtracting 10 (since D is 10 units right of A along the same row): so P becomes (3−10, 4) = (−7, 4).
4. OR: Taking A as origin, R (from the figure) is located at R = (6, 3).

1. (i) B(3,4) and C(6,4) have the same y-value, so distance BC = |6−3| = 3 miles.
2. (ii) Since the walkway turns left at C and travels vertically (same direction pattern as AB), D shares the same x-coordinate as C. So the abscissa of D = 6.
3. (iii) Total distance = AB + BC + CD = 8 miles. AB = |4−1| = 3 miles, BC = 3 miles, so CD = 8 − 3 − 3 = 2 miles. Since D is above C, ordinate of D = 4 + 2 = 6.
4. OR: Combining both results, D = (6, 6).

1. (i) Deepak's path moves through negative-x, negative-y regions (Temple, Zoo, Home) — his drive covers Quadrant III and Quadrant IV.
2. (ii) Sanjay's path moves through positive-x, positive-y regions on the way to Hospital and School — his drive stays in Quadrant I.
3. (iii) Tracing Sanjay's turns from O(0,0): right 1 km ⇒ (1,0); left turn, up 2 km ⇒ (1,2); right turn, right 2 km ⇒ reaches the Hospital at (3, 2).
4. OR: Continuing from the Hospital (3,2), Sanjay drives 3 km more (up) to reach School at (3,5). Tracing Deepak's path similarly gives Zoo at (3,−2) and Home at (3,−4). All four locations — School(3,5), Hospital(3,2), Zoo(3,−2), Home(3,−4) — share the same abscissa, x = 3.