Ganita Manjari Class 9 Maths Chapter 1: Orienting Yourself: The Use of Coordinates — NCERT Solutions

Explanation: From the figure, A(3, 4), D(7, 1) and M(9, 6). For each side, form a right triangle whose legs are parallel to the axes: the horizontal leg is the difference of x-coordinates and the vertical leg is the difference of y-coordinates, and then apply the Pythagorean theorem, i.e. the distance formula sqrt((x2 - x1)² + (y2 - y1)²). AD: legs 7 - 3 = 4 and 4 - 1 = 3, so AD = sqrt(4² + 3²) = sqrt(25) = 5 units. DM: legs 9 - 7 = 2 and 6 - 1 = 5, so DM = sqrt(2² + 5²) = sqrt(29), about 5.39 units. MA: legs 9 - 3 = 6 and 6 - 4 = 2, so MA = sqrt(6² + 2²) = sqrt(40) = 2 sqrt(10), about 6.32 units.

Explanation: The sketch of Reiaan's room is a floor plan, i.e. the room seen from directly above, so it can record only two measurements: positions along the length and the breadth of the floor. Doors can be shown because their openings start at floor level, but windows are fixed on the vertical walls at some height above the floor. Specifying a window's position needs a third coordinate, its height above the floor, which a 2-D floor map has no way to represent, so the windows' positions cannot be marked on it.

Explanation: The horizontal distance is the difference of the x-coordinates: 7 - 3 = 4 units, moved parallel to the x-axis. The vertical distance is the difference of the y-coordinates: 4 - 1 = 3 units, moved parallel to the y-axis (downwards, since D is lower than A). So going from A(3, 4) to D(7, 1) means 4 units across and 3 units down.

Explanation: A floor plan shows the room as seen from above, so every object is located using only two measurements, along the length and the breadth of the floor. A window sits on a vertical wall at some height above the floor, so fixing its position needs a third measurement, the height. Since a 2-D map cannot show this third dimension, the positions of the windows cannot be indicated on it.

Explanation: Complete the right triangle by taking the point C(3, 1): AC is vertical with length 4 - 1 = 3 units and CD is horizontal with length 7 - 3 = 4 units, and the angle at C is 90 degrees since the segments are parallel to the axes. By the Pythagorean theorem, AD² = AC² + CD² = 3² + 4² = 9 + 16 = 25. Therefore AD = sqrt(25) = 5 units, so the two axis-parallel distances do determine the direct distance.

Explanation: Reflection in the y-axis maps A(3, 4) to A'(-3, 4), D(7, 1) to D'(-7, 1) and M(9, 6) to M'(-9, 6): each y-coordinate is unchanged while each x-coordinate changes sign. Applying the distance formula to the image gives A'D' = sqrt(4² + 3²) = 5, D'M' = sqrt(2² + 5²) = sqrt(29) and M'A' = sqrt(6² + 2²) = sqrt(40), exactly the lengths of AD, DM and MA, so the size, shape, angles and area are preserved. What changes is the position of the triangle (it now lies in the second quadrant) and its orientation: going round A' to D' to M' is the reverse sense of going round A to D to M, as expected for a mirror image.

Explanation: Read the hinge points off the plan: B1(0, 1.5) and B2(0, 4), so the door width is 4 - 1.5 = 2.5 ft. Hinged at B1 and swinging into the bedroom, every point of the door stays within 2.5 ft of (0, 1.5). The wardrobe occupies x from 3 to 7 and y from 0 to 2, so its closest point to the hinge is (3, 1.5), at a distance of 3 - 0 = 3 ft. Since the swing radius 2.5 ft is less than 3 ft, the door stops 0.5 ft short and never reaches the wardrobe. Widening the door only matters past 3 ft; beyond that the wardrobe must move further from the y-axis, or the door be re-hinged at B2, opened into the bathroom, or made a sliding door.

Explanation: (i) The bathroom is 6 ft x 9 ft and lies to the left of the y-axis, so its corners are O(0, 0), F(0, 9), R(-6, 9) and P(-6, 0). (ii) The showering area has corners S(-6, 6), H(-3, 6), W(-2, 9) and R(-6, 9); SH and RW are both horizontal, hence parallel, while HW is slanted, so exactly one pair of opposite sides is parallel and SHWR is a trapezium. (iii) Any placement inside the bathroom that avoids the showering area and the door B1B2 works: for example a 3 ft x 2 ft washbasin along the bottom wall with corners (-4, 0), (-1, 0), (-1, 2), (-4, 2), and a 2 ft x 3 ft toilet along the left wall with corners (-6, 2), (-4, 2), (-4, 5), (-6, 5); these regions do not overlap each other or the showering area.

Explanation: This is a measurement activity: measure the clear width of several doors at home and at school with a measuring tape and tabulate the results. Typical findings are that interior doors of living rooms and bedrooms are around 3 ft (about 90 cm) wide, bathroom and toilet doors are slightly narrower at about 2.5 ft (75 cm), and main doors are wider, about 3.5 ft to 4 ft, so that furniture and people can pass easily. So most room doors fall in the range of roughly 2.5 ft to 3.5 ft.

Explanation: Under reflection in the x-axis, each point (x, y) maps to (x, -y), so A(3, 4), D(7, 1), M(9, 6) go to (3, -4), (7, -1), (9, -6). The differences of coordinates between any two vertices keep the same absolute values, so by the distance formula every side length is unchanged (AD = 5, DM = sqrt(29), MA = sqrt(40)), and hence the angles, shape, size and area are all preserved, while the figure is again a mirror image with reversed orientation. The observations therefore still hold, with the roles of x and y swapped: previously the x-coordinates changed sign and the triangle moved to the second quadrant; now the y-coordinates change sign and the triangle moves to the fourth quadrant.

Explanation: This is a measurement-and-comparison activity. A typical wheelchair is about 60 cm to 70 cm wide, and accessibility guidelines recommend a clear doorway width of at least about 90 cm (roughly 3 ft) so that the chair can pass with room for the user's hands on the wheels. Measure the clear width of classroom, laboratory and toilet doors at school with a tape: doors of 3 ft (90 cm) or more can accommodate wheelchairs comfortably, while doors of about 2.5 ft (75 cm) or less are too narrow and would need widening.

Explanation: Plot Z(5, -6), which lies in the fourth quadrant, 5 units right of the y-axis and 6 units below the x-axis. Drop a perpendicular from Z to the x-axis to meet it at N(5, 0), and a perpendicular to the y-axis to meet it at I(0, -6); since IZ is horizontal and ZN is vertical, the angle at Z is 90 degrees, giving right triangle IZN. Then IZ = 5 - 0 = 5 units and ZN = 0 - (-6) = 6 units, and by the Pythagorean theorem the hypotenuse IN = sqrt(5² + 6²) = sqrt(25 + 36) = sqrt(61), about 7.81 units.

Explanation: The x-coordinate of a point measures its perpendicular distance from the y-axis (positive to the right, negative to the left). A point lying on the y-axis is at zero distance from it, so its x-coordinate must be 0. Hence every point on the y-axis has coordinates of the form (0, y).

Explanation: Method: compute the three pairwise distances with the distance formula; the points are collinear exactly when the sum of the two smaller distances equals the largest. MA = sqrt((0 - (-3))² + (0 - (-4))²) = sqrt(9 + 16) = sqrt(25) = 5 units; AG = sqrt(6² + 8²) = sqrt(36 + 64) = sqrt(100) = 10 units; MG = sqrt((6 - (-3))² + (8 - (-4))²) = sqrt(81 + 144) = sqrt(225) = 15 units. Since MA + AG = 5 + 10 = 15 = MG, the point A lies on the segment MG, so the three points are on the same straight line. (Equivalently, for both M and G the ratio y/x equals 4/3, so both lie on the same line through the origin A.)