Telangana SCERT Class 9 Math Solution Chapter 5 Co-Ordinate Geometry Exercise 5.3

Telangana SCERT Solution Class IX (9) Math Chapter 5 Co-Ordinate Geometry Exercise 5.3

(1) Given,

X 2 3 -1 0 -1 4
Y -3 -3 4 11 0 -6
(x ,y) (2 , -3) (3, – 3) (-1 , 4) (0, 11) (-1 , 0) (4 , -6)

The above points are plotted on the graph

(2) Are the positions of (5, −8) and (−8, 5) same? Justify your answer

No, (5, -8) and (-8, 5) are not same because (5, – 8) lies on the 4th Quadrant and (-8, 5) lies on the 2nd Quadrant.

(3) What can you say about the position of the points (1, 2), (1, 3), (1, −4), (1, 0), and (1, 8). Locate on a graph sheet

Plotting positions (1, 2),(1, 3), (1, – 4), (1, 0) and (1, 8) on graph sheet.

By plotting all the points we see that,

(1, 2) (1,3) and (1, 8) and the three point that lies on the first quadrant

Point (1, -4) lies on the 4th quadrant

Point(1, 0) lies on the x – axis

Now, if we connect all the paints we get a line segment connecting all the points from 1st quadrant to the 4th quadrant parallel to the y axis

Therefore, all the abscissa of the given paints are same that is 1.

(4) What can you say about the position of the points (5, 4), (8, 4), (3, 4), (0, 4), (−4, 4), (−2, 4)? Locate the points on a graph sheet. Justify your answer.

Plotting the given, points on the graph sheet.

(5, 4) (8, 4) (3, 4) (0, 4) (-4, 4) (-2, 4)

From the graph sheet we can say that that,

(i) points (3, 4) (5, 4) and (8, 4) lies on the 1st quadrant

(ii) point (0, 4) lies on the Y axis

(iii) points (-4, 4) and (-2, 4) lies on the 2nd quadrant

(iv) when we connect all the point we get a line segment – connecting point (8, 4) to point (-4, 4) parallel to the x – axis

(v) Therefore, all the ordinates of the points are same that’s 4

(5) Plot the points (0, 0) (0, 3) (4, 3) (4, 0) in graph sheet. Join the points in order with straight lines to make a rectangle. Find the area of the rectangle.

Points to plot are (0, 0) (0, 3) (4, 3) (4, 0)

From the graph we can see that,

On joining all the points in a graph line we get a rectangle.

Length of the rectangle is 4 units

Breadth of the rectangle is 3 units

Therefore, Area of the rectangle = length x breadth

= 4 x 3

= 12 units2

(6) Plot the points (2, 3), (6, 3) and (4, 7) in a graphsheet. Join them to make it a triangle. Find the area of the triangle.

Points to plot (2, 3) (6, 3) and (4, 7)

From the graph we can see that

By connectivity all the points we get a triangle

Now, on producing a parallel line from point (4, 7) to the box of the triangle we get a point (4, 3) of the base.  Which is also perpendicular to the base of the triangle.

The (h height of the triangle is the difference between the ordinates of point (4, 7) and point (4, 3) at the base

= 7 – 3

= 4 units

(b) Base of the triangle is the difference between the abscissa of points (2, 3) and (6, 3)

= 6 – 2

= 4 units

Therefore, Area of the triangle = ½ x b x h

= ½ x 4 x 4

= 8 units2

(7) Plot at least six points in a graph sheet, each having the sum of its coordinates equal to 5.

Six points whose seem of the coordinates is 5 are: –

(-2, 7) (1, 4) (3, 2) (-1, 6) (2, 3) (4, 1)

Now, plotting these points on graph

On plotting all the point on the graph we can see that all the points connects with each other through a straight line segment passing from 2nd quadrant to the 1st quadrant

(8) From the given graph,

Coordinates of A = (-3, 4)

Coordinates of B = (0, 5)

Coordinates of C = (3, 4)

Coordinates of D = (2, 4)

Coordinates of E = (2, 0)

Coordinates of F = (3, 0)

Coordinates of G = (3, -1)

Coordinates of H = (0, -1)

Coordinates of I = (-3, -1)

Coordinates of J = (-3, 0)

Coordinates of K = (-2, 0)

Coordinates of L = (-2, 4)

Coordinates of M = (-1, 0)

Coordinates of N = (-1, 3)

Coordinates of O = (0, 0)

Coordinates of Q = (1, 0)

(9) To plot,

(i) (2, 5) (4, 7)

(ii) (-3, 5) (-1, 7)

(iii) (-3, -4) (2, -4)

(iv) -3, -5) (2, -5)

(v) (4, -2) (4, -3)

(vi) (-2, 4), (-2, 3)

(vii) (-2, 1) (-2, 0)

(viii) (4, 7) (4,  -3)

(ix) (4, -2) (2, -4)

(x) (4, -3) (2, -5)

(xi) (2, 5) (2, -5)

(xii) (-3, 5) (-3, -5)

(xiii) (-3, 5) (2, 5)

(xiv) (-1, 7) (4, 7)

On plotting above points the figure is a cuboid

The absent figure also seems to resemble a refrigerator with lines (vi)  and (vii) representing its handles.

(10) Points to plot

(1, 0), (0,4); (2, 0), (0 , 8); (3, 0), (-, 7); (4, 0), (0, 6); (5, 0) (0, 5); (6, 0), (0, 4); (7, 0) (0, 3); (8, 0) (0, 2): (9, 0), (0, 1)

The figure upon plotting these points is known as a Quadratic Bezier curve which is a segment of a parabola.

Updated: September 21, 2021 — 4:36 pm

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