Co-ordinate Geometry — Exercise 23(A)

In equation 2x - 5 = 0; ordinate is :

1. 0

2. 2

3. -5

4. 1

Answer

In the equation 2x − 5 = 0, we need to find the ordinate, which means the y-coordinate of a point on the graph of this equation.

2x - 5 = 0

⇒ 2x = 5

⇒ x = $\dfrac{5}{2}$

⇒ x = 2.5

Notice that this equation only involves x and has no y-term. This tells us that the equation represents a vertical line parallel to y-axis at x = 2.5. Hence, ordinate = 0 (i.e., y = 0).

Hence, option 1 is the correct option.

In equation $y = 3 + \dfrac{4x}{5}$, the independent variable is :

1. y

2. x

3. $\dfrac{4x}{5}$

4. $3 + \dfrac{4x}{5}$

Answer

Given:

$y = 3 + \dfrac{4x}{5}$

In the above equation, the value of y depends on the value of x, so y is said to be dependent variable and x is said to be independent variable.

Hence, option 2 is the correct option.

If (2x - 1, y + 5) = (3, 10), the values of x and y are :

1. x = 4, y = 5

2. x = 8, y = 5

3. x = 2, y = 5

4. x = 1, y = 5

Answer

Since two ordered pairs are equal, their first components are equal and their second components are separately equal.

(2x - 1, y + 5) = (3, 10)

⇒ 2x - 1 = 3 and y + 5 = 10

⇒ 2x = 3 + 1 and y = 10 - 5

⇒ 2x = 4 and y = 5

⇒ x = $\dfrac{4}{2}$ and y = 5

⇒ x = 2 and y = 5

Hence, option 3 is the correct option.

(5, -7) belongs to :

1. 1st quadrant

2. 2nd quadrant

3. 3rd quadrant

4. 4th quadrant

Answer

The point (5, -7) has a positive x-coordinate and a negative y-coordinate. This places it in the 4th quadrant.

Hence, option 4 is the correct option.

Abscissa of a point is the solution of equation 3x - 2 = 7 and its ordinate is the solution of equation 8 - 3y = 2. The point is :

1. (3, 2)

2. (-3, 2)

3. (3, -2)

4. (-3, -4)

Answer

Given:

3x - 2 = 7

⇒ 3x = 7 + 2

⇒ 3x = 9

⇒ x = $\dfrac{9}{3}$

⇒ x = 3

And, 8 - 3y = 2

⇒ 3y = 8 - 2

⇒ 3y = 6

⇒ y = $\dfrac{6}{3}$

⇒ y = 2

Therefore, the co-ordinates of the point = (3, 2).

Hence, option 1 is the correct option.

For each equation given below; name the dependent and independent variables.

(i) $y = \dfrac{4}{3}x - 7$

(ii) x = 9y + 4

(iii) $x = \dfrac{5y + 3}{2}$

(iv) $y = \dfrac{1}{7}(6x + 5)$

Answer

(i) $y = \dfrac{4}{3}x - 7$

In the above equation, the value of y depends on the value of x, so y is said to be dependent variable and x is said to be independent variable.

Hence, x = independent variable and y = dependent variable.

(ii) x = 9y + 4

In the above equation, the value of x depends on the value of y, so x is said to be dependent variable and y is said to be independent variable.

Hence, x = dependent and y = independent.

(iii) $x = \dfrac{5y + 3}{2}$

In the above equation, the value of x depends on the value of y, so x is said to be dependent variable and y is said to be independent variable.

Hence, x = dependent and y = independent.

(iv) $y = \dfrac{1}{7}(6x + 5)$

In the above equation, the value of y depends on the value of x, so y is said to be dependent variable and x is said to be independent variable.

Hence, x = independent and y = dependent.

Plot the following points on the same graph paper :

(i) (8, 7)

(ii) (3, 6)

(iii) (0, 4)

(iv) (0, -4)

(v) (3, -2)

(vi) (-2, 5)

(vii) (-3, 0)

(viii) (5, 0)

(ix) (-4, -3)

Answer

Find the values of x and y if :

(i) (x - 1, y + 3) = (4, 4)

(ii) (3x + 1, 2y - 7) = (9, - 9)

(iii) (5x - 3y, y - 3x) = (4, - 4)

Answer

Since two ordered pairs are equal, their first components are equal and their second components are separately equal.

(i) (x - 1, y + 3) = (4, 4)

⇒ x - 1 = 4 and y + 3 = 4

⇒ x = 4 + 1 and y = 4 - 3

⇒ x = 5 and y = 1

Hence, x = 5 and y = 1.

(ii) (3x + 1, 2y - 7) = (9, - 9)

⇒ 3x + 1 = 9 and 2y - 7 = -9

⇒ 3x = 9 - 1 and 2y = -9 + 7

⇒ 3x = 8 and 2y = -2

⇒ x = $\dfrac{8}{3}$ and y = -$\dfrac{2}{2}$

⇒ x = $2\dfrac{2}{3}$ and y = -1

Hence, x = $2\dfrac{2}{3}$ and y = -1.

(iii) (5x - 3y, y - 3x) = (4, - 4)

⇒ 5x - 3y = 4 ................(1)

And, y - 3x = - 4

⇒ y = -4 + 3x ................(2)

Putting the value of y in (1), we get

⇒ 5x - 3(-4 + 3x) = 4

⇒ 5x + 12 - 9x = 4

⇒ 5x - 9x = 4 - 12

⇒ - 4x = - 8

⇒ x = $\dfrac{8}{4}$

⇒ x = 2

Putting the value of x in (2),

y = - 4 + 3x

⇒ y = - 4 + 3 x 2

⇒ y = - 4 + 6

⇒ y = 2

Hence, x = 2 and y = 2.

Use the graph given below, to find the co-ordinates of the point (s) satisfying the given conditions :

(i) the abscissa is 2.

(ii) the ordinate is 0.

(iii) the ordinate is 3.

(iv) the ordinate is - 4.

(v) the abscissa is 5.

(vi) the abscissa is equal to the ordinate.

(vii) the ordinate is half of the abscissa.

Answer

(i) The abscissa is 2.

The abscissa represents the x-coordinate, so x = 2.

Observing the graph, the y-coordinate for this point is also 2.

Hence, the point is A = (2, 2).

(ii) The ordinate is 0.

The ordinate represents the y-coordinate, so y = 0.

Observing the graph, the x-coordinate for this point is 5.

Hence, the point is B = (5, 0).

(iii) The ordinate is 3.

The ordinate represents the y-coordinate, so y = 3.

Observing the graph, the x-coordinates for this point are -4 and 6.

Hence, the points are C = (-4, 3) and E = (6, 3).

(iv) The ordinate is - 4.

The ordinate represents the y-coordinate, so y = - 4.

Observing the graph, the x-coordinate for this point is 2.

Hence, the point is D = (2, -4).

(v) The abscissa is 5.

The abscissa represents the x-coordinate, so x = 5.

Observing the graph, the y-coordinates for this point are 0, 5 and -3.

Hence, the points are B = (5, 0), H = (5, 5) and G = (5, -3).

(vi) The abscissa is equal to the ordinate.

The abscissa represents the x-coordinate, and ordinate represents the y-coordinate so x = y.

Observing the graph, the points are (2,2), (5, 5) and (4, 4).

Hence, the points are A = (2, 2), H = (5, 5) and I = (4, 4).

(vii) The ordinate is half of the abscissa.

The abscissa represents the x-coordinate, and ordinate represents the y-coordinate so y = $\dfrac{\text{x}}{2}$.

Observing the graph, the point is (6, 3).

Hence, the point is E = (6, 3).

State, true or false :

(i) The ordinate of a point is its x-co-ordinate.

(ii) The origin is in the first quadrant.

(iii) The y-axis is the vertical number line.

(iv) Every point is located in one of the four quadrants.

(v) If the ordinate of a point is equal to its abscissa; the point lies either in the first quadrant or in the second quadrant.

(vi) The origin (0, 0) lies on the x-axis.

(vii) The point (a, b) lies on the y-axis if b = 0.

Answer

(i) False

Reason :

The ordinate is the y-co-ordinate of a point.

(ii) False

Reason :

The origin (0, 0) is the point where the x-axis and y-axis intersect.

(iii) True

Reason :

The y-axis is the vertical number line and the x-axis is the horizontal number line in the coordinate plane.

(iv) False

Reason :

Points that lie on the x-axis or y-axis are not located in any of the four quadrants.

(v) False

Reason :

If the ordinate (y-coordinate) equals the abscissa (x-coordinate), (x = y), the point lies either in the first quadrant (x > 0, y > 0) or the third quadrant (x < 0, y < 0).

(vi) True

Reason :

The origin (0, 0) lies on both the x-axis and y-axis, as it is the intersection point of the two axes.

(vii) False

Reason :

The point (a, b) lies on the y-axis if the x-coordinate a = 0, not b = 0.

In each of the following, find the co-ordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation :

(i) 3 - 2x = 7; 2y + 1 = 10 - $2 \dfrac{1}{2}y$.

(ii) $\dfrac{2a}{3} - 1 = \dfrac{a}{2}; \dfrac{15 - 4b}{7} = \dfrac{2b - 1}{3}$

(iii) $5x -(5 - x) = \dfrac{1}{2}(3 - x); 4 - 3y = \dfrac{4 + y}{3}$

Answer

(i) 3 - 2x = 7

⇒ 2x = 3 - 7

⇒ 2x = - 4

⇒ x = - $\dfrac{4}{2}$

⇒ x = - 2

2y + 1 = 10 - $2 \dfrac{1}{2}y$

⇒ 2y + $\dfrac{5}{2}y$ = 10 - 1

⇒ $\dfrac{4 + 5}{2}y$ = 9

⇒ $\dfrac{9}{2}y$ = 9

⇒ y = $\dfrac{2 \times 9}{9}$

⇒ y = 2

Hence, the co-ordinates of the point = (-2, 2).

(ii)

$\dfrac{2a}{3} - 1 = \dfrac{a}{2}\\[1em] ⇒ \dfrac{2a}{3} - \dfrac{a}{2} = 1\\[1em] ⇒ \dfrac{4a - 3a}{6} = 1\\[1em] ⇒ \dfrac{a}{6} = 1\\[1em] ⇒ a = 6\\[1em]$

$\dfrac{15 - 4b}{7} = \dfrac{2b - 1}{3}\\[1em] ⇒ 3(15 - 4b) = 7(2b - 1)\\[1em] ⇒ 45 - 12b = 14b - 7\\[1em] ⇒ 45 + 7 = 14b + 12b\\[1em] ⇒ 52 = 26b\\[1em] ⇒ b = \dfrac{52}{26}\\[1em] ⇒ b = 2\\[1em]$

Hence, the co-ordinates of the point = (6, 2).

(iii)

$5x -(5 - x) = \dfrac{1}{2}(3 - x)\\[1em] ⇒ 5x - 5 + x = \dfrac{1}{2}(3 - x) \\[1em] ⇒ 6x - 5 = \dfrac{1}{2}(3 - x) \\[1em] ⇒ 2(6x - 5) = 3 - x\\[1em] ⇒ 12x + x = 3 + 10 \\[1em] ⇒ 13x = 13 \\[1em] ⇒ x = \dfrac{13}{13}\\[1em] ⇒ x = 1 \\[1em]$

$4 - 3y = \dfrac{4 + y}{3}\\[1em] ⇒ 3(4 - 3y) = 4 + y\\[1em] ⇒ 12 - 9y = 4 + y\\[1em] ⇒ 12 - 4 = 9y + y\\[1em] ⇒ 8 = 10y\\[1em] ⇒ y = \dfrac{8}{10}\\[1em] ⇒ y = \dfrac{4}{5}\\[1em]$

Hence, the co-ordinates of the point = $\Big(1, \dfrac{4}{5}\Big)$.

In each of the following, the co-ordinates of the three vertices of a rectangle ABCD are given. By plotting the given points; find, in each case, the co-ordinates of the fourth vertex :

(i) A (2, 0), B (8, 0) and C (8, 4).

(ii) A (4, 2), B (-2, 2) and D (4, -2).

(iii) A (- 4, - 6), C (6, 0) and D (- 4, 0)

(iv) B (10, 4), C (0, 4) and D (0, - 2).

Answer

(i) After plotting the given points A(2, 0), B(8, 0) and C(8, 4) on a graph paper, join A with B and B with C. From the graph, it is clear that the vertical distance between the points B(8, 0) and C(8, 4) is 4 units. Therefore, the vertical distance between points A(2, 0) and D must be 4 units.

Now, complete the rectangle ABCD and read the coordinates of point D. As shown on the graph, D = (2, 4).

Hence, D = (2, 4).

(ii) After plotting the given points A(4, 2), B(-2, 2) and D(4, -2) on a graph paper, join A with B and A with D.

From the graph it is clear that the vertical distance between the points A(4, 2) and D(4, -2) is 4 units and the horizontal distance between the points A(4, 2) and B(-2, 2) is 6 units. Therefore, the vertical distance between points B(-2, 2) and C must be 4 units and the horizontal distance between points B(-2, 2) and C must be 6 units.

Now, complete the rectangle ABCD and read the coordinates of point C. As shown on the graph, C = (-2, -2).

Hence, C = (-2, -2).

(iii) After plotting the given points A(-4, -6), C(6, 0) and D(-4, 0) on a graph paper, join A with D and C with D.

From the graph it is clear that the vertical distance between the points A(-4, -6) and D(-4, 0) is 6 units and the horizontal distance between the points C(6, 0) and D(-4, 0) is 10 units. Therefore, the vertical distance between the points A(-4, -6) and B must be 10 units and the horizontal distance between the points C(6, 0) and B must be 10 units.

Now complete the rectangle ABCD and read the coordinates of point B. As shown on the graph, B = (6, -6).

Hence, B = (6, -6).

(iv) After plotting the given points B (10, 4), C (0, 4) and D (0, - 2) on a graph paper, join B with C and C with D.

From the graph it is clear that the vertical distance between the points C(0, 4) and D(0, -2) is 6 units and the horizontal distance between the points C(0, 4) and D(0, -2) is 10 units. Therefore, the vertical distance between the points B(10, 4) and A must be 6 units and the horizontal distance between the points D(0, -2) and B must be 10 units.

Now complete the rectangle ABCD and read the coordinates of point A. As shown on the graph, A = (10, -2).

Hence, A = (10, -2).

A (-2, 2), B (8, 2) and C (4, -4) are the vertices of a parallelogram ABCD. By plotting the given points on a graph paper; find the co-ordinates of the fourth vertex D.

Also, from the same graph, state the co-ordinates of the mid-points of the sides AB and CD.

Answer

Plot the points A (-2, 2), B (8, 2) and C (4, -4) on the graph paper. Join point A with B and B with C.

From the graph, it is clear that the horizontal distance between the points A (-2, 2) and B (8, 2) is 10 units and the vertical distance between the points B (8, 2) and C (4, -4) is 6 units. Therefore, the vertical distance between the points A (-2, 2) and D must be 6 units and the horizontal distance between the points C (4, -4) and D must be 10 units.

Now, complete the parallelogram ABCD and read the coordinates of point D. As shown on the graph, D = (-6, -4).

The midpoint of AB lies exactly halfway between A(-2, 2) and B(8, 2). On the graph, this midpoint is at (3, 2), as it is 5 units from both A and B.

The midpoint of CD lies exactly halfway between C (4, -4) and B (-6, -4). On the graph, this midpoint is at (-1, -4), as it is 5 units from both C and D.

Hence, D = (-6, -4) and mid point of AB = (3, 2) and CD = (-1, -4).

A (- 2, 4), C (4, 10) and D (- 2, 10) are the vertices of a square ABCD. Use the graphical method to find the co-ordinates of the fourth vertex B. Also, find :

(i) the co-ordinates of the mid-point of BC;

(ii) the co-ordinates of the mid-point of CD and

(iii) the co-ordinates of the point of intersection of the diagonals of the square ABCD.

Answer

Plot the points A (- 2, 4), C (4, 10) and D (- 2, 10) on the graph paper. Join point A with D and D with C.

From the graph, it is clear that the horizontal distance between the points C (4, 10) and D (-2, 10) is 6 units and the vertical distance between the points A (-2, 4) and D (-2, 10) is 6 units. Therefore, the vertical distance between the points C (4, 10) and B must be 6 units and the horizontal distance between the points A (-2, 4) and B must be 6 units.

Now, complete the square ABCD and read the coordinates of point B, as shown on the graph, B = (4, 4).

The midpoint of BC lies exactly halfway between B(4, 4) and C(4, 10). On the graph, this midpoint is at E(4, 7), as it is 3 units from both B and C.

The midpoint of CD lies exactly halfway between C(4, 10) and D(-2, 10). On the graph, this midpoint is at F(1, 10), as it is 3 units from both C and D.

The coordinates of the midpoint of diagonals of the square is G = (1, 7).

Hence, the co-ordinates of the mid-point of BC = (4, 7), the co-ordinates of the mid-point of CD = (1, 10) and the co-ordinates of the point of intersection of the diagonals of the square ABCD = (1, 7).