Co-ordinate Geometry — Test Yourself

For point P(-5, 4):

1. abscissa = 4, ordinate = -5

2. ordinate = 4, abscissa = -5

3. abscissa = ordinate = -5

4. abscissa = ordinate = 4

Answer

Given, point P(-5, 4)

Abscissa (x-coordinate): -5

Ordinate (y-coordinate): 4

Hence, option 2 is the correct option.

If the perpendicular distance of a point P from the x-axis is 5 unit then the point P has

1. x-co-ordinate = -5

2. y-co-ordinate = 5 only

3. y-co-ordinate = -5 only

4. y-co-ordinate = 5 or -5

Answer

As we know that the distance of a point from the X-axis is the absolute value of its y-coordinate.

So, if perpendicular distance of a point from the x-axis is 5 units. Then the y-coordinate can be 5 or -5.

Hence, option 4 is the correct option.

A point lies on y-axis at a distance of 2 unit from x-axis. Its co-ordinates are :

1. (2, 0) only

2. (0, 2) only

3. (2, 2)

4. (0, 2) or (0, -2)

Answer

If a point lies on the y‑axis, its x‑coordinate is 0.

If it’s at a distance of 2 units from the x‑axis, that means its y-coordinate is either +2 or −2 (since distance ignores sign).

Its co-ordinates are (0, 2) or (0, -2).

Hence, option 4 is the correct option.

Three vertices of a square ABCD are A(2, 0), B(-3, 0) and C(-3, -5). Its fourth vertex D is:

1. (2, 5)

2. (2, -5)

3. (-2, 5)

4. (-2, -5)

Answer

The points are shown on the graph below:

Steps of construction :

1. Plot points A(2, 0), B(-3, 0) and C(-3, -5) on graph.

2. Measure AB.

3. Mark point D such that it is at a distance AB from points A and C.

4. Join AB, BC, CD and DA.

From graph,

The co-ordinates of fourth vertex D is (2, -5).

Hence, option 2 is the correct option.

Statement 1: In the given diagram, OAB is an equilateral triangle.

Co-ordinates of the vertex B = (4, 4)

Statement 2: B = (4, $4\sqrt{3}$)

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given equilateral triangle OAB.

OA = OB = AB = 8 units.

Draw BD ⊥ OA.

In an equilateral triangle, a perpendicular drawn from one of the vertices to the opposite side bisects the side.

∴ OD = $\dfrac{1}{2}$ x OA = ​$\dfrac{1}{2}$ x 8 = 4.

In right angle triangle OBD,

By pythagoras theorem,

⇒ OB² = OD² + BD²

⇒ 8² = 4² + BD²

⇒ 64 = 16 + BD²

⇒ BD² = 64 - 16

⇒ BD² = 48

⇒ BD = $\sqrt{48}$

⇒ BD = $4\sqrt{3}$

From graph,

Co-ordinates of O = (0, 0)

Co-ordinates of A = (8, 0)

As, OD = 4 units and BD = $4\sqrt{3}$ units.

Co-ordinates of B = (4, $4\sqrt{3}$).

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Statement 1: The vertex B of square OABC with each side 4 units lies in the fourth quadrant and its side are along the co-ordinate axes. The co-ordinate of vertex B are (4, -4).

Statement 2: B = (4, 4)

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Vertex B lies in the fourth quadrant. The square's sides are along the coordinate axes, so one vertex, O, is at the origin (0, 0).

The side length is 4 units.

With O at (0, 0), the other two vertices on the axes must be A (4, 0) on the positive x-axis and C (0, -4) on the negative y-axis.

The fourth quadrant is where the x-coordinates are positive and the y-coordinates are negative.

Vertex B is located at the point where the line extending 4 units to the right from C (0, -4) intersects with the line extending 4 units down from A (4, 0), which is (4, -4).

⇒ B = (4, –4).

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Assertion (A): PQR is an equilateral triangle. The co-ordinates of point Q are (0, $2\sqrt{2}$).

Reason (R): In ΔOPQ,

OQ² = PQ² - OP² = 4² - 2² = 12.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

The coordinates of R(0, 2) and P(0, -2).

Using distance formula,

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

PR = $\sqrt{(0 - 0)^2 + (-2 - 2)^2} = \sqrt{(-4)^2} = \sqrt{16} = 4$ units.

Given equilateral triangle PQR.

∴ PQ = QR = PR = 4 units.

In an equilateral triangle, a perpendicular drawn from one of the vertices to the opposite side bisects the side.

∴ OP = $\dfrac{1}{2}$ x PR = ​$\dfrac{1}{2}$ x 4 = 2 units.

In right angle triangle OPQ,

By pythagoras theorem,

⇒ QP² = OP² + OQ²

⇒ OQ² = QP² - OP²

⇒ OQ² = 4² - 2²

⇒ OQ² = 16 - 4

⇒ OQ² = 12

⇒ OQ = $\sqrt{12}$

⇒ OQ = $2\sqrt{3}$

Since, OQ = $2\sqrt{3}$ units and Q lies on x-axis.

Co-ordinates of Q = ($2\sqrt{3}$, 0).

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Assertion (A): (2x - 3y, 8) = (2, x + 2y)

⇒ x = 1 and y = -2

Reason (R): 2x - 3y = 2 and 8 = x + 2y

which on solving give x = 4 and y = 2.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, (2x - 3y, 8) = (2, x + 2y)

Thus,

2x - 3y = 2 ........(1)

x + 2y = 8 ........(2)

So, reason (R) is true.

Multiplying equation (2) by 2, we get :

⇒ 2(x + 2y) = 2 × 8

⇒ 2x + 4y = 16 ........(3)

Subtracting equation (1) from (3), we get :

⇒ 2x + 4y - (2x - 3y) = 16 - 2

⇒ 2x - 2x + 4y + 3y = 14

⇒ 7y = 14

⇒ y = $\dfrac{14}{7}$ = 2.

Substituting the value y = 2 in first equation,

⇒ 2x - 3 x 2 = 2

⇒ 2x - 6 = 2

⇒ 2x = 2 + 6

⇒ 2x = 8

⇒ x = $\dfrac{8}{2}$ = 4.

Hence, the value of x = 4 and y = 2.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

By plotting the following points on the same graph paper, check whether they are collinear or not :

(3, 5), (1, 1) and (0, -1)

Answer

(3, 5), (1, 1) and (0, -1)

As shown in the figure, all three points lie on the same straight line.

Hence, the points are collinear.

By plotting the following points on the same graph paper, check whether they are collinear or not :

(-2, -1), (-1, -4) and (-4, 1)

Answer

(-2, -1), (-1, -4) and (-4, 1)

As shown in the figure, all three points do not lie on the same straight line.

Hence, the points are not collinear.

Plot the point A (5, -7). From point A, draw AM perpendicular to x-axis and AN perpendicular to y-axis. Write the co-ordinates of points M and N.

Answer

Given point : A (5, -7)

AM is perpendicular to the x-axis.

AN is perpendicular to the y-axis.

The graph intersects the x-axis at M(5, 0).

The graph intersects the y-axis at N(0, -7).

Hence, the coordinates of the points are M = (5, 0) and N = (0, -7).

In square ABCD; A = (3, 4), B = (-2, 4) and C = (-2, -1). By plotting these points on a graph paper, find the co-ordinates of vertex D. Also, find the area of the square.

Answer

Plot the points A = (3, 4), B = (-2, 4) and C = (-2, -1) 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 (3, 4) and B (-2, 4) is 5 units and the vertical distance between the points B (-2, 4) and C (-2, -1) is 5 units. Therefore, the vertical distance between the points A (3, 4) and D must be 5 units and the horizontal distance between the points C (-2, -1) and D must be 5 units.

Now, complete the square ABCD and read the coordinates of point D, as shown on the graph, D = (3, -1).

Area of the square ABCD = AD x CD

= 5 x 5 square units

= 25 square units

Hence, the coordinates of D = (3, -1) and the area of the square = 25 square units.

In rectangle OABC; point O is the origin, OA = 10 units along x-axis and AB = 8 units. Find the co-ordinates of vertices A, B and C.

Answer

Given that in rectangle OABC, point O is the origin and OA = 10 units along x-axis, therefore we have O(0, 0) and A(10, 0).

It is also given that AB = 8 units. Since AB is perpendicular to OA (as OABC is a rectangle), the line through A is a vertical line and it meets B at B(10, 8).

Through point B, draw a horizontal line (parallel to x-axis) that meets the y-axis at y = 8. This gives the coordinates of point C: (0, 8).

Thus, the vertices of rectangle OABC are: O(0, 0), A(10, 0), B(10, 8) and C(0, 8)

Hence, the coordinates of the vertices are A = (10, 0), B = (10, 8) and C = (0, 8).

Draw the graph of equation x + 2y - 3 = 0. From the graph, find :

(i) x₁, the value of x, when y = 3

(ii) x₂, the value of x, when y = -2.

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) + 2y - 3 = 0 ⇒ y = 2

Let x = 0, then 0 + 2y - 3 = 0 ⇒ y = 1.5

Let x = 1, then 1 + 2y - 3 = 0 ⇒ y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line AB passing through the points plotted on the graph.

(i) To find x₁, the value of x, when y = 3:

Through the point y = 3, draw a horizontal straight line which meets the line AB at point C. Through point C, draw a vertical line which meets the x-axis at x = -3.

Hence, the value of x, when y = 3 is -3 , i.e, x₁ = -3.

(ii) To find x₂, the value of x, when y = -2:

Through the point y = -2, draw a horizontal straight line which meets the line AB at point D. Now, through point D, draw a vertical line which meets the y-axis at x = 7.

Hence, the value of x, when y = -2 is 7 , i.e, x₂ = 7.

Draw the graph of equation 3x - 4y = 12. Use the graph drawn to find :

(i) y₁, the value of y, when x = 4

(ii) y₂, the value of y, when x = 0.

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then 3 $\times$ (-2) - 4y = 12 ⇒ y = -4.5

Let x = 1, then 3 $\times$ 1 - 4y = 12 ⇒ y = -2.2

Let x = 2, then 3 $\times$ 2 - 4y = 12 ⇒ y = -1.5

Let x = 4, then 3 $\times$ 4 - 4y = 12 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

(i) To find y₁, the value of y, when x = 4:

Through the point x = 4, draw a horizontal line which meets the y-axis at y = 0.

Hence, the value of y, when x = 4 is 0 , i.e, y₁ = 0.

(ii) To find y₂, the value of y, when x = 0:

Through the point x = 0, draw a vertical line which meets the y-axis at x = -3.

Hence, the value of y, when x = 0 is -3 , i.e, y₂ = -3.

Draw the graph of equation $\dfrac{x}{4} + \dfrac{y}{5} = 1$. Use the graph drawn to find :

(i) x₁, the value of x, when y = 10

(ii) y₁, the value of y, when x = 8.

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then $\dfrac{-4}{4} + \dfrac{y}{5} = 1$ ⇒ y = 10

Let x = 0, then $\dfrac{0}{4} + \dfrac{y}{5} = 1$ ⇒ y = 5

Let x = 4, then $\dfrac{4}{4} + \dfrac{y}{5} = 1$ ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line AB passing through the points plotted on the graph.

(i) To find x₁, the value of x, when y = 10:

Through the point y = 10, draw a horizontal straight line which meets the line AB at point C.

Through point C, draw a vertical line which meets the x-axis at x = -4.

Hence, the value of x, when y = 10 is -4 , i.e, x₁ = -4.

(ii) To find y₁, the value of y, when x = 8:

Through the point x = 8, draw a vertical line which meets the line AB at point D.

Now, through point D, draw a horizontal line which meets the y-axis at y = -5.

Hence, the value of y, when x = 8 is -5 , i.e, y₁ = -5.

Use the graphical method to show that the straight lines given by the equations $x + y = 2, x - 2y = 5$ and $\dfrac{x}{3} + y = 0$ pass through the same point.

Answer

First equation : x + y = 2

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) + y = 2 ⇒ y = 3

Let x = 0, then 0 + y = 2 ⇒ y = 2

Let x = 1, then 1 + y = 2 ⇒ y = 1

Let x = 3, then 3 + y = 2 ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation : x - 2y = 5

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) - 2y = 5 ⇒ y = -3

Let x = 0, then 0 - 2y = 5 ⇒ y = -2.5

Let x = 1, then 1 - 2y = 5 ⇒ y = -2

Let x = 3, then 3 - 2y = 5 ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Third equation : $\dfrac{x}{3} + y = 0$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -3, then $\dfrac{-3}{3} + y = 0$ ⇒ y = 1

Let x = 0, then $\dfrac{0}{3} + y = 0$ ⇒ y = 0

Let x = 3, then $\dfrac{3}{3} + y = 0$ ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the three straight lines pass through the same point.

Draw the graph of line x + y = 5. Use the graph paper drawn to find the inclination and the y-intercept of the line.

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = (-1), then (-1) + y = 5 ⇒ y = 6

Let x = 0, then 0 + y = 5 ⇒ y = 5

Let x = 1, then 1 + y = 5 ⇒ y = 4

Let x = 5, then 5 + y = 5 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph, inclination of the line = θ = 135° and y-intercept of the line = OB = 5.

Hence, inclination = 135° and y-intercept of the line = 5.

Draw the graph of line 2x + y = 5.

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 2 $\times$ (-1) + y = 5 ⇒ y = 7

Let x = 0, then 2 $\times$ 0 + y = 5 ⇒ y = 5

Let x = 1, then 2 $\times$ 1 + y = 5 ⇒ y = 3

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Draw the graph of line 4x - y = 5. Use this graph to find :

(i) x₁, the value of x when y = 3.

(ii) y₁, the value of y, when x = 3.

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 4, then 4 $\times$ (4) - y = 5 ⇒ y = 11

Let x = 3.5, then 4 $\times$ 3.5 - y = 5 ⇒ y = 9

Let x = 1, then 4 $\times$ 1 - y = 5 ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line AB passing through the points plotted on the graph.

(i) To find x₁, the value of x, when y = 3:

Through the point y = 3, draw a horizontal straight line which meets the line AB at point C.

Through point C, draw a vertical line which meets the x-axis at x = 2.

Hence, the value of x, when y = 3 is 2 , i.e, x₁ = 2.

(ii) To find y₁, the value of y, when x = 3:

Through the point x = 3, draw a vertical line which meets the line AB at point D.

Through point D, draw a horizontal line which meets the y-axis at y = 7.

Hence, the value of y, when x = 3 is 7 , i.e, y₁ = 7.