Coordinate Geometry — Exercise 18.1

Find the coordinates of points whose

(i) abscissa is 3 and ordinate -4.

(ii) abscissa is $-\dfrac{3}{2}$ and ordinate 5.

(iii) whose abscissa is $-1\dfrac{2}{3}$ and ordinate $-2\dfrac{1}{4}$.

(iv) whose ordinate is 5 and abscissa is -2.

(v) whose abscissa is -2 and lies on x-axis.

(vi) whose ordinate is $\dfrac{3}{2}$ and lies on y-axis.

Answer

We know that,

Abscissa is the x-coordinate and ordinate is the y-coordinate of a point.

(i) The coordinate of the point whose abscissa is 3 and ordinate is -4 is (3, -4).

(ii) The coordinate of the point whose abscissa is $-\dfrac{3}{2}$ and ordinate is 5 is $\Big(-\dfrac{3}{2}, 5)$.

(iii) The coordinate of the point whose whose abscissa is $-1\dfrac{2}{3}$ and ordinate $-2\dfrac{1}{4}$ is $\Big(-1\dfrac{2}{3}, -2\dfrac{1}{4}\Big)$.

(iv) The coordinate of the point whose ordinate is 5 and abscissa is -2 is (-2, 5).

(v) Since, point lies on x-axis. So, ordinate of point will be = 0.

The coordinate of the point whose abscissa is -2 and lies on x-axis is (-2,0).

(vi) Since, point lies on y-axis. So, abscissa of point will be = 0.

The coordinate of the point whose ordinate is $\dfrac{3}{2}$ and lies on y-axis is $\Big(0, \dfrac{3}{2}\Big)$.

In which quadrant or on which axis each of the following points lie?

(-3, 5), (4, -1) (2, 0), (2, 2), (-3, -6)

Answer

From figure,

(-3, 5) lies in second quadrant.

(4, -1) lies in fourth quadrant.

(2, 0) lies on x-axis.

(2, 2) lies in first quadrant.

(-3, -6) lies in third quadrant.

Which of the following points lie on (i) x-axis? (ii) y-axis?

A(0, 2), B(5, 6), C(23, 0), D(0, 23), E(0, -4), F(-6, 0), G ($\sqrt{3}$,0).

Answer

Given points are A (0, 2), B (5, 6), C (23, 0), D (0, 23), E (0, -4), F (-6, 0), G ($\sqrt{3}$,0)

(i) If y-coordinate of a point is zero, then the point lies on x-axis.

Hence, C(23, 0), F(-6, 0) and G($\sqrt{3}$, 0) lies on x-axis.

(ii) If x-coordinate of a point is zero, then the point lies on y-axis.

Hence, A(0, 2), D(0, 23) and E(0, -4) lies on y-axis.

Plot the following points on the graph paper :

A(3, 4), B(-3, 1), C(1, -2), D(-2, -3), E(0, 5), F(5, 0), G(0, -3), H(-3, 0).

Answer

The points are shown in the below graph:

Write the co-ordinates of the points A, B, C, D, E, F, G and H shown in the adjacent figure.

Answer

From figure,

The co-ordinates of the points are:

In which quadrants are the points A, B, C and D of problem 5 located ?

Answer

In point A(2, 2), both x and y coordinates are positive. So it lies in the first quadrant.

In point B(-3, 0), y-coordinate is zero. So it lies on x-axis.

In point C(-2, -4), both x and y coordinates are negative. So it lies in the third quadrant.

In point D(3, -1), x coordinate is positive and y coordinate is negative. So it lies in the fourth quadrant.

Plot the following points on the same graph paper :

$A\Big(2, \dfrac{5}{2}\Big), B\Big(-\dfrac{3}{2}, 3\Big), C\Big(\dfrac{1}{2}, -\dfrac{3}{2}\Big)\text{ and } D\Big(-\dfrac{5}{2}, -\dfrac{1}{2}\Big).$

Answer

The points are shown on the graph below:

Plot the following points on the same graph paper.

$A\Big(\dfrac{4}{3}, -1\Big), B\Big(\dfrac{7}{2}, \dfrac{5}{3}\Big), C\Big(\dfrac{13}{6}, 0\Big)\text{ and } D\Big(-\dfrac{5}{3}, -\dfrac{5}{2}\Big).$

Answer

The points are shown on the graph below:

Plot the following points and check whether they are collinear or not :

(i) (1, 3), (-1, -1) and (-2, -3)

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

(iii) (0, 1), (2, -2) and $(\dfrac{2}{3} ,0)$.

Answer

(i)

Points (1, 3), (-1, -1) and (-2, -3) lie on a straight line, so they are collinear.

(ii)

Points (1, 2), (2, -1) and (-1, 4) do not lie on a line. So they are non-collinear.

(iii)

Points (0, 1), (2, -2) and $(\dfrac{2}{3} ,0)$ lie on a straight line, so they are collinear.

Plot the point P(-3, 4). Draw PM and PN perpendiculars to x-axis and y-axis respectively. State the co-ordinates of the points M and N.

Answer

The points are shown on the graph below:

Steps of construction :

1. Plot point P(-3, 4) on graph.

2. Draw PM and PN which are perpendiculars to x-axis and y-axis respectively.

From figure,

Coordinates of point M = (-3, 0) and N = (0, 4).

Plot the points A(1, 2), B(-4, 2), C(-4, -1) and D (1, -1). What kind of quadrilateral is ABCD? Also find the area of the quadrilateral ABCD.

Answer

The points are shown on the graph below:

Steps of construction :

1. Plot the points A (1, 2), B (-4, 2), C (-4, -1) and D (1, -1) on graph.

2. Join AB, BC, CD and AD.

From graph,

ABCD is a rectangle.

1 block = 1 unit

AB = 5 units, AD = 3 units

Area of rectangle ABCD = length × breadth

= AB × AD

= 5 × 3

= 15 sq. units.

Hence, ABCD is a rectangle and its area = 15 sq. units.

Plot the points (0, 2), (3, 0), (0, -2) and (-3, 0) on a graph paper. Join these points (in order). Name the figure so obtained and find the area of the figure obtained.

Answer

The points are shown on the graph below:

Steps of construction :

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

2. Join AB, BC, CD and DA. ABCD is a rhombus.

3. Join BD and AC.

BD and AC are the diagonals of the rhombus.

Area of a rhombus = $\dfrac{1}{2}$ × d₁ × d₂

From graph,

1 block = 1 unit.

AC = 4 units

BD = 6 units.

Area of rhombus ABCD = $\dfrac{1}{2}$ × BD × AC

= $\dfrac{1}{2}$ × 6 × 4

= 12 sq. units.

Hence, figure obtained form graph is a rhombus with area = 12 sq. units.

Three vertices of a square are A(2, 3), B (-3, 3) and C (-3, -2). Plot these points on a graph paper and hence use it to find the co-ordinates of the fourth vertex. Also find the area of the square.

Answer

The points are shown on the graph below:

Steps of construction :

1. Plot points A(2, 3), B(-3, 3) and C(-3, -2) 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.

On measuring,

AB = 5 units [As, 1 block = 1 unit]

Area of the square = side × side

Area of the square ABCD = AB × AB

= 5 × 5 = 25 sq. units.

Hence, the coordinates of D = (2, -2) and area of the square is 25 sq. units.

Write the co-ordinates of the vertices of a rectangle which is 6 units long and 4 units wide if the rectangle is in the first quadrant, its longer side lies on the x-axis and one vertex is at the origin.

Answer

The points are shown on the graph below:

Given,

The rectangle which is 6 units long and 4 units wide is shown in the graph.

Rectangle is in the first quadrant.

Longer side lies on x-axis and one vertex is at origin A(0, 0).

Steps of construction :

1. Mark point A(0, 0).

2. At a distance of 6 units on x-axis mark point B(6, 0).

3. From B draw a line segment parallel to y-axis and mark point C on it at distance of 4 units.

4. From C draw a line segment parallel to x-axis and mark point D on it at distance of 6 units.

5. Join DA.

Hence, coordinates of the rectangle are A(0, 0), B(6, 0), C(6, 4) and D(0, 4).

In the adjoining figure, ABCD is a rectangle with length 6 units and breadth 3 units. If O is the mid-point of AB, find the coordinates of A, B, C and D.

Answer

Given,

Length = 6 units.

Breadth = 3 units

AB = 6 units.

As, 1 block = 1 unit

From graph,

A = (-3, 0)

B = (3, 0)

C = (3, 3)

D = (-3, 3).

The adjoining figure shows an equilateral triangle OAB with each side = 2a units. Find the coordinates of the vertices.

Answer

Given equilateral triangle OAB.

OA = OB = AB = 2a units.

Draw AD ⊥ OB.

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

∴ OD = $\dfrac{1}{2}$ x OB = $\dfrac{1}{2}$ x 2a = a.

In right angle triangle OAD,

⇒ OA² = OD² + AD²

⇒ (2a)² = a² + AD²

⇒ 4a² = a² + AD²

⇒ AD² = 4a² - a²

⇒ AD² = 3a²

⇒ AD = $\sqrt{3a}$ units.

⇒ AD = $\sqrt{3}$a units.

From graph,

Co-ordinates of O = (0, 0)

Co-ordinates of B = (2a, 0)

As, OD = a units and AD = $\sqrt{3}$a units.

Co-ordinates of A = (a, $\sqrt{3}$a).

Hence, co-ordinates of O = (0, 0), B = (2a, 0) and A = (a, $\sqrt{3}$a).

In the given figure, △PQR is equilateral. If the coordinates of the points Q and R are (0, 2) and (0, -2) respectively, find the coordinates of the point P.

Answer

Given, PQR is an equilateral triangle in which Q(0, 2) and R(0, -2) and O = (0, 0).

Let (x, 0) be the coordinates of P. [As, P lies on x axis, so y-coordinate is zero.]

By distance formula,

$\Rightarrow QR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(0 - 0)^2 + (-2 - 2)^2} \\[1em] = \sqrt{0 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] \Rightarrow OQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(0 - 0)^2 + (2 - 0)^2} \\[1em] = \sqrt{0 + (2)^2} \\[1em] = \sqrt{4} \\[1em] = 2 \text{ units}.$

PQ = PR = QR = 4.

OQ = 2

In right angle triangle POQ,

⇒ PQ² = OP² + OQ² [By pythagoras theorem]

⇒ 4² = OP² + 2²

⇒ 16 = OP² + 4

⇒ OP² = 16 - 4

⇒ OP² = 12

⇒ OP = $\sqrt{12} = 2\sqrt{3}$.

Since, P lies on x-axis and OP = $2\sqrt{3}$.

Hence, the coordinates of P are ($2\sqrt{3}$, 0).