Coordinate Geometry — Exercise 18.3

Solve the following equations graphically :

3x - 2y = 4, 5x - 2y = 0.

Answer

Given,

Equation :

⇒ 3x - 2y = 4

⇒ 2y = 3x - 4

⇒ y = $\dfrac{3}{2}$x - 2 .........(1)

When, x = -2, y = $\dfrac{3}{2} \times -2 - 2$ = -3 - 2 = -5,

x = 0, y = $\dfrac{3}{2} \times 0 - 2$ = 0 - 2 = -2,

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

Tables of values for equation (1)

Steps of construction :

1. Plot the points (-2, -5), (0, -2) and (2, 1) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 5x - 2y = 0

⇒ 2y = 5x

⇒ y = $\dfrac{5}{2}x$ ..............(2)

When, x = -2, y = $\dfrac{5}{2} \times -2$ = -5,

x = 0, y = $\dfrac{5}{2} \times 0$ = 0,

x = 2, y = $\dfrac{5}{2} \times 2$ = 5.

Tables of values for equation (2)

Steps of construction :

1. Plot the points (-2, -5), (0, 0) and (2, 5) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(-2, -5).

Hence, the solution of the given equations is x = -2, y = -5.

Solve the following pair of equations graphically. Plot atleast 3 points for each straight line.

2x - 7y = 6, 5x - 8y = -4.

Answer

Given,

Equation :

⇒ 2x - 7y = 6

⇒ 2x = 6 + 7y

⇒ x = $\dfrac{6 + 7y}{2}$ ..........(1)

When y = 0, x = $\dfrac{6 + 7 \times 0}{2} = \dfrac{6}{2}$ = 3,

y = -1, x = $\dfrac{6 + 7 \times -1}{2} = \dfrac{6 - 7}{2}$ = -0.5,

y = -2, x = $\dfrac{6 + 7 \times -2}{2} = \dfrac{6 - 14}{2} = -\dfrac{8}{2}$ = -4.

Table of values for equation (1)

Steps of construction :

1. Plot the points (3, 0), $(-\dfrac{1}{2}, -1)$ and (-4, -2) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 5x - 8y = -4

⇒ 5x = 8y - 4

⇒ x = $\dfrac{8}{5}$y - $\dfrac{4}{5}$ ..............(2)

When, y = 0, x = $\dfrac{8}{5} \times 0 - \dfrac{4}{5} = 0 - \dfrac{4}{5} =$ -0.8,

y = 3, x = $\dfrac{8}{5} \times 3 - \dfrac{4}{5} = \dfrac{24}{5} - \dfrac{4}{5} = \dfrac{20}{5}$ = 4,

y = -2, x = $\dfrac{8}{5} \times -2 - \dfrac{4}{5} = -\dfrac{16}{5} - \dfrac{4}{5} = -\dfrac{20}{5}$ = -4.

Table of values for equation (2)

Steps of construction :

1. Plot the points (-0.8, 0), (4, 3) and (-4, -2) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(-4, -2).

Hence, the solution of the given equations is x = -4, y = -2.

Using the same axes of coordinates and the same unit, solve graphically.

x + y = 0, 3x - 2y = 10.

Answer

Given,

Equation :

⇒ x + y = 0

⇒ y = -x .............(1)

When, x = 0, y = 0,

x = 1, y = -(1) = -1,

x = 2, y = -(2) = -2.

Table of values for equation (1)

Steps of construction :

1. Plot the points (0, 0), (1, -1) and (2, -2) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 3x - 2y = 10

⇒ 2y = 3x - 10

⇒ y = $\dfrac{3}{2}x - 5$ ............(2)

When, x = 0, y = $\dfrac{3}{2} \times 0 - 5$ = 0 - 5 = -5,

x = 2, $y = \dfrac{3}{2} \times 2 - 5$ = 3 - 5 = -2,

x = 4, $y = \dfrac{3}{2} \times 4 - 5$ = 6 - 5 = 1.

Table of values for equation (2)

Steps of construction :

1. Plot the points (0, -5), (2, -2) and (4, 1) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(2, -2).

Hence, the solution of the given equations is x = 2, y = -2.

Take 1 cm to represent 1 unit on each axis to draw the graphs of the equations 4x - 5y = -4 and 3x = 2y - 3 on the same graph sheet (same axes). Use your graph to find the solution of the above simultaneous equations.

Answer

Given,

Equation :

⇒ 4x - 5y = -4

⇒ 5y = 4x + 4

⇒ y = $\dfrac{4x + 4}{5}$.............(1)

When, x = -1, y = $\dfrac{4 \times -1 + 4}{5} = \dfrac{0}{5}$ = 0,

x = 1.5, y = $\dfrac{4 \times 1.5 + 4}{5} = \dfrac{10}{5}$ = 2,

x = 4, y = $\dfrac{4 \times 4 + 4}{5} =\dfrac{20}{5}$ = 4.

Table of values for equation (1)

Steps of construction :

1. Plot the points (-1, 0), (1.5, 2) and (4, 4) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 3x = 2y - 3

⇒ 2y = 3x + 3

⇒ y = $\dfrac{3x + 3}{2}$ ............(2)

When, x = -1, y = $\dfrac{3 \times -1 + 3}{2} = \dfrac{-3 + 3}{2} = \dfrac{0}{2}$ = 0,

x = 0, $y = \dfrac{3 \times 0 + 3}{2} = \dfrac{3}{2}$ = 1.5,

x = 1, $y = \dfrac{3 \times 1 + 3}{2} = \dfrac{6}{2}$ = 3.

Table of values for equation (2)

Steps of construction :

1. Plot the points (-1, 0), (0, 1.5) and (1, 3) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(-1, 0).

Hence, the solution of the given equations is x = -1, y = 0.

Solve the following simultaneous equations graphically :

x + 3y = 8, 3x = 2 + 2y.

Answer

Given,

Equation :

x + 3y = 8

x = 8 - 3y ..........(1)

When, y = 1, x = 8 - 3(1) = 8 - 3 = 5,

y = 2, x = 8 - 3(2) = 8 - 6 = 2,

y = 3, x = 8 - 3(3) = 8 - 9 = -1.

Table of values for equation (1)

Steps of construction :

1. Plot the points (-1, 3), (2, 2) and (5, 1) on graph paper.

2. Connect points by straight line.

Given,

Equation :

⇒ 3x = 2 + 2y

⇒ x = $\dfrac{2 + 2y}{3}$ ..........(2)

When, y = -1, x = $\dfrac{2 + 2 \times -1}{3} = \dfrac{2 - 2}{3}$ = 0.

y = 2, x = $\dfrac{2 + 2 \times 2}{3} = \dfrac{6}{3}$ = 2,

y = 5, x = $\dfrac{2 + 2 \times 5}{3} = \dfrac{12}{3}$ = 4.

Table of values for equation (2)

Steps of construction :

1. Plot the points (0, -1), (2, 2) and (4, 5) on graph paper.

2. Connect points by straight line.

The graphs of both the straight lines are shown in the figure.

The lines intersect at point A(2, 2).

Hence, the solution of the given equations is x = 2, y = 2..

Solve graphically the simultaneous equations 3y = 5 - x, 2x = y + 3.

Answer

Given,

⇒ 3y = 5 - x

⇒ y = $\dfrac{5 - x}{3}$ .......(1)

When x = -4, y = $\dfrac{5 - (-4)}{3} = \dfrac{9}{3}$ = 3,

x = -1, y = $\dfrac{5 - (-1)}{3} = \dfrac{6}{3}$ = 2,

x = 2, y = $\dfrac{5 - 2}{3} = \dfrac{3}{3}$ = 1.

Table of values for equation (1)

Steps of construction :

1. Plot the points (-4, 3), (-1, 2) and (2, 1) on graph paper.

2. Connect points by straight line.

Given,

⇒ 2x = y + 3

⇒ y = 2x - 3 ...........(2)

When x = 0, y = 2 × 0 - 3 = 0 - 3 = -3,

x = 1, y = 2 × 1 - 3 = 2 - 3 = -1,

x = 2, y = 2 × 2 - 3 = 4 - 3 = 1.

Table of values for equation (2)

Steps of construction :

1. Plot the points (0, -3), (1, -1) and (2, 1) on graph paper.

2. Connect points by straight line.

From graph,

The two lines meet at point P(2, 1).

Hence, the solution of the given equations is x = 2, y = 1.

Use graph paper for this question. Take 2 cm = 1 unit on both axes.

(i) Draw the graphs of x + y + 3 = 0 and 3x - 2y + 4 = 0. Plot three points per line.

(ii) Write down the coordinates of the point of intersection of the lines.

(iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.

Answer

(i) Given,

⇒ x + y + 3 = 0

⇒ y = -(3 + x) .........(1)

When x = -1, y = -[3 + (-1)] = -(3 - 1) = -2,

x = 0, y = -(3 + 0) = -3,

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

Table of values for equation (1)

Steps of construction :

1. Plot the points (-1, -2), (0, -3) and (1, -4) on graph paper.

2. Connect points by straight line.

Given,

⇒ 3x - 2y + 4 = 0

⇒ 2y = 3x + 4

⇒ y = $\dfrac{3x + 4}{2}$

When x = -2, y = $\dfrac{3 \times -2 + 4}{2} = \dfrac{-6 + 4}{2} = \dfrac{-2}{2}$ = -1,

x = 0, y = $\dfrac{3 \times 0 + 4}{2} = \dfrac{4}{2}$ = 2,

x = 2, y = $\dfrac{3 \times 2 + 4}{2} = \dfrac{6 + 4}{2} = \dfrac{10}{2}$ = 5.

Table of values for equation (2)

Steps of construction :

1. Plot the points (-2, -1), (0, 2) and (2, 5) on graph paper.

2. Connect points by straight line.

(ii) From graph,

P(-2, -1) is the intersection of lines.

Hence, coordinates of point of intersection = (-2, -1).

(iii) From P, draw a perpendicular to y-axis.

As, 1 unit = 2 cm.

So, PQ = 2 unit = 4 cm, OQ = 1 unit = 2 cm.

In right angle triangle,

By pythagoras theorem,

⇒ OP² = PQ² + OQ²

⇒ OP² = 4² + 2²

⇒ OP² = 16 + 4

⇒ OP² = 20

⇒ OP = $\sqrt{20}$ = 4.5 cm

Hence, distance of the point of intersection of the lines from the origin = 4.5 cm.

Solve the following simultaneous equations, graphically :

2x - 3y + 2 = 4x + 1 = 3x - y + 2.

Answer

Considering,

⇒ 2x - 3y + 2 = 4x + 1

⇒ 3y = 2x - 4x + 2 - 1

⇒ 3y = -2x + 1

⇒ y = $\dfrac{1 - 2x}{3}$ .........(1)

When, x = -4, y = $\dfrac{1 - 2 \times (-4)}{3} = \dfrac{1 + 8}{3} = \dfrac{9}{3}$ = 3,

x = -1, y = $\dfrac{1 - 2 \times (-1)}{3} = \dfrac{1 + 2}{3} = \dfrac{3}{3}$ = 1,

x = 2, y = $\dfrac{1 - 2 \times 2}{3} = \dfrac{1 - 4}{3} = \dfrac{-3}{3}$ = -1.

Table of values for equation (1)

Steps of construction :

1. Plot the points (-4, 3), (-1, 1) and (2, -1) on graph paper.

2. Connect points by straight line.

Considering,

⇒ 4x + 1 = 3x - y + 2

⇒ y = 3x - 4x + 2 - 1

⇒ y = -x + 1

⇒ y = 1 - x ............(2)

When, x = 0, y = 1 - 0 = 1,

x = 1, y = 1 - 1 = 0,

x = 2, y = 1 - 2 = -1.

Table of values for equation (2)

Steps of construction :

1. Plot the points (0, 1), (1, 0) and (2, -1) on graph paper.

2. Connect points by straight line.

From graph,

The two lines intersect at P(2, -1).

Hence, the solution of the given equations is x = 2, y = -1.

Use graph paper for this question :

(i) Draw the graphs of 3x - y - 2 = 0 and 2x + y - 8 = 0. Take 1 cm = 1 unit on both axes and plot three points per line.

(ii) Write down the coordinates of the point of intersection and the area of the triangle formed by the lines and the x-axis.

Answer

(i) Given,

⇒ 3x - y - 2 = 0

⇒ y = 3x - 2 ............(1)

When x = 0, y = 3 × 0 - 2 = 0 - 2 = -2,

x = 1, y = 3 × 1 - 2 = 3 - 2 = 1,

x = 2, y = 3 × 2 - 2 = 6 - 2 = 4.

Table of values for equation (1)

Steps of construction :

1. Plot the points (0, -2), (1, 1) and (2, 4).

2. Join the points.

Given,

⇒ 2x + y - 8 = 0

⇒ y = 8 - 2x ...........(2)

When x = 2, y = 8 - 2 × 2 = 8 - 4 = 4,

x = 3, y = 8 - 2 × 3 = 8 - 6 = 2,

x = 4, y = 8 - 2 × 4 = 8 - 8 = 0.

Table of values for equation (2)

Steps of construction :

1. Plot the points (2, 4), (3, 2) and (4, 0).

2. Join the points.

(ii) From the graph,

P(2, 4) is the point of intersection of lines.

BC = $3\dfrac{2}{3}$ cm

PV = 4 cm.

Area of triangle = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × PV

= $\dfrac{1}{2} \times 3\dfrac{2}{3}$ × 4

= $6\dfrac{2}{3}$ cm².

Hence, P(2, 4) is the point of intersection of lines and area = $6\dfrac{2}{3}$ cm².

Solve the following system of linear equations graphically :

2x - y - 4 = 0, x + y + 1 = 0.

Hence, find the area of the triangle formed by these lines and the y-axis.

Answer

Given,

⇒ 2x - y - 4 = 0

⇒ y = 2x - 4 ............(1)

When x = 1, y = 2(1) - 4 = 2 - 4 = -2,

x = 2, y = 2(2) - 4 = 4 - 4 = 0,

x = 3, y = 2(3) - 4 = 6 - 4 = 2.

Table of values for equation (1)

Steps of construction :

1. Plot the points (1, -2), (2, 0) and (3, 2).

2. Join the points.

Given,

⇒ x + y + 1 = 0

⇒ y = -(x + 1) .............(2)

When, x = -1, y = -(-1 + 1) = 0,

x = 0, y = -(0 + 1) = -1,

x = 1, y = -(1 + 1) = -2.

Table of values for equation (2)

Steps of construction :

1. Plot the points (-1, 0), (0, -1) and (1, -2).

2. Join the points.

From graph,

A(1, -2) is the point of intersection of lines.

ABC are the vertices of triangle.

From A, draw AD perpendicular to BC.

AD = 1 unit and BC = 3 units.

Area of △ABC = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × AD

= $\dfrac{1}{2}$ × 3 × 1

= $\dfrac{3}{2}$ sq. units

Hence, point of intersection = (1, -2) and area of triangle = $\dfrac{3}{2}$ sq. units

Solve graphically the following equations: x + 2y = 4, 3x - 2y = 4.

Take 2 cm = 1 unit on each axis. Write down the area of the triangle formed by the lines and the y-axis. Also, find the area of the triangle formed by the lines and the x-axis.

Answer

Given,

⇒ x + 2y = 4

⇒ 2y = 4 - x

⇒ y = $\dfrac{4 - x}{2}$ ......................(1)

When x = 0, y = $\dfrac{4 - 0}{2} = \dfrac{4}{2}$ = 2,

x = 2, y = $\dfrac{4 - 2}{2} = \dfrac{2}{2}$ = 1,

x = 4, y = $\dfrac{4 - 4}{2} = \dfrac{0}{2}$ = 0.

Table of values for equation (1)

Steps of construction :

1. Plot the points (0, 2), (2, 1) and (4, 0).

2. Join the points.

Given,

⇒ 3x - 2y = 4

⇒ 2y = 3x - 4

⇒ y = $\dfrac{3x - 4}{2}$ ..................(2)

When x = 0, y = $\dfrac{3 \times 0 - 4}{2} = \dfrac{0 - 4}{2} = \dfrac{-4}{2}$ = -2,

x = 2, y = $\dfrac{3 \times 2 - 4}{2} = \dfrac{6 - 4}{2} = \dfrac{2}{2}$ = 1,

x = 4, y = $\dfrac{3 \times 4 - 4}{2} = \dfrac{12 - 4}{2} = \dfrac{8}{2}$ = 4.

x = $\dfrac{4}{3}, y = \dfrac{3 \times \dfrac{4}{3} - 4}{2} = \dfrac{4 - 4}{2} = \dfrac{0}{2}$ = 0.

Table of values for equation (2)

Steps of construction :

1. Plot the points (0, -2), (2, 1), (4, 4) and $\Big(\dfrac{4}{3}, 0\Big)$.

2. Join the points.

From graph,

A(2, 1) is the point of intersection of lines.

AEF is the triangle between lines and y-axis.

From A, draw AG perpendicular to EF.

Using distance formula, distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance between E (0, 2) and F (0, -2)

= $\sqrt{(0 - 0)^2 + \Big(2 - (-2)\Big)^2} = \sqrt{(2 + 2)^2} = \sqrt{4^2} = 4$ units

From graph,

AG = 2 units

EF = 4 units

Area of triangle = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x EF x AG

= $\dfrac{1}{2}$ x 4 x 2

= 1 x 4

= 4 sq. units.

And, ABC is the triangle between lines and x-axis.

Using distance formula, distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance between B $\Big(\dfrac{4}{3}, 0\Big)$ and C (4, 0)

= $\sqrt{\Big(4 - \dfrac{4}{3}\Big)^2 + (0 - 0)^2} = \sqrt{\Big(\dfrac{12 - 4}{3}\Big)^2} = \sqrt{\Big(\dfrac{8}{3}\Big)^2} = \dfrac{8}{3}$

From A, draw AD perpendicular to BC.

AD = 1 unit

BC = $\dfrac{8}{3}$ units

Area of triangle = $\dfrac{1}{2}$ x base x height

$\text{Area of triangle ABC} = \dfrac{1}{2} \times BC \times AD \\[1em] = \dfrac{1}{2} \times \dfrac{8}{3} \times 1 \\[1em] = \dfrac{4}{3} \text{sq. units}$

Hence, point of intersection of lines is x = 2, y = 1 and area of the triangle formed by the lines and the y-axis = 4 sq. unit and area of the triangle formed by the lines and the x-axis = $\dfrac{4}{3}$ sq units.

On graph paper, take 2 cm to represent one unit on both axes, draw the lines :

x + 3 = 0, y - 2 = 0, 2x + 3y = 12.

Write down the coordinates of the vertices of the triangle formed by these lines.

Answer

Given,

1st equation :

⇒ x + 3 = 0

⇒ x = -3 ............(1)

2nd equation :

⇒ y - 2 = 0

⇒ y = 2 .............(2)

3rd equation :

⇒ 2x + 3y = 12

⇒ 3y = 12 - 2x

⇒ y = $\dfrac{12 - 2x}{3}$ .........(3)

When, x = -3, y = $\dfrac{12 - 2 \times (-3)}{3} = \dfrac{12 + 6}{3} = \dfrac{18}{3}$ = 6,

x = 0, y = $\dfrac{12 - 2 \times 0}{3} = \dfrac{12}{3}$ = 4,

x = 3, y = $\dfrac{12 - 2 \times 3}{3} = \dfrac{12 - 6}{3} = \dfrac{6}{3}$ = 2.

Table of values of equation (3) :

Steps of construction :

1. Plot the points (-3, 6), (0, 4) and (3, 2).

2. Join the points.

From graph,

ABC is the required triangle.

Hence, coordinates of vertices of triangle formed by these lines = (-3, 6), (-3, 2) and (3, 2).

Find graphically the coordinates of the vertices of the triangle formed by the lines y = 0, y = x and 2x + 3y = 10. Hence, find the area of the triangle formed by these lines.

Answer

Given,

1st equation :

y = 0,

2nd equation :

y = x,

When, x = 0, y = 0,

x = 1, y = 1,

x = 2, y = 2.

Table of values of equation (2) :

Steps of construction :

1. Plot the points (0, 0), (1, 1) and (2, 2).

2. Join the points.

3rd equation :

⇒ 2x + 3y = 10

⇒ 3y = 10 - 2x

⇒ y = $\dfrac{10 - 2x}{3}$ ..........(3)

When, x = -1, y = $\dfrac{10 - 2 \times (-1)}{3} = \dfrac{10 + 2}{3} = \dfrac{12}{3}$ = 4,

x = 2, y = $\dfrac{10 - 2 \times 2}{3} = \dfrac{10 - 4}{3} = \dfrac{6}{3}$ = 2,

x = 5, y = $\dfrac{10 - 2 \times 5}{3} = \dfrac{10 - 10}{3}$ = 0.

Table of values of equation (3) :

Steps of construction :

1. Plot the points (-1, 4), (2, 2) and (5, 0).

2. Join the points.

From graph,

ABC is the triangle.

From A, draw AD perpendicular to x-axis.

AD = 2 units and BC = 5 units.

Area of triangle = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2} \times BC \times AD$

= $\dfrac{1}{2} \times 5 \times 2$

= 5 sq. units.

Hence, coordinates of triangle are (0, 0), (5, 0) and (2, 2) and area = 5 sq. units.