Coordinate Geometry

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).

Draw the graph of the following linear equation :

2x + y + 3 = 0

Answer

The given equation 2x + y + 3 = 0, can be written as :

⇒ y = -2x - 3

When x = 0, y = -2(0) - 3 = -3,

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

x = 2, y = -2(2) - 3 = -4 - 3 = -7.

Table of values :

Steps of construction :

1. Plot the points (0, -3), (1, -5) and (2, -7) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

Draw the graph of the following linear equation :

x - 5y - 4 = 0

Answer

The given equation x - 5y - 4 = 0, can be written as :

⇒ 5y = x - 4

⇒ y = $\dfrac{1}{5}$(x - 4)

When x = -6, y = $\dfrac{1}{5}(-6 - 4) = -\dfrac{10}{5}$ = -2,

x = -1, y = $\dfrac{1}{5}(-1 - 4) = -\dfrac{5}{5}$ = -1,

x = 4, y = $\dfrac{1}{5}(4 - 4) = \dfrac{0}{5}$ = 0.

Table of values :

Steps of construction :

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

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

Draw the graph of 3y = 12 - 2x. Take 2 cm = 1 unit on both axes.

Answer

The above equation, 3y = 12 - 2x can be written as :

⇒ 3y = 12 - 2x

⇒ y = $\dfrac{12}{3} - \dfrac{2x}{3}$

⇒ y = 4 - $\dfrac{2x}{3}$.

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

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

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

Table of values :

Steps of construction :

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

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

Draw the graph of 5x + 6y - 30 = 0 and use it to find the area of the triangle formed by the line and coordinate axes.

Answer

The above equation, 5x + 6y - 30 = 0 can be written as :

⇒ 6y = -5x + 30

⇒ y = $\dfrac{1}{6}(-5x + 30)$

⇒ y = $-\dfrac{5x}{6} + 5$.

When x = 0, y = -$\dfrac{(5 \times 0)}{6} + 5$ = 0 + 5 = 5,

x = 6, y = -$\dfrac{5 \times 6}{6} + 5$ = -5 + 5 = 0,

x = 12, y = -$\dfrac{5 \times 12}{6}$ + 5 = -10 + 5 = -5.

Table of values :

Steps of construction :

1. Plot the points (0, 5), (6, 0) and (12, -5) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

By formula,

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

From graph,

Base = 6 units, Height = 5 units.

Area = $\dfrac{1}{2} \times 6 \times 5 = 15$ sq. units.

Hence, the graph of the given equation is shown in the adjoining figure and area of triangle = 15 sq. units.

Draw the graph of 4x - 3y + 12 = 0 and use it to find the area of the triangle formed by the line and co-ordinate axes. Take 2 cm = 1 unit on both axes.

Answer

The above equation, 4x - 3y + 12 = 0 can be written as :

⇒ 3y = 4x + 12

⇒ y = $\dfrac{4}{3}x + 4$

When x = -6, y = $\dfrac{4}{3} \times -6 + 4$ = -8 + 4 = -4,

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

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

Table of values :

Steps of construction :

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

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

By formula,

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

From graph,

Base = 3 units, Height = 4 units.

Area = $\dfrac{1}{2} \times 3 \times 4 = 6$ sq. units.

Hence, the graph of the given equation is shown in the adjoining figure and area of triangle = 6 sq. units.

Draw the graph of the equation y = 3x - 4. Find graphically

(i) the value of y when x = -1

(ii) the value of x when y = 5.

Answer

Equation :

⇒ y = 3x - 4

When x = 0, y = 3(0) - 4 = 0 - 4 = -4,

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

Table of values :

Steps of construction :

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

2. Connect the two points by a straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

(i) Steps of construction :

1. From point, P (x = -1) draw a line parallel to y-axis touching the graph. Mark the point as Q.

2. From Q draw a line parallel to x-axis, touching y-axis at point R (y = -7).

Hence, y = -7, when x = -1.

(ii) Steps of construction :

1. From point, S (y = 5) draw a line parallel to x-axis touching the graph. Mark the point as T.

2. From T draw a line parallel to y-axis, touching x-axis at point U (x = 3).

Hence, x = 3, when y = 5.

The graph of a linear equation in x and y passes through (4, 0) and (0, 3). Find the value of k if the graph passes through (k, 1.5).

Answer

Steps of construction :

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

2. Connect the points through a straight line.

3. Take a point Q (y = 1.5) and draw a line parallel to x-axis touching the graph at point R.

4. From point R draw a line parallel to y-axis and touching x-axis at point S (x = 2).

From graph R = (2, 1.5)

Comparing point R with (k, 1.5), we get :

k = 2.

Hence, the value of k = 2.

Use the table given alongside to draw the graph of a straight line. Find, graphically, the values of a and b.

Answer

Steps of construction :

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

2. Connect the points through a straight line.

3. Take a point C (x = 2) and draw a line parallel to y-axis touching the graph at point B.

4. From point B draw a line parallel to x-axis and touching y-axis at point A (y = 1).

From graph,

B = (2, 1)

Comparing it with (2, b) we get :

b = 1.

Also, the graph touches y-axis at (0, -5).

Comparing it with (a, -5) we get :

a = 0.

Hence, a = 0 and b = 1.

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.

Find the distance between the following pair of points :

(2, 3), (4, 1)

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (2, 3)

Co-ordinates of Q = (4, 1)

$PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2} \\[1em] = \sqrt{2^2 + (-2)^2} \\[1em] = \sqrt{4 + 4}\\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.$

Hence, distance between (2, 3) and (4, 1) = $2\sqrt{2}$ units.

Find the distance between the following pair of points :

(0, 0), (36, 15)

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (0, 0)

Co-ordinates of Q = (36, 15)

$PQ = \sqrt{(36 - 0)^2 + (15 - 0)^2} \\[1em] = \sqrt{36^2 + 15^2} \\[1em] = \sqrt{1296 + 225}\\[1em] = \sqrt{1521} \\[1em] = 39.$

Hence, distance between (0, 0) and (36, 15) = 39 units.

Find the distance between the following pair of points :

(a, b), (-a, -b)

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (a, b)

Co-ordinates of Q = (-a, -b)

$PQ = \sqrt{(-a - a)^2 + (-b - b)^2} \\[1em] = \sqrt{(-2a)^2 + (-2b)^2} \\[1em] = \sqrt{4a^2 + 4b^2}\\[1em] = \sqrt{4(a^2 + b^2)} \\[1em] = 2\sqrt{a^2 + b^2}.$

Hence, distance between (a, b) and (-a, -b) = $2\sqrt{a^2 + b^2}$ units.

A is a point on y-axis whose ordinate is 4 and B is a point on x-axis whose abscissa is -3. Find the length of the line segment AB.

Answer

Since, point A is on y-axis so its x-coordinate = 0.

Given,

Ordinate of A = 4.

∴ A = (0, 4).

Since, point B is on x-axis so its y-coordinate = 0.

Given,

Abscissa of B = -3.

∴ B = (-3, 0).

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(-3 - 0)^2 + (0 - 4)^2} \\[1em] = \sqrt{(-3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.$

Hence, AB = 5 units.

Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\Rightarrow AB = \sqrt{[a - (-3)]^2 + [(-5) - (-14)]^2} \\[1em] \Rightarrow 9 = \sqrt{(a + 3)^2 + (-5 + 14)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 9 + 6a + (9)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 6a + 9 + 81}$

On squaring both sides,

$\Rightarrow a^2 + 6a + 90 = (9)^2 \\[1em] \Rightarrow a^2 + 6a + 90 = 81 \\[1em] \Rightarrow a^2 + 6a + 90 - 81 = 0 \\[1em] \Rightarrow a^2 + 6a + 9 = 0 \\[1em] \Rightarrow a^2 + 3a + 3a + 9 = 0 \\[1em] \Rightarrow a(a + 3) + 3(a + 3) = 0 \\[1em] \Rightarrow (a + 3)(a + 3) = 0 \\[1em] \Rightarrow a + 3 = 0 \\[1em] \Rightarrow a = -3.$

Hence, value of a = -3.

Find points on the x-axis which are at a distance of 5 units from the point (5, -4).

Answer

We know that,

y-coordinate of any point on x-axis = 0.

Let point on x-axis which is at a distance of 5 units from (5, -4) be P(x, 0).

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 5 = \sqrt{(x - 5)^2 + [0 - (-4)]^2} \\[1em] \Rightarrow 5 = \sqrt{x^2 + 25 - 10x + 16} \\[1em] \Rightarrow (5)^2 = x^2 - 10x + 41$

On squaring both sides,

$\Rightarrow 25 = x^2 - 10x + 41 \\[1em] \Rightarrow x^2 - 10x + 41 - 25 = 0 \\[1em] \Rightarrow x^2 - 10x + 16 = 0 \\[1em] \Rightarrow x^2 - 8x - 2x + 16 = 0 \\[1em] \Rightarrow x(x - 8) - 2(x - 8) = 0 \\[1em] \Rightarrow (x - 2)(x - 8) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x - 8 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = 8.$

∴ P = (x, 0) = (2, 0) or (8, 0).

Hence, points on the x-axis which are at a distance of 5 units from the point (5, -4) are (2, 0) or (8, 0).

Find points on the y-axis which are at a distance of 10 units from the point (8, 8).

Answer

We know that,

x-coordinate of any point on y-axis = 0.

Let point on y-axis which is at a distance of 10 units from (8, 8) be P(0, y).

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 10 = \sqrt{(0 - 8)^2 + (y - 8)^2} \\[1em] \Rightarrow 10 = \sqrt{(-8)^2 + y^2 + 8^2 - 16y} \\[1em] \Rightarrow 10 = \sqrt{64 + y^2 + 64 - 16y} \\[1em]$

On squaring both sides,

$\Rightarrow 10^2 = 64 + y^2 + 64 - 16y \\[1em] \Rightarrow 100 = y^2 + 128 - 16y \\[1em] \Rightarrow y^2 - 16y + 128 - 100 = 0 \\[1em] \Rightarrow y^2 - 16y + 28 = 0 \\[1em] \Rightarrow y^2 - 14y - 2y + 28 = 0 \\[1em] \Rightarrow y(y - 14) - 2(y - 14) = 0 \\[1em] \Rightarrow (y - 2)(y - 14) = 0 \\[1em] \Rightarrow y - 2 = 0 \text{ or } y - 14 = 0 \\[1em] \Rightarrow y = 2 \text{ or } y = 14.$

∴ P = (0, y) = (0, 2) or (0, 14).

Hence, points on the y-axis which are at a distance of 10 units from the point (8, 8) are (0, 2) or (0, 14).

Find point (or points) which are at distance of $\sqrt{10}$ units from the point (4, 3) given that the ordinate of the point (or points) is twice the abscissa.

Answer

Given,

Ordinate of the point is twice the abscissa.

Let abscissa of point be k, then ordinate = 2k

Let the required point be P.

∴ P = (k, 2k).

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

Distance between (4, 3) and (k, 2k) = $\sqrt{10}$ units.

$\Rightarrow \sqrt{10} = \sqrt{(k - 4)^2 + (2k - 3)^2} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + (4)^2 - 8k + (2k)^2 + (3)^2 - 12k} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + 16 - 8k + 4k^2 + 9 - 12k}$

On squaring both sides

$\Rightarrow 10 = k^2 + 16 - 8k + 4k^2 + 9 - 12k \Rightarrow 10 = 5k^2 - 20k + 25 \\[1em] \Rightarrow 10 = 5(k^2 - 4k + 5) \\[1em] \Rightarrow k^2 - 4k + 5 = 2 \\[1em] \Rightarrow k^2 - 4k + 5 - 2 = 0 \\[1em] \Rightarrow k^2 - 4k + 3 = 0 \\[1em] \Rightarrow k^2 - 3k - k + 3 = 0 \\[1em] \Rightarrow k(k - 3) - 1(k - 3) = 0 \\[1em] \Rightarrow (k - 1)(k - 3) = 0 \\[1em] \Rightarrow k - 1 = 0 \text{ or } k - 3 = 0 \\[1em] \Rightarrow k = 1 \text{ or } k = 3.$

When k = 1,

P = (k, 2k) = (1, 2).

When k = 3,

P = (k, 2k) = (3, 6).

Hence, required points are (1, 2) or (3, 6).

Find the point on the x-axis which is equidistant from the points (2, -5) and (-2, 9).

Answer

We know that,

y-coordinate of any point on x-axis = 0.

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let point on x-axis which is equidistant from the points (2, -5) and (-2, 9) be P(x, 0).

∴ Distance between (2, -5) and (x, 0) = Distance between (-2, 9) and (x, 0).

$\Rightarrow \sqrt{(x - 2)^2 + [0 - (-5)]^2} = \sqrt{[x - (-2)]^2 + (0 - 9)^2} \\[1em] \Rightarrow \sqrt{x^2 + 4 - 4x + 5^2} = \sqrt{(x + 2)^2 + (-9)^2} \\[1em] \Rightarrow \sqrt{x^2 - 4x + 25 + 4} = \sqrt{x^2 + 4 + 4x + 81} \\[1em]$

On squaring both sides,

$\Rightarrow x^2 - 4x + 29 = x^2 + 4x + 85 \\[1em] \Rightarrow x^2 - x^2 + 4x + 4x = 29 - 85 \\[1em] \Rightarrow 8x = -56 \\[1em] \Rightarrow x = -7.$

P = (x, 0) = (-7, 0).

Hence, required point is (-7, 0).

Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given, PQ = QR.

∴ Distance between (6, -1) and (1, 3) = Distance between (1, 3) and (x, 8).

$\Rightarrow \sqrt{(1 - 6)^2 + [3 - (-1)]^2} = \sqrt{(x - 1)^2 + (8 - 3)^2} \\[1em] \Rightarrow \sqrt{(-5)^2 + [4]^2} = \sqrt{x^2 + 1 - 2x + (5)^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 - 2x + 1 + 25} \\[1em] \Rightarrow \sqrt{41} = \sqrt{x^2 - 2x + 26}$

On squaring both sides,

$\Rightarrow 41 = x^2 - 2x + 26 \\[1em] \Rightarrow x^2 - 2x + 26 - 41 = 0 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3(x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0 \\[1em] \Rightarrow x = -3 \text{ or } x = 5.$

Hence, x = -3 or 5.

If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x.

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

∴ Distance between Q(0, 1) and P(5, -3) = Distance between Q(0, 1) and R(x, 6).

$\Rightarrow \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{(x - 0)^2 + (6 - 1)^2} \\[1em] \Rightarrow \sqrt{5^2 + (-4)^2} = \sqrt{x^2 + 5^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 + 25}$

On squaring both sides,

$\Rightarrow 25 + 16 = x^2 + 25 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x = \sqrt{16} \\[1em] \Rightarrow x = \pm 4.$

Hence, x = ±4.

Find a relation between x and y such that point (x, y) is equidistant from the points (7, 1) and (3, 5).

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

∴ Distance between (x, y) and (7, 1) = Distance between (x, y) and (3, 5).

$\therefore \sqrt{(7 - x)^2 + (1 - y)^2} = \sqrt{(3 - x)^2 + (5 - y)^2} \\[1em] \Rightarrow \sqrt{49 + x^2 - 14x + 1 + y^2 - 2y} = \sqrt{9 + x^2 - 6x + 25 + y^2 - 10y}$

On squaring both sides,

$\Rightarrow 49 + x^2 - 14x + 1 + y^2 - 2y = 9 + x^2 - 6x + 25 + y^2 - 10y \\[1em] \Rightarrow x^2 + y^2 - 14x - 2y + 50 = x^2 + y^2 - 6x - 10y + 34 \\[1em] \Rightarrow x^2 + y^2 - x^2 - y^2 - 6x + 14x = 10y - 2y + 50 - 34 \\[1em] \Rightarrow 8x = 8y + 16 \\[1em] \Rightarrow 8x = 8(y + 2) \\[1em] \Rightarrow x = y + 2 \\[1em] \Rightarrow x - y = 2.$

Hence, relation between x and y is x - y = 2.

The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from the points Q(2, -5) and R(-3, 6), then find the coordinates of P.

Answer

Let y-coordinate of point P be k.

∴ x-coordinate = 2k.

P = (2k, k)

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

P is equidistant from the points Q(2, -5) and R(-3, 6).

$\Rightarrow \sqrt{(2 - 2k)^2 + (-5 - k)^2} = \sqrt{(-3 - 2k)^2 + (6 - k)^2} \\[1em] \Rightarrow \sqrt{4 + 4k^2 - 8k + 25 + k^2 + 10k} = \sqrt{9 + 4k^2 + 12k + 36 + k^2 - 12k} \\[1em] \Rightarrow \sqrt{5k^2 + 2k + 29} = \sqrt{5k^2 + 45}$

On squaring both sides,

$\Rightarrow 5k^2 + 2k + 29 = 5k^2 + 45 \\[1em] \Rightarrow 5k^2 - 5k^2 + 2k = 45 - 29 \\[1em] \Rightarrow 2k = 16 \\[1em] \Rightarrow k = \dfrac{16}{2} \\[1em] \Rightarrow k = 8.$

P = (2k, k) = (16, 8).

Hence, coordinates of P = (16, 8).

If the points A(4, 3) and B(x, 5) are on a circle with center C(2, 3), find the values of x.

Answer

Given,

A(4, 3) and B(x, 5) are on a circle with center C(2, 3).

∴ AC = BC [Radius of same circle]

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting values in above equation, we get :

$\Rightarrow \sqrt{(2 - 4)^2 + (3 - 3)^2} = \sqrt{(2 - x)^2 + (3 - 5)^2} \\[1em] \Rightarrow \sqrt{(-2)^2} = \sqrt{4 + x^2 - 4x + (-2)^2} \\[1em] \Rightarrow \sqrt{4} = \sqrt{4 + x^2 - 4x + 4}$

On squaring both sides,

$\Rightarrow 4 = x^2 - 4x + 8 \\[1em] \Rightarrow x^2 - 4x + 4 = 0 \\[1em] \Rightarrow x^2 - 2x - 2x + 4 = 0 \\[1em] \Rightarrow x(x - 2) - 2(x - 2) = 0 \\[1em] \Rightarrow (x - 2)(x - 2) = 0 \\[1em] \Rightarrow x - 2 = 0 \\[1em] \Rightarrow x = 2.$

Hence, value of x = 2.

If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p.

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

AB = AC.

$\Rightarrow \sqrt{(3 - 0)^2 + (p - 2)^2} = \sqrt{(p - 0)^2 + (5 - 2)^2} \\[1em] \Rightarrow \sqrt{3^2 + p^2 + 4 - 4p} = \sqrt{p^2 + 3^2} \\[1em] \Rightarrow \sqrt{9 + p^2 + 4 - 4p} = \sqrt{p^2 + 3^2}$

On squaring both sides,

$\Rightarrow p^2 + 13 - 4p = p^2 + 9 \\[1em] \Rightarrow p^2 - p^2 + 4p = 13 - 9 \\[1em] \Rightarrow 4p = 4 \\[1em] \Rightarrow p = 1.$

Hence, p = 1.

Using distance formula, show that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).

Answer

Given,

Center (O) = (3, 3)

Let, A = (6, 2), B = (0, 4) and C = (4, 6).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow AO = \sqrt{(6 - 3)^2 + (2 - 3)^2} \\[1em] = \sqrt{(3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow BO = \sqrt{(0 - 3)^2 + (4 - 3)^2} \\[1em] = \sqrt{(-3)^2 + 1^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow CO = \sqrt{(4 - 3)^2 + (6 - 3)^2} \\[1em] = \sqrt{1^2 + 3^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10}. \\[1em]$

Since, AO = BO = CO.

Hence, proved that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).

The center of a circle is C(2α - 1, 3α + 1) and it passes through the point A(-3, -1). If a diameter of the circle is of length 20 units, find the value(s) of α.

Answer

Given,

Diameter = 20 units

Radius = $\dfrac{\text{Diameter}}{2} = \dfrac{20}{2}$ = 10 units.

∴ AC = 10 units

$\Rightarrow \sqrt{[-3 - (2α - 1)]^2 + [-1 - (3α + 1)]^2} = 10 \\[1em] \Rightarrow \sqrt{(-3 - 2α + 1)^2 + (-1 - 3α - 1)^2} = 10 \\[1em] \Rightarrow \sqrt{(-2 - 2α)^2 + (-3α - 2)^2} = 10 \\[1em] \Rightarrow \sqrt{4 + 4α^2 + 8α + 9α^2 + 4 + 12α} = 10 \\[1em] \Rightarrow \sqrt{13α^2 + 20α + 8} = 10$

On squaring both sides,

$\Rightarrow 13α^2 + 20α + 8 = 10^2 \\[1em] \Rightarrow 13α^2 + 20α + 8 = 100 \\[1em] \Rightarrow 13α^2 + 20α + 8 - 100 = 0 \\[1em] \Rightarrow 13α^2 + 20α - 92 = 0 \\[1em] \Rightarrow 13α^2 + 46α - 26α - 92 =0 \\[1em] \Rightarrow α(13α + 46) - 2(13α + 46) = 0 \\[1em] \Rightarrow (13α + 46)(α - 2) = 0 \\[1em] \Rightarrow 13α + 46 = 0 \text{ or } α - 2 = 0 \\[1em] \Rightarrow 13α = -46 \text{ or } α = 2 \\[1em] \Rightarrow α = -\dfrac{46}{13} \text{ or } α = 2.$