Coordinate Geometry — Chapter Test

Three vertices of a rectangle are A(2, -1), B(2, 7) and C(4, 7). Plot these points on a graph and hence use it to find the co-ordinates of the fourth vertex D. Also find the co-ordinates of

(i) the mid-point of BC

(ii) the mid-point of CD

(iii) the point of intersection of the diagonals.

What is the area of the rectangle ?

Answer

In graph,

1 block = 1 unit.

Steps of construction :

1. Plot the points A(2, -1), B(2, 7) and C(4, 7) on graph.

2. Join AB and BC.

3. Measure AB. Draw a line segment CD, from point C parallel to y-axis.

4. Measure BC. Draw a line segment AD, from point A parallel to x-axis.

5. Mark F and G mid-point of BC and CD respectively.

6. Join AC and BD diagonals of rectangle.

7. Mark E as the point of intersection of diagonals.

From graph,

Coordinates of D = (4, -1).

(i) From graph,

F is the mid-point of BC and F = (3, 7).

Hence, coordinates of mid-point of BC = (3, 7).

(ii) From graph,

G is the mid-point of CD and G = (4, 3).

Hence, coordinates of mid-point of CD = (4, 3).

(iii) From graph,

E is the point of intersection of diagonals.

Hence, point of intersection of the diagonals = (3, 3).

From graph,

AB = 8 units and BC = 2 units.

Area of the rectangle ABCD = length × breadth

= AB × BC

= 8 × 2

= 16 sq. units.

Hence, the area of the rectangle = 16 sq. units.

Three vertices of a parallelogram are A(3, 5), B(3, -1) and C(-1, -3). Plot these points on a graph paper and hence use it to find the coordinates of the fourth vertex D. Also find the coordinates of the mid-point of the side CD. What is the area of the parallelogram?

Answer

Steps of construction :

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

2. Join AB and BC.

3. From C draw a line CD parallel to AB, such that CD = AB.

4. From A draw a line AD parallel to BC, such that AD = BC.

5. Mark E, the mid-point of CD.

From graph,

The coordinates of fourth vertex D are (-1, 3).

The coordinates of the midpoint of CD i.e. E are (-1, 0).

As, 1 block = 1 unit

EF = 4 units

CD = 6 units

Area of parallelogram ABCD = Base × height

= CD × EF

= 6 × 4

= 24 sq. units.

Hence, D = (-1, 3), coordinates of mid-point of CD = (-1, 0) and the area of the parallelogram is 24 sq. units.

Draw the graphs of the following linear equation.

y = 2x - 1

Also find slope and y-intercept of this line.

Answer

Given,

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

When x = 1, y = 2 × 1 - 1 = 2 - 1 = 1

x = 2, y = 2 × 2 - 1 = 4 - 1 = 3

x = 3, y = 2 × 3 - 1 = 6 - 1 = 5.

Steps of construction :

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

2. Join the points.

Comparing equation (1) with y = mx + c, we get :

m = 2 and c = -1.

Hence, slope of the line = 2 and y-intercept = -1.

Draw the graphs of the following linear equation.

2x + 3y = 6

Also find slope and y-intercept of this line.

Answer

Given,

⇒ 2x + 3y = 6

⇒ 3y = 6 - 2x

⇒ y = $\dfrac{6}{3} - \dfrac{2x}{3}$

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

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

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

x = 6, y = $2 - \dfrac{2 \times 6}{3} = 2 - \dfrac{12}{3}$ = 2 - 4 = -2.

Steps of construction :

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

2. Join the points.

Comparing equation (1) with y = mx + c, we get :

m = $-\dfrac{2}{3}$ and c = 2.

Hence the slope of the line = $-\dfrac{2}{3}$ and y intercept = 2.

Draw the graphs of the following linear equation.

2x - 3y = 4

Also find slope and y-intercept of this line.

Answer

Given,

⇒ 2x - 3y = 4

⇒ 3y = 2x - 4

⇒ y = $\dfrac{2x - 4}{3}$

⇒ y = $\dfrac{2}{3}x - \dfrac{4}{3}$ ........(1)

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

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

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

Steps of construction :

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

2. Join the points.

Comparing equation (1) with y = mx + c.

m = $\dfrac{2}{3}$ and c = $-\dfrac{4}{3}$.

Hence, the slope of line = $\dfrac{2}{3}$ and y intercept = $-\dfrac{4}{3}$.

Draw the graph of the equation 3x - 4y = 12. From the graph, find :

(i) the value of y when x = -4

(ii) the value of x when y = 3.

Answer

Given,

⇒ 3x - 4y = 12

⇒ 4y = 3x - 12

⇒ y = $\dfrac{3x - 12}{4}$

⇒ y = $\dfrac{3}{4}x - 3$.

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

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

x = 8, y = $\dfrac{3}{4} \times 8 - 3 = \dfrac{24}{4} - 3$ = 6 - 3 = 3.

Steps of construction :

1. Plot the points (0, -3), (4, 0) and (8, 3) on graph.

2. Join the points.

(i) Steps of construction :

1. Take a point N (x = -4) and draw a line parallel to y-axis, touching the graph at point M.

2. From M draw a line parallel to x-axis touching y-axis at point O (y = -6).

Hence, when x = -4, the value of y is -6.

(ii) Steps of construction :

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

2. From R draw a line parallel to y-axis touching x-axis at point Q (x = 8).

Hence, when y = 3, the value of x is 8.

Solve graphically, the simultaneous equations : 2x - 3y = 7; x + 6y = 11.

Answer

Given,

⇒ 2x - 3y = 7

⇒ 3y = 2x - 7

⇒ y = $\dfrac{2x - 7}{3}$ .........(1)

When, x = -1, y = $\dfrac{2 \times -1 - 7}{3} = \dfrac{-2 - 7}{3} = -\dfrac{9}{3}$ = -3,

x = 2, y = $\dfrac{2 \times 2 - 7}{3} = \dfrac{4 - 7}{3} = -\dfrac{3}{3}$ = -1,

x = 5, y = $\dfrac{2 \times 5 - 7}{3} = \dfrac{10 - 7}{3} = \dfrac{3}{3}$ = 1.

Table of values of equation (1) :

Steps of construction :

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

2. Join the points.

Given,

⇒ x + 6y = 11

⇒ 6y = 11 - x

⇒ y = $\dfrac{11 - x}{6}$ ..........(2)

When x = -7, y = $\dfrac{11 - (-7)}{6} = \dfrac{18}{6}$ = 3,

x = -1, y = $\dfrac{11 - (-1)}{6} = \dfrac{12}{6}$ = 2,

x = 5, y = $\dfrac{11 - 5}{6} = \dfrac{6}{6}$ = 1.

Table of values of equation (2) :

Steps of construction :

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

2. Join the points.

From graph,

The two lines intersect at P(5, 1).

Hence x = 5, y = 1.

Solve the following system of equations graphically :

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

Answer

Given,

⇒ x - 2y - 4 = 0

⇒ 2y = x - 4

⇒ y = $\dfrac{x - 4}{2}$ .........(1)

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

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

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

Table of values of equation (1) :

Steps of construction :

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

2. Join the points.

Given,

⇒ 2x + y - 3 = 0

⇒ y = 3 - 2x .........(2)

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

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

x = 2, y = 3 - 2(2) = 3 - 4 = -1.

Table of values of equation (2) :

Steps of construction :

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

2. Join the points.

From graph,

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

Hence, x = 2, y = -1.

Using a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations: 6y = 5x + 10, y = 5x - 15. From the graph, find

(i) the coordinates of the point where the two lines intersect.

(ii) the area of the triangle between the lines and the x-axis.

Answer

(i) Given,

⇒ 6y = 5x + 10

⇒ y = $\dfrac{5x + 10}{6}$ .........(1)

When x = 1, y = $\dfrac{5 \times 1 + 10}{6} = \dfrac{15}{6}$ = 2.5

x = -2, y = $\dfrac{5 \times -2 + 10}{6} = \dfrac{-10 + 10}{6} = \dfrac{0}{6}$ = 0,

x = 4, y = $\dfrac{5 \times 4 + 10}{6} = \dfrac{20 + 10}{6} = \dfrac{30}{6}$ = 5.

Table of values of equation (1) :

Steps of construction :

1. Plot the points (1, 2.5), (-2, 0) and (4, 5) on graph.

2. Join the points.

Given,

y = 5x - 15

When x = 2.5, y = 5 × 2.5 - 15 = 12.5 - 15 = -2.5,

x = 3, y = 5 × 3 - 15 = 15 - 15 = 0,

x = 4, y = 5 × 4 - 15 = 20 - 15 = 5.

Table of values of equation (2) :

Steps of construction :

1. Plot the points (2.5, -2.5), (3, 0) and (4, 5) on graph.

2. Join the points.

From graph,

The lines intersect at point P(4, 5).

Hence, x = 4, y = 5.

(ii) From graph,

Triangle = PQR.

Draw a line PJ, from P perpendicular to x-axis.

PJ = 5 units

QR = 5 units

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

= $\dfrac{1}{2}$ × QR × PJ

= $\dfrac{1}{2}$ × 5 × 5

= $\dfrac{25}{2}$

= 12.5 sq. units.

Hence, area = 12.5 sq. units.

Find, graphically, the coordinates of the vertices of the triangle formed by the lines:

8y - 3x + 7 = 0, 2x - y + 4 = 0 and 5x + 4y = 29.

Answer

Given,

⇒ 8y - 3x + 7 = 0

⇒ 8y = 3x - 7

⇒ y = $\dfrac{3x - 7}{8}$ .........(1)

When x = -3, y = $\dfrac{3 \times -3 - 7}{8} = \dfrac{-9 - 7}{8} = -\dfrac{16}{8}$ = -2,

x = 1, y = $\dfrac{3 \times 1 - 7}{8} = \dfrac{3 - 7}{8} = -\dfrac{4}{8}$ = -0.5

x = 5, y = $\dfrac{3 \times 5 - 7}{8} = \dfrac{15 - 7}{8} = \dfrac{8}{8}$ = 1.

Table of values of equation (1) :

Steps of construction :

1. Plot the points (-3, -2), (1, -0.5), (5, 1) on graph.

2. Join the points.

Given,

⇒ 2x - y + 4 = 0

⇒ y = 2x + 4

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

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

x = 0, y = 2(0) + 4 = 0 + 4 = 4.

Table of values of equation (2) :

Steps of construction :

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

2. Join the points.

Given,

⇒ 5x + 4y = 29

⇒ 4y = 29 - 5x

⇒ y = $\dfrac{29 - 5x}{4}$ ..........(3)

When x = 1, y = $\dfrac{29 - 5 \times 1}{4} = \dfrac{29 - 5}{4} = \dfrac{24}{4}$ = 6,

x = 3, y = $\dfrac{29 - 5 \times 3}{4} = \dfrac{29 - 15}{4} = \dfrac{14}{4}$ = 3.5,

x = 5, y = $\dfrac{29 - 5 \times 5}{4} = \dfrac{29 - 25}{4} = \dfrac{4}{4}$ = 1.

Table of values of equation (3) :

Steps of construction :

1. Plot the points (1, 6), (3, 3.5), (5, 1) on graph.

2. Join the points.

From graph,

These three lines intersect each other at (-3, -2), (5, 1) and (1, 6).

Hence, the coordinates of the vertices of the triangle formed by these lines are (-3, -2), (5, 1) and (1, 6).

Find graphically the coordinates of the vertices of the triangle formed by the lines y - 2 = 0, 2y + x = 0 and y + 1 = 3 (x - 2). Hence, find the area of the triangle formed by these lines.

Answer

Given,

⇒ y - 2 = 0

⇒ y = 2 ........(1)

Given,

⇒ 2y + x = 0

⇒ 2y = -x

⇒ y = -$\dfrac{x}{2}$ ............(2)

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

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

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

Table of equation (2) :

Steps of construction :

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

2. Join the points.

Given,

⇒ y + 1 = 3(x - 2)

⇒ y + 1 = 3x - 6

⇒ y = 3x - 6 - 1

⇒ y = 3x - 7 ...........(3)

When x = 1, y = 3 × 1 - 7 = 3 - 7 = -4,

x = 2, y = 3 × 2 - 7 = 6 - 7 = -1,

x = 3, y = 3 × 3 - 7 = 9 - 7 = 2.

Table of equation (3) :

Steps of construction :

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

2. Join the points.

From graph,

A(-4, 2), B(3, 2), C(2, -1) are the vertices of the triangle.

From C, draw CD perpendicular to AB.

As, 1 block = 1 unit.

AB = 7 units and CD = 3 units

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

= $\dfrac{1}{2}$ × AB × CD

= $\dfrac{1}{2}$ × 7 × 3

= $\dfrac{21}{2}$ = 10.5 sq. units

Hence, coordinates of the vertices of the triangle are (-4, 2), (3, 2), (2, -1) and area = 10.5 sq. units.

A line segment is of length 10 units and one of its end is (-2, 3). If the ordinate of the other end is 9, find the abscissa of the other end.

Answer

Given,

Ordinate of the point on the other end = 9.

Let abscissa = x.

Given,

Distance between the two ends (-2, 3) and (x, 9) = 10 units.

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore \sqrt{[x - (-2)]^2 + (9 - 3)^2} = 10 \\[1em] \Rightarrow \sqrt{[x + 2]^2 + 6^2} = 10 \\[1em] \Rightarrow \sqrt{x^2 + 4 + 4x + 36} = 10 \\[1em] \Rightarrow x^2 + 4x + 40 = 100$

On squaring both sides,

$\Rightarrow x^2 + 4x + 40 - 100 = 0 \\[1em] \Rightarrow x^2 + 4x - 60 = 0\\[1em] \Rightarrow x^2 + 10x - 6x - 60 = 0 \\[1em] \Rightarrow x(x + 10) - 6(x + 10) = 0 \\[1em] \Rightarrow (x - 6)(x + 10) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x + 10 = 0 \\[1em] \Rightarrow x = 6 \text{ or } x = -10.$

Hence, abscissa of other end = 6 or -10.

A(-4, -1), B(-1, 2) and C(α, 5) are the vertices of an isosceles triangle. Find the value of α given that AB is the unequal side.

Answer

It is given that

A(-4, -1), B(-1, 2) and C(α, 5) are the vertices of an isosceles triangle and AB is the unequal side.

∴ AC = BC.

By distance formula,

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

Since, AC = BC

$\Rightarrow \sqrt{[α - (-4)]^2 + [5 - (-1)]^2} = \sqrt{[α - (-1)]^2 + [5 - 2]^2} \\[1em] \Rightarrow \sqrt{[α + 4]^2 + [5 + 1]^2} = \sqrt{[α + 1]^2 + [3]^2} \\[1em] \Rightarrow \sqrt{α^2 + 16 + 8α + 6^2} = \sqrt{α^2 + 1 + 2α + 9}$

On squaring both sides we get,

$\Rightarrow α^2 + 16 + 8α + 6^2 = α^2 + 1 + 2α + 9 \\[1em] \Rightarrow α^2 + 8α + 16 + 36 = α^2 + 2α + 10 \\[1em] \Rightarrow α^2 - α^2 + 8α - 2α = 10 - 16 - 36 \\[1em] \Rightarrow 6α = -42 \\[1em] \Rightarrow α = -\dfrac{42}{6} \\[1em] \Rightarrow α = -7.$

Hence, α = -7.

If A(-3, 2), B(α, β) and C(-1, 4) are the vertices of an isosceles triangle, prove that α + β = 1, given AB = BC.

Answer

Given,

A(-3, 2), B(α, β) and C(-1, 4) are the vertices of an isosceles triangle

By distance formula,

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

As, AB = BC

$\therefore \sqrt{[α - (-3)]^2 + (β - 2)^2} = \sqrt{(-1 - α)^2 + (4 - β)^2}$

On squaring both sides,

$\Rightarrow [α - (-3)]^2 + (β - 2)^2 = (-1 - α)^2 + (4 - β)^2 \\[1em] \Rightarrow [α + 3]^2 + (β - 2)^2 = (-1 - α)^2 + (4 - β)^2 \\[1em] \Rightarrow α^2 + 3^2 + 6α + β^2 + 4 - 4β = 1 + α^2 + 2α + 16 + β^2 - 8β \\[1em] \Rightarrow α^2 + 9 + 6α + β^2 + 4 - 4β = α^2 + 2α + 17 + β^2 - 8β \\[1em] \Rightarrow α^2 - α^2 + β^2 - β^2 + 6α - 2α - 4β + 8β = 17 - 9 - 4 \\[1em] \Rightarrow 4α + 4β = 4 \\[1em] \Rightarrow 4(α + β) = 4 \\[1em] \Rightarrow α + β = 1.$

Hence, proved that α + β = 1.

Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.

Answer

Let the points be A(3, 0), B(6, 4) and C(-1, 3).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(6 - 3)^2 + (4 - 0)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. \\[1em] \therefore BC = \sqrt{(-1 - 6)^2 + (3 - 4)^2} \\[1em] = \sqrt{(-7)^2 + (-1)^2} \\[1em] = \sqrt{49 + 1} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.\\[1em] \therefore AC = \sqrt{(-1 - 3)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5.$

∴ AB = AC = 5

∴ ΔABC is an isosceles triangle

AB² + AC² = 5² + 5²

= 25 + 25

= 50.

BC² = $(5\sqrt{2})^2$ = 50.

Since, AB² + AC² = BC².

Hence, proved that (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.

Show that the points (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square. Also find the area of the square.

Answer

Let A(2, 1), B(0, 3), C(-2, 1) and D(0, -1) be the four points.

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(0 - 2)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore BC = \sqrt{(-2 - 0)^2 + (1 - 3)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.\\[1em] \therefore CD = \sqrt{[0 - (-2)]^2 + [-1 - 1]^2} \\[1em] = \sqrt{[0 + 2]^2 + [-2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore AD = \sqrt{(0 - 2)^2 + (-1 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] =\sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.$

Since, AB = BC = CD = AD i.e. all sides are equal so, ABCD can be a rhombus or a square.

Calculating diagonals,

$AC = \sqrt{(-2 - 2)^2 + (1 - 1)^2} \\[1em] =\sqrt{(-4)^2 + 0^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] BD = \sqrt{(0 - 0)^2 + (-1 - 3)^2} \\[1em] = \sqrt{0^2 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = \sqrt{4} \text{ units}.$

Since, diagonals are equal.

∴ ABCD is a square.

Area of square = (side)²

= (AB)²

= $(\sqrt{8})^2$

= 8 sq. units.

Hence, proved that (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square and area = 8 sq. units.

Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4), taken in order, are the vertices of rhombus. Also, find its area. Do the given points form a square?

Answer

Let A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{[-5 - (-3)]^2 + [-5 - 2]^2} \\[1em] = \sqrt{[-5 + 3]^2 + [-7]^2} \\[1em] = \sqrt{[-2]^2 + [-7]^2} \\[1em] = \sqrt{4 + 49} \\[1em] = \sqrt{53}. \\[1em] \therefore BC = \sqrt{[2 - (-5)]^2 + [-3 - (-5)]^2} \\[1em] = \sqrt{[2 + 5]^2 + [-3 + 5]^2} \\[1em] = \sqrt{7^2 + 2^2} \\[1em] = \sqrt{49 + 4} \\[1em] = \sqrt{53}. \\[1em] \therefore CD = \sqrt{(4 - 2)^2 + [4 - (-3)]^2} \\[1em] = \sqrt{2^2 + 7^2} \\[1em] = \sqrt{4 + 49} \\[1em] = \sqrt{53}. \\[1em] \therefore AD = \sqrt{[4 - (-3)]^2 + (4 - 2)^2} \\[1em] = \sqrt{7^2 + 2^2} \\[1em] = \sqrt{49 + 4} \\[1em] = \sqrt{53}.$

Calculating diagonals,

$AC = \sqrt{[2 - (-3)]^2 + [-3 - 2]^2} \\[1em] = \sqrt{[2 + 3]^2 + [-5]^2} \\[1em] = \sqrt{5^2 + [-5]^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2} \text{ units}. \\[1em] BD = \sqrt{[4 - (-5)]^2 + [4 - (-5)]^2} \\[1em] = \sqrt{[4 + 5]^2 + [4 + 5]^2} \\[1em] = \sqrt{9^2 + 9^2} \\[1em] = \sqrt{81 + 81} \\[1em] = \sqrt{162} \\[1em] = 9\sqrt{2} \text{ units}.$

Since, all sides are equal and diagonals are not equal.

∴ ABCD is a rhombus.

Area of rhombus = $\dfrac{1}{2} \times d_1 \times d_2$

$= \dfrac{1}{2} \times 5\sqrt{2} \times 9\sqrt{2} \\[1em] = \dfrac{1}{2} \times 45 \times 2 \\[1em] = 45 \text{ sq. units}.$

Hence, ABCD is a rhombus and area = 45 sq. units.

The ends of a diagonal of a square have co-ordinates (-2, p) and (p, 2). Find p if the area of the square is 40 sq. units.

Answer

Given,

Ends of a diagonal of a square are (-2, p) and (p, 2).

Area of square = 40 sq. units

By formula,

Area of square = (side)²

∴ (side)² = 40

⇒ side = $\sqrt{40} = 2\sqrt{10}$ units.

By formula,

Diagonal of a square = $\sqrt{2}$ × side = $\sqrt{2} \times 2\sqrt{10} = 2\sqrt{20}$.

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{[p - (-2)]^2 + [2 - p]^2} \\[1em] \Rightarrow 2\sqrt{20} =\sqrt{[p + 2]^2 + [2 - p]^2} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{p^2 + 4 + 4p + 4 + p^2 - 4p} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{2p^2 + 8}$

On squaring both sides,

$\Rightarrow 4 \times 20 = 2p^2 + 8 \\[1em] \Rightarrow 2p^2 = 80 - 8 \\[1em] \Rightarrow 2p^2 = 72 \\[1em] \Rightarrow p^2 = \dfrac{72}{2} \\[1em] \Rightarrow p^2 = 36 \\[1em] \Rightarrow p = \sqrt{36} \\[1em] \Rightarrow p = \pm 6.$

Hence, p = ±6.

What type of quadrilateral do the points A(2, -2), B(7, 3), C(11, -1) and D(6, -6), taken in that order, form?

Answer

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(7 - 2)^2 + [3 - (-2)]^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50}. \\[1em] \therefore BC = \sqrt{(11 - 7)^2 + (-1 - 3)^2} \\[1em] = \sqrt{4^2 + (-4)^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32}. \\[1em] \therefore CD = \sqrt{(6 - 11)^2 + [-6 - (-1)]^2} \\[1em] = \sqrt{(-5)^2 + (-5)^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50}. \\[1em] \therefore AD = \sqrt{(6 - 2)^2 + [-6 - (-2)]^2} \\[1em] = \sqrt{4^2 + (-4)^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32}.$

Hence,

AB = CD and BC = AD

Since, opposite sides are equal,

Hence, proved that ABCD is a rectangle.

Find the coordinates of the centre of the circle passing through the three given points A(5, 1), B(-3, -7) and C(7, -1).

Answer

Let O(x, y) be the coordinates of the centre of the circle. Points A(5, 1), B(-3, -7), and C(7, -1) are on the circle.

By distance formula,

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

As, OA = OB [∵ Both are radius of the same circle]

$\therefore \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{[x - (-3)]^2 + [y - (-7)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 25 - 10x + y^2 + 1 - 2y} = \sqrt{[x + 3]^2 + [y + 7]^2} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 10x - 2y + 26} = \sqrt{x^2 + 9 + 6x + y^2 + 49 + 14y} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 10x - 2y + 26} = \sqrt{x^2 + y^2 + 6x + 14y + 58}$

On squaring both sides,

$\Rightarrow x^2 + y^2 - 10x - 2y + 26 = x^2 + y^2 + 6x + 14y + 58 \\[1em] \Rightarrow x^2 - x^2 + y^2 - y^2 + 6x + 10x + 14y + 2y = 26 - 58 \\[1em] \Rightarrow 16x + 16y = -32 \\[1em] \Rightarrow 16(x + y) = -32 \\[1em] \Rightarrow x + y = \dfrac{-32}{16} \\[1em] \Rightarrow x + y = -2 \\[1em] \Rightarrow x = -2 - y \space .........(1).$

As, OC = OB [∵ Both are radius of the same circle]

$\therefore \sqrt{(x - 7)^2 + [y - (-1)]^2} = \sqrt{[x - (-3)]^2 + [y - (-7)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 49 - 14x + [y + 1]^2} = \sqrt{[x + 3]^2 + [y + 7]^2} \\[1em] \Rightarrow \sqrt{x^2 + 49 - 14x + y^2 + 1 + 2y} = \sqrt{x^2 + 9 + 6x + y^2 + 49 + 14y} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 14x + 2y + 50} = \sqrt{x^2 + y^2 + 6x + 14y + 58}$

On squaring both sides,

$\Rightarrow x^2 + y^2 - 14x + 2y + 50 = x^2 + y^2 + 6x + 14y + 58 \\[1em] \Rightarrow x^2 + y^2 - x^2 - y^2 + 6x + 14x + 14y - 2y = 50 - 58 \\[1em] \Rightarrow 20x + 12y = -8 \\[1em] \Rightarrow 5x + 3y = -2 \\[1em]$

Substituting value of x in above equation from equation 1 :

⇒ 5(-2 - y) + 3y = -2

⇒ -10 - 5y + 3y = -2

⇒ -2y = -2 + 10

⇒ -2y = 8

⇒ y = -4.

⇒ x = -2 - y = -2 - (-4) = -2 + 4 = 2.

C = (x, y) = (2, -4).

Hence, the coordinates of the centre of the circle are (2, -4).