Two perpendicular number lines intersecting at point zero are called coordinate axes. The point of intersection is called origin and denoted by ‘O’. The horizontal number line is the x-axis (denoted by X’OX) and the vertical one is the y-axis (denoted by Y’OY).

The axes divide the Cartesian plane into four parts called the quadrants (one fourth part), numbered I, II, III and IV anticlockwise from OX.

Sign of coordinates depicts the quadrant in which it lies.

The coordinates of a point lying on the x-axis are of the form (x, 0) and that of the point on the y-axis are of the form (0, y).

The distance formula is used to find the distance between two any points say P(x₁, y₁) and Q(x₂, y₂) which is given by: $PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Distance between Two Points on the Same Coordinate Axes

The distance between two points that are on the same axis (x-axis or y-axis), is given by the difference between their ordinates if they are on the y-axis, else by the difference between their abscissa if they are on the x-axis.

Distance AB = 6 – (-2) = 8 units

Distance CD = 4 – (-8) = 12 units

Distance between Two Points Using Pythagoras Theorem

Finding distance between 2 points using Pythagoras Theorem

Let P(x₁, y₁) and Q(x₂, y₂) be any two points on the cartesian plane.

Draw lines parallel to the axes through P and Q to meet at T.

ΔPTQ is right-angled at T.

By Pythagoras Theorem,

PQ² = PT² + QT²

= (x₂ – x₁)² + (y₂ – y₁)²

PQ = √[x₂ – x₁)² + (y₂ – y₁)²]

For the given four points A, B, C and D, if:

The point of intersection of the perpendicular bisectors of the sides of a triangle is called the circumcentre. In the figure, O is the circumcentre of the triangle ABC.

Circumcentre of a triangle is equidistant from the vertices of the triangle. That is, P is the circumcentre of ∆ ABC, if PA = PB = PC.

If P is a point lying on the line segment joining the points A and B such that AP: BP = m: n. Then, we say that the point P divides the line segment AB internally in the ratio m: n.

Coordinates of a point which divides the line segment joining the points (x₁, y₁) and (x₂, y₂) in the ratio m: n internally are given by: $\left(\frac{{mx}_2+{nx}_1}{m+n},\frac{{my}_2+{my}_1}{m+n}\right)$This is known as the section formula.

11.If P is a point lying on AB produced such that AP: BP = m: n, then point P divides AB externally in the ratio m: n.

If P divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n externally, then the coordinates of point P are given by $\left(\frac{{mx}_2-{nx}_1}{m-n},\frac{{my}_2-{my}_1}{m-n}\right)$

Mid-point divides the line segment in the ratio 1:1. Coordinates of the mid-point of a line segment joining the points (x₁, y₁) and (x₂, y₂) are $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

The point of intersection of the three medians of a triangle is called the centroid.

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In the figure, G is the centroid of the triangle ABC where AD, BF and CE are the medians through A, B and C respectively.

Centroid divides the median in the ratio of 2:1.

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of a triangle ABC, then the coordinates of the centroid are given by $G\left(x,y\right)=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

The point of intersection of all the internal bisectors of the angles of a triangle is called the incentre.

It is also the centre of a circle which touches all the sides of a triangle (such type of a circle is named as the incircle).

In the figure, I is the incentre of the triangle ABC.

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of a triangle, then the coordinates of incentre are given by $\left(\frac{{ax}_1+{bx}_2+{cx}_3}{a+b+c},\frac{{ay}_1+{by}_2+{cy}_3}{a+b+c}\right)$

The point of intersection of all the perpendiculars drawn from the vertices on the opposite sides (called altitudes) of a triangle is called the Orthocentre which can be obtained by solving the equations of any two of the altitudes.

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In the figure, O is the orthocentre of the triangle ABC.

If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a triangle, then the area of triangle ABC is given by $\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

Area of a quadrilateral can be calculated by dividing it into two triangles.

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Area of quadrilateral ABCD = Area of ∆ABC+ Area of ∆ACD

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Note: To find the area of a polygon, divide it into triangular regions having no common area, then add the areas of these regions.

Short Questions :

(i) (-5, 7), (-1, 3)

(ii) (a, b), (-a, -b)  

⇒ Given points are collinear

Then 2(3) – 3(a) = 5

– 3a = 5 – 6

– 3a = -1

⇒ a = $\frac{1}{3}$

x₂ = 0, y₂ = 0

⇒ 1(0 – b) + 0 (b – 2) + a(2 – 0) = 0

⇒ -b + 2a = 0 or 2a = b

Hence, the point P divides AB in the ratio 2 : 7.

Since, (x, 0) is equidistant from the points (-3, 4) and (2, 5).

Given that this point lies on the x-axis

Thus, the required ratio is 2 : 1.

∴ Given statement is true.

AP ≠ BP

∴ P(0, 2) does not lie on the perpendicular bisector of AB. So, given statement is false.

Clearly, AB + BC ≠ AC

∴ A, B, C are not collinear.

Thus, we have x₁ = -5 and x₂ = -1

y₁ = 7 and y₂ = 3

A (x₁, y₁) = (7, -2), B (x₂, Y₂) = (5, 1) and C (x₃, y₃) = (3, k)

Since these points are collinear therefore area (∆ABC) = 0

⇒ 12 [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0

⇒ x1(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0

⇒ 7(1 – k) + 5(k + 2) + 3(-2 -1) = 0

⇒ 7 – 7k + 5k + 10 – 9 = 0

⇒ -2k + 8 = 0

⇒ 2k = 8

⇒ k = 4

Hence, given points are collinear for k = 4.

Now, let P, Q, R be the mid-points of BC, CA and AB, respectively.

So, coordinates of P, Q, R are

Ratio of ar (∆PQR) to the ar (∆ABC) = 1 : 4.

Let A(4, -2), B(-3, -5), C(3, -2) and D(2, 3) be the vertices of the quadrilateral ABCD.

Now, area of quadrilateral ABCD

= area of ∆ABC + area of ∆ADC

Hence, the median divides it into two triangles of equal areas.

The ratio in which p divides the line segment is $\frac{3}{5}$, i.e., 3 : 5.

⇒ BD = DC

The coordinates of midpoint D are given by.

Hence, AD divides ∆ABC into two equal areas.

According to the question,

PA = QA

∴ Coordinates of point B are (0, 3). So,

BC = 6 units Let the coordinates of point A be (x, 0).

Using distance formula,

∴ Coordinates of point A = (x, 0) = (3√3, 0)

Since BACD is a rhombus.

∴ AB = AC = CD = DB

∴ Coordinates of point D = (-3√3, 0).

Area of the triangle = $\frac{1}{2}$[t + 2 – t) + (t + 2) (t – t + 2) + (t + 3) (t – 2 – t – 2)]

= $\frac{1}{2}$ [2t + 2t + 4 – 4t – 12 ]

= 4 sq. units

which is independent of t.

Hence proved.

It can be observed that the coordinates of point P, Q and R are (4, 6), (3, 2), and (6, 5) respectively.

Assertion Reason Answer-