Chapter: Coordinate Geometry — Class 9 Maths (NCERT)

🔹 1. Introduction

Coordinate Geometry (also called Cartesian Geometry) is the branch of mathematics in which the position of a point, line, or shape is described using ordered pairs (x, y) on a plane.

It forms a bridge between Algebra and Geometry.

🔹 2. Cartesian System

(a) Coordinate Axes

• Two perpendicular lines intersecting at a point O (origin) form the coordinate axes.

• The horizontal line is called the x-axis.

• The vertical line is called the y-axis.

(b) Quadrants

• The plane is divided into four quadrants by the axes.

Diagram: Cartesian plane showing the four quadrants and origin.

🔹 3. Coordinates of a Point

Each point on the plane is represented by an ordered pair $(x, y)$:

• $x$: the abscissa (distance along the x-axis)

• $y$: the ordinate (distance along the y-axis)

Example:
Point $A(3, 2)$ means $x = 3$, $y = 2$.

🔹 4. Signs of Coordinates

🔹 5. Coordinates on the Axes

• On x-axis: $y = 0$→ Points like $(2, 0)$
  

• On y-axis: $x = 0$→ Points like $(0, -3)$
  

• At origin: $(0, 0)$
  

🔹 6. Distance Formula

To find the distance between two points
$A(x_1, y_1)$ and $B(x_2, y_2)$:

$$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^{\mathrm{2<span\; style="font-size:\; revert;\; font-family:\; Manrope,\; sans-serif;\; letter-spacing:\; -0.1px;"></span>}}}$$

Example:

Find the distance between $P(2, 3)$ and $Q(5, 7)$.

$$AB = \sqrt{(5 – 2)^2 + (7 – 3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

🔹 7. Section Formula

If a point $P(x, y)$ divides the line joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m : n$, then:

$$x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}$$

Special case:
When $m = n$, $P$ is the mid-point of $AB$.

🔹 8. Mid-Point Formula

$$\text{If } P(x, y) \text{ is the mid-point of } A(x_1, y_1) \text{ and } B(x_2, y_2),$$

then

$$x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{\mathrm{2<span\; style="font-family:\; Manrope,\; sans-serif;\; font-size:\; revert;\; letter-spacing:\; -0.1px;"></span><span\; class="vlist-s"\; style="font-size:\; revert;\; font-family:\; Manrope,\; sans-serif;\; letter-spacing:\; -0.1px;"></span>}}$$Example:
Find the midpoint of $A(2, 3)$ and $B(4, 7)$.

$$x = \frac{2 + 4}{2} = 3, \quad y = \frac{3 + 7}{2} = 5$$∴ Mid-point = $(3, 5)$

🔹 9. Area of a Triangle (using coordinates)

For vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$:

$$\text{Area}=\frac{1}{2}\mid x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\mid$$

$$$$Example: Find the area of the triangle formed by $A(2, 3)$, $B(4, 5)$, $C(6, 3)$.

$$\text{Area}=\frac{1}{2}\mathrm{\mid }2(5-3)+4(3-3)+6(3-5)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }4+0-12\mathrm{\mid }=\frac{1}{2}\times 8=4$$

Area = 4 square units.

🔹 10. Collinearity of Points

Three points $A$, $B$, and $C$ are collinear if:

$$\text{Area of } \triangle ABC = 0$$

or equivalently, the slopes between each pair are equal.

🔹 11. Important Observations

• Distance formula derives from the Pythagoras theorem.

• Coordinates help in proving geometrical properties algebraically.

• Used widely in graphs, geometry, and physics (motion, forces).

✏️ Key Formulas at a Glance

✅ Summary

• Coordinate geometry combines algebra with geometry.

• Helps locate points, find distances, midpoints, areas.

• Fundamental for higher topics like graphs, slopes, equations of lines, and geometry proofs.

ANSWER KEY — Coordinate Geometry (Class 9 Mathematics)

All solutions are written in full-sentence, CBSE-style format and show step-by-step calculations.

Q1. What do you understand by the terms abscissa, ordinate, and origin in the Cartesian coordinate system?

Answer:
The abscissa of a point is the x-coordinate which gives the horizontal distance of the point from the y-axis. The ordinate of a point is the y-coordinate which gives the vertical distance of the point from the x-axis. The origin is the point where the x-axis and y-axis meet; its coordinates are $(0,0)$.

Q2. Write the coordinates of:

(a) A point lying on the x-axis at a distance of 4 units from the origin on the right-hand side.
(b) A point lying on the y-axis at a distance of 3 units above the origin.

Answer:
(a) A point on the x-axis 4 units to the right of the origin has x = +4 and y = 0, so its coordinates are $(4,0)$.
(b) A point on the y-axis 3 units above the origin has x = 0 and y = +3, so its coordinates are $(0,3)$.

Q3. Identify the quadrant in which each of the following points lies and state the sign of their abscissa and ordinate: (a) $(5,7)$ (b) $(−3,6)$ (c) $(−4,−2)$ (d) $(6,−3)$.

Answer:
(a) Point $(5,7)$ lies in the first quadrant because $x>0$ and $y>0$. The abscissa is positive and the ordinate is positive.
(b) Point $(−3,6)$ lies in the second quadrant because $x<0$ and $y>0$. The abscissa is negative and the ordinate is positive.
(c) Point $(−4,−2)$ lies in the third quadrant because $x<0$ and $y<0$. The abscissa is negative and the ordinate is negative.
(d) Point $(6,−3)$ lies in the fourth quadrant because $x>0$ and $y<0$. The abscissa is positive and the ordinate is negative.

Q4. Fill in the blanks:

(a) The coordinates of the origin are __________.
(b) If a point lies on the y-axis, its x-coordinate is __________.
(c) If a point lies on the x-axis, its y-coordinate is __________.

Answer:
(a) The coordinates of the origin are $(0,0)$.
(b) If a point lies on the y-axis, its x-coordinate is $0$.
(c) If a point lies on the x-axis, its y-coordinate is $0$.

Q5. Find the distance between the two points $A(2,3)$ and $B(5,7)$ using the distance formula. Show all steps.

Solution and Answer:
The distance formula for points $A(x_1,y_1)$ and $B(x_2,y_2)$ is

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

Here $x_1=2,\; y_1=3,\; x_2=5,\; y_2=7$. Compute step by step:

• $x_2-x_1 = 5-2 = 3$.

• $(x_2-x_1)^2 = 3^2 = 9$.

• $y_2-y_1 = 7-3 = 4$.

• $(y_2-y_1)^2 = 4^2 = 16$.

• Sum $= 9 + 16 = 25$.

• $AB = \sqrt{25} = 5$.

Therefore, the distance between $A$ and $B$ is $5$ units.

Q6. Find the coordinates of the midpoint of the line segment joining the points $P(−2,4)$ and $Q(6,−8)$. Also explain the formula used.

Solution and Answer:
The midpoint $M$ of the line segment joining $P(x_1,y_1)$ and $Q(x_2,y_2)$ is given by

$$M\!\left(\frac{x_1+x_2}{2},\,\frac{y_1+y_2}{2}\right).$$

Here $x_1=-2,\; y_1=4,\; x_2=6,\; y_2=-8$. Compute step by step:

• $\dfrac{x_1+x_2}{2} = \dfrac{-2 + 6}{2} = \dfrac{4}{2} = 2.$

• $\dfrac{y_1+y_2}{2} = \dfrac{4 + (-8)}{2} = \dfrac{-4}{2} = -2.$

Thus the midpoint is $M(2,-2)$.

Q7. The point $R(x,y)$ divides the line segment joining the points $A(2,3)$ and $B(8,5)$ in the ratio $2:1$. Find the coordinates of the point $R$ using the section formula.

Solution and Answer:
Using the internal section formula: if a point divides $AB$ in the ratio $m:n$ (where the ratio corresponds to $AP:PB = m:n$), then

$$x=\frac{m x_2 + n x_1}{m+n},\qquad y=\frac{m y_2 + n y_1}{m+n}.$$

Here $x_1=2,\; y_1=3,\; x_2=8,\; y_2=5,\; m=2,\; n=1$. Compute each coordinate:

• $x=\dfrac{2\cdot 8 + 1\cdot 2}{2+1}=\dfrac{16+2}{3}=\dfrac{18}{3}=6.$

• $y=\dfrac{2\cdot 5 + 1\cdot 3}{2+1}=\dfrac{10+3}{3}=\dfrac{13}{3}.$

Therefore, $R$ has coordinates $\displaystyle R\!\left(6,\;\frac{13}{3}\right)$.

Q8. Verify whether the points $P(1,2)$, $Q(4,6)$, and $R(7,10)$ are collinear by using the concept of slope.

Solution and Answer:
Three points are collinear if the slope of $PQ$ equals the slope of $QR$.

Slope $m_{PQ}=\dfrac{y_Q-y_P}{x_Q-x_P}=\dfrac{6-2}{4-1}=\dfrac{4}{3}.$
Slope $m_{QR}=\dfrac{y_R-y_Q}{x_R-x_Q}=\dfrac{10-6}{7-4}=\dfrac{4}{3}.$

Since $m_{PQ}=m_{QR}=\dfrac{4}{3}$, the three given points are collinear.

Q9. Find the area of the triangle whose vertices are $A(2,3)$, $B(4,5)$, and $C(6,3)$ using the coordinate-geometry area formula.

Solution and Answer:
The area of $\triangle ABC$ is

$$\text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$

Substitute $x_1=2,y_1=3;\; x_2=4,y_2=5;\; x_3=6,y_3=3$:

Compute each term:

• $x_1(y_2-y_3)=2(5-3)=2\times2=4.$

• $x_2(y_3-y_1)=4(3-3)=4\times0=0.$

• $x_3(y_1-y_2)=6(3-5)=6\times(-2)=-12.$

Sum $=4 + 0 -12 = -8$. Absolute value $|-8| = 8$. Then area $=\dfrac{1}{2}\times 8 = 4.$

Therefore, the area of the triangle is $4$ square units.

Q10. The points $A(2,3)$, $B(4,k)$, and $C(6,-3)$ are collinear. Find the value of $k$.

Solution and Answer:
If three points are collinear, area of the triangle formed by them is zero. Using the area formula:

$$\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|=0.$$

Substitute $x_1=2,y_1=3;\; x_2=4,y_2=k;\; x_3=6,y_3=-3$:

Compute inside absolute value:

• $x_1(y_2-y_3)=2\big(k – (-3)\big)=2(k+3)=2k+6.$

• $x_2(y_3-y_1)=4\big(-3-3\big)=4(-6)=-24.$

• $x_3(y_1-y_2)=6\big(3-k\big)=18-6k.$

Sum $=(2k+6)+(-24)+(18-6k)=2k+6-24+18-6k=(2k-6k)+(6-24+18)=-4k+0=-4k.$

So $\tfrac12\lvert -4k\rvert=0\Rightarrow |-4k|=0\Rightarrow -4k=0\Rightarrow k=0.$

Therefore, $k=0$.

Q11. The vertices of a quadrilateral are $A(2,3)$, $B(6,7)$, $C(10,3)$, and $D(6,-1)$. Show that the given quadrilateral is a square by using the distance formula and verifying that adjacent sides are perpendicular.

Solution and Answer:
We will compute the lengths of all four sides and show that adjacent sides are equal in length and one pair of adjacent sides is perpendicular.

1. Compute side lengths using the distance formula.

• $AB=\sqrt{(6-2)^2+(7-3)^2}=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}.$

• $BC=\sqrt{(10-6)^2+(3-7)^2}=\sqrt{4^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}.$

• $CD=\sqrt{(6-10)^2+(-1-3)^2}=\sqrt{(-4)^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}.$

• $DA=\sqrt{(2-6)^2+(3-(-1))^2}=\sqrt{(-4)^2+4^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}.$

All four sides are equal in length, each equal to $4\sqrt{2}$.

2. Verify that an adjacent pair of sides is perpendicular by computing the dot product of vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$.

• $\overrightarrow{AB}=(6-2,\;7-3)=(4,4).$

• $\overrightarrow{BC}=(10-6,\;3-7)=(4,-4).$

• Dot product: $(4,4)\cdot(4,-4)=4\cdot4 + 4\cdot(-4)=16-16=0.$

A zero dot product shows that $\overrightarrow{AB}$ is perpendicular to $\overrightarrow{BC}$. Since all sides are equal and one pair of adjacent sides are perpendicular, the quadrilateral is a square.

Therefore, the given quadrilateral is a square.

Q12. The line segment joining the points $A(1,2)$ and $B(7,4)$ is extended beyond $B$ to a point $C$ such that $AC = 3\times AB$. Find the coordinates of point $C$.

Solution and Answer:
Vector $\overrightarrow{AB}=(7-1,\;4-2)=(6,2).$
If $AC=3\times AB$ as vectors, then $\overrightarrow{AC}=3\cdot\overrightarrow{AB}=(18,6).$ Since $\overrightarrow{AC}=\overrightarrow{A C}=(x_C-1,\; y_C-2)$, we have

• $x_C – 1 = 18 \Rightarrow x_C = 19.$

• $y_C – 2 = 6 \Rightarrow y_C = 8.$

Therefore, the coordinates of $C$ are $(19,8)$.

(Remark: this places $C$ beyond $B$ on the same straight line, at a distance three times $AB$ from $A$.)

Q13. A point $P$ divides the line segment joining $A(4,-3)$ and $B(-2,6)$ internally in the ratio $3:2$. Find the coordinates of $P$ using the section formula.

Solution and Answer:
Using the internal section formula with $m:n = 3:2$, where $m$ corresponds to the portion toward $B$ in the chosen convention $x=\dfrac{m x_2 + n x_1}{m+n}$, we set $x_1=4,y_1=-3,\;x_2=-2,y_2=6,\;m=3,n=2$:

Compute $x$ coordinate:

$$x=\frac{3\cdot(-2)+2\cdot 4}{3+2}=\frac{-6+8}{5}=\frac{2}{5}.$$

Compute $y$ coordinate:

$$y=\frac{3\cdot 6+2\cdot (-3)}{3+2}=\frac{18-6}{5}=\frac{12}{5}\mathrm{.}$$

Therefore, $P\left(\dfrac{2}{5},\;\dfrac{12}{5}\right)$.

Q14. Write the coordinates of the reflection of the point $A(5,-4)$:

(a) In the x-axis
(b) In the y-axis

Answer:
Reflection rules: reflection in the x-axis changes the sign of the y-coordinate only; reflection in the y-axis changes the sign of the x-coordinate only.

(a) Reflection of $A(5,-4)$ in the x-axis is $(5,4)$.
(b) Reflection of $A(5,-4)$ in the y-axis is $(-5,-4)$.

Q15. If the midpoint of the line segment joining the points $A(2,x)$ and $B(4,6)$ is $(3,5)$, find the value of $x$.

Solution and Answer:
Midpoint formula gives

$$\left(\frac{2+4}{2},\; \frac{x+6}{2}\right)=(3,5).$$

Compute the x-coordinate check: $\dfrac{2+4}{2}=\dfrac{6}{2}=3$ (consistent). Now equate the y-coordinate:

$$\frac{x+6}{2}=5 \Rightarrow x+6=10 \Rightarrow x=4.$$

Therefore, $x=4$.

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