Class 10 Mathematics Chapter 7 Coordinate Geometry

Key Concept: Basic Formula

a) $\frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$
[Solution Description] The formula for calculating the area of a triangle using coordinates is given by:

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

This formula calculates the absolute value of a determinant divided by 2 which represents the area.

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Key Concept: Simple Distance Check

a) Yes
[Solution Description] Calculate the distances CD and CE:

Distance $CD = \sqrt{(3 + 3)^2 + (0 – 4)^2} = \sqrt{36 + 16} = \sqrt{52}$

Distance $CE = \sqrt{(3 – 9)^2 + (0 + 4)^2} = \sqrt{36 + 16} = \sqrt{52}$

Since both distances are equal, point C is equidistant from D and E.

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Key Concept: Midpoint Concept

b) (1, 5)
[Solution Description] Use the midpoint formula $P\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$:

For $x_1 = -1$, $y_1 = 2$, $x_2 = 3$, $y_2 = 8$,

$P = \left(\frac{-1 + 3}{2}, \frac{2 + 8}{2}\right) = (1, 5).$

Hence, the midpoint is (1, 5).

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Key Concept: Distance Formula

b) $2\sqrt{13}$
[Solution Description] To find the distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$, we use the distance formula: $AB = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
Substitute $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (-1, 4)$ into the formula:
$AB = \sqrt{((-1) – 3)^2 + (4 – (-2))^2}$
$\sqrt{(-4)^2 + (6)^2}$
$\sqrt{16 + 36}$
$= \sqrt{52}$
$= \sqrt{4 \times 13}$
$= 2\sqrt{13}$
Thus, the distance is $2\sqrt{13}$

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Key Concept: Ratio Identification

b) 1:5
[Solution Description] Let the point (3, 4) divide the line segment in the ratio $m_1:m_2$. Using the section formula for x-coordinates:

$3 = \frac{m_1 \cdot 8 + m_2 \cdot 2}{m_1 + m_2} \\$
$3(m_1 + m_2) = 8m_1 + 2m_2 \\$
$3m_1 + 3m_2 = 8m_1 + 2m_2 \\$
$3m_2 – 2m_2  = 8m_1 – 3m_1 \\$
$m_2  = 5m_1$

Therefore, the point divides the segment in the ratio 1:5.

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Key Concept: Collinearity Check

a) Yes
[Solution Description] Three points are collinear if the area of the triangle formed by them is zero or they lie on a straight line. We can check using the distance formula to see if one segment equals the sum of the other two.

Calculate distances between the points: $d_1 = \sqrt{(5 – 3)^2 + (11 – 7)^2} = \sqrt{4 + 16} = \sqrt{20}$

$d_2 = \sqrt{(9 – 5)^2 + (19 – 11)^2} = \sqrt{16 + 64} = \sqrt{80}$

$d_2 = \sqrt{(9 – 5)^2 + (19 – 11)^2} = \sqrt{16 + 64} = \sqrt{80}$

Since d_3 = d_1 + d_2​, the points are collinear.

Notes:

Collinearity Check in Points

To check if three points are collinear (lie on the same straight line), we can use the area method or slope method.

1. Area Method:
   The area of the triangle formed by three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$
   is given by: $\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|$
   If the area is zero, the points are collinear.
2. Slope Method:
   Calculate the slope between any two pairs of points. If the slopes are equal, the points are collinear.
   Slope between two points $\text{slope} = \frac{y_2 – y_1}{x_2 – x_1}$

Key Concept: Coordinate Calculation

a) (5.6, 5)
[Solution Description] By using the section formula for internal division:
$x = \frac{{m_1 x_2 + m_2 x_1}}{{m_1 + m_2}} = \frac{{3 \times 8 + 2 \times 2}}{3+2}$

$= \frac{24 + 4}{5} = \frac{28}{5} = 5.6$
$y = \frac{{m_1 y_2 + m_2 y_1}}{{m_1 + m_2}} = \frac{{3 \times 9 + 2 \times (-1)}}{3+2}$

$= \frac{27 – 2}{5} = \frac{25}{5} = 5$
Thus, the point is $ (5.6, 5) $.

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Key Concept: Simple Section Formula

a) (9.2, 12.8)
[Solution Description] Using the section formula, the coordinates of the point dividing the line segment in the ratio m_1:m_2 are given by:
$P(x, y) = \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right).$

Here, $x_1 = 6, y_1 = 8, x_2 = 14, y_2 = 20, m_1 = 2, and m_2 = 3$

Calculating the x-coordinate: $x = \frac{2 \cdot 14 + 3 \cdot 6}{2 + 3} = \frac{28 + 18}{5} = \frac{46}{5} = 9.2.$

Calculating the y-coordinate: $y = \frac{2 \cdot 20 + 3 \cdot 8}{2 + 3} = \frac{40 + 24}{5} = \frac{64}{5} = 12.8.$

Hence, the coordinates of the point are (9.2, 12.8).

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Key Concept: Origin Distance

c) 10 units
[Solution Description] To find the distance of a point from the origin, use the formula $OI = \sqrt{x^2 + y^2}$. For point I(6, 8), the distance is $OI = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ units.

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Key Concept: Area of Triangle

a) 6
[Solution Description] The area of a triangle with vertices at $\frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|$
Here,$(x_1, y_1) = (1, 2), \quad (x_2, y_2) = (3, 4), \quad (x_3, y_3) = (5, 0)$

Calculate the area: $\text{Area} = \frac{1}{2} \left| 1(4-0) + 3(0-2) + 5(2-4) \right|$
$= \frac{1}{2} \left| 4 – 6 – 10 \right|$
$= \frac{1}{2} \left| -12 \right|$
$= \frac{1}{2} \times 12 = 6$
So, the area is 6 square units.

Notes:

• Area of Triangle Formula: The area of a triangle with vertices at coordinates
  $\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|$
  
• Significance of the Formula:
  • This formula is derived from the determinant of a matrix formed by the coordinates of the triangle’s vertices.
  • The area is always positive, and the absolute value ensures the result is non-negative.
• Application:
  • Can be used to find the area of any triangle when the coordinates of its three vertices are known.
  • It is also useful in geometric problems involving coordinate geometry.

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Key Concept: Coordinate Division

a) $(0, \frac{2}{3})$
[Solution Description] Use the section formula for the point dividing internally a line segment in the ratio $m_1:m_2$:

$P = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$,

here $x_1 = 2$, $y_1 = -2$, $x_2 = -4$, $y_2 = 6$, $m_1 = 1$, and $m_2 = 2$. Substituting the values:

$x = \frac{1(-4) + 2(2)}{1 + 2} = \frac{-4 + 4}{3} = 0$,

$y = \frac{1(6) + 2(-2)}{1 + 2} = \frac{6 – 4}{3} = \frac{2}{3}$.

So, the coordinates of the point are $(0, \frac{2}{3})$.

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Key Concept: Origin Distance

b) 10 units
[Solution Description] Use the distance formula from the origin:
$\text{Distance} = \sqrt{(6 – 0)^2 + (8 – 0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. So, the distance is 10 units.

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Key Concept: Quadrant Distance
b) $\sqrt{193}$ units
[Solution Description] To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the formula:
$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
Substituting the given points A(3, -5) and B(-4, 7), we have:
$PQ = \sqrt{((-4) – 3)^2 + (7 – (-5))^2} = \sqrt{(-7)^2 + (12)^2}$
Simplifying further: $PQ = \sqrt{49 + 144} = \sqrt{193}$
Therefore, the distance is $\sqrt{193}$ units.

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Key Concept: Coordinate Identification

b) Equilateral triangle
[Solution Description] For an equilateral triangle, all sides must be equal in length. Calculate each side using the distance formula

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

Distance AB: $d_{AB} = \sqrt{(3 – 0)^2 + (0 – 0)^2} = \sqrt{9} = 3$

Distance BC: $d_{BC} = \sqrt{(1.5 – 3)^2 + (\sqrt{6.75} – 0)^2} = \sqrt{(-1.5)^2 + 6.75} = \sqrt{9} = 3$

Distance CA:$d_{CA} = \sqrt{(1.5 – 0)^2 + (\sqrt{6.75} – 0)^2} = \sqrt{2.25 + 6.75} = \sqrt{9} = 3$

All sides are 3 units, hence it is an equilateral triangle.

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Key Concept: Distance Formula Application

b) $8\sqrt{2}$​
[Solution Description] To find the distance between two points $(x_1, y_1) \text{ and } (x_2, y_2)$
we use the distance formula: $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

Substituting the given points: $d = \sqrt{(-5) – 3)^2 + (7 – (-1)^2}$

Simplifying inside the square root: $d = \sqrt{(-8)^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128}$ Finally, simplifying further:

$d = 128 = 64 \times 2 = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$

Therefore, the distance is $8\sqrt{2}$

Notes

• • Combines algebra and geometry to study shapes and figures using coordinates on a plane.
  • Introduces tools to calculate distances, midpoints, and areas.
• Distance Formula: Used to find the distance between two points $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
  Application of Distance Formula: To calculate the length of line segments.
  • To check if points are collinear (sum of two segment lengths equals the third).
  • To verify geometric properties, e.g., diagonals in squares or rectangles.
• Real-Life Usage: Used in map navigation, design, and construction for measuring distances accurately.
• Benefits: Provides an accurate, algebraic method to solve geometric problems.

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Key Concept: Simple Ratio Division

a) (5, 5)
[Solution Description] Using the section formula

$\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$ with $m:n = 1:3$,
$\text{Coordinates} = \left(\frac{1 \times 2 + 3 \times 6}{1+3}, \frac{1 \times 8 + 3 \times 4}{1+3}\right)$

$= \left(\frac{2 + 18}{4}, \frac{8 + 12}{4}\right) = \left(\frac{20}{4}, \frac{20}{4}\right) = (5, 5).$
Therefore, the coordinates are $(5, 5)$.

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Key Concept: Basic Distance Formula

b) 5 units

[Solution Description] The distance between two points $H(2,3)$ and $I(5,7)$ can be found using the distance formula:

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

Substituting the given coordinates: $d = \sqrt{(5 – 2)^2 + (7 – 3)^2}$

$ \sqrt{3^2 + 4^2}$

$ \sqrt{9 + 16} = \sqrt{25} = 5$

Thus, the distance between points H and I is 5 units. This formula is crucial in coordinate geometry for measuring spatial relationships.

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