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Coordinate Geometry - Section Formula and Mid-point Formula

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Cartesian Plane and Coordinates: A point in a 2D plane is identified by an ordered pair (x,y)(x, y). The horizontal line is the X-axis and the vertical line is the Y-axis. They intersect at the origin O(0,0)O(0, 0). When points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) are plotted, they form a line segment ABAB that can be analyzed using coordinate geometry formulas.

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Internal Division (Section Formula): If a point P(x,y)P(x, y) lies on the line segment joining A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) such that it divides the segment into two parts APAP and PBPB in the ratio m1:m2m_1 : m_2, the coordinates of PP are found using the section formula. Visually, imagine PP being a marker on the line ABAB that is closer to AA if m1<m2m_1 < m_2 and closer to BB if m1>m2m_1 > m_2.

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Mid-point Formula: The mid-point MM of a line segment ABAB is the point that divides the segment into two equal halves (ratio 1:11:1). Geometrically, it represents the average of the x-coordinates and the average of the y-coordinates of the endpoints. On a graph, MM is the exact center of the distance between AA and BB.

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Determining the Ratio: To find the ratio in which a point P(x,y)P(x, y) divides the segment ABAB, we often assume the ratio to be k:1k : 1. By substituting the known coordinates of P,A,P, A, and BB into the section formula, we can solve for kk. If kk is positive, the point PP lies between AA and BB (internal division).

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Centroid of a Triangle: The centroid G(x,y)G(x, y) is the point where the three medians of a triangle meet. A median is a line segment connecting a vertex to the mid-point of the opposite side. The centroid divides each median in the ratio 2:12:1 starting from the vertex. Visually, if you have a triangle with vertices A,B,A, B, and CC, the centroid is the point where the triangle would perfectly balance if cut out of a flat sheet.

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Trisection of a Line Segment: Points of trisection are two points that divide a line segment ABAB into three equal parts. To find these points, we use the section formula twice: first for the point that divides ABAB in the ratio 1:21:2, and second for the point that divides it in the ratio 2:12:1.

📐Formulae

Section Formula (Internal): x=m1x2+m2x1m1+m2,y=m1y2+m2y1m1+m2x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}

Mid-point Formula: x=x1+x22,y=y1+y22x = \frac{x_1 + x_2}{2}, y = \frac{y_1 + y_2}{2}

Centroid Formula: G(x,y)=(x1+x2+x33,y1+y2+y33)G(x, y) = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})

Ratio k:1k:1 formula: x=kx2+x1k+1,y=ky2+y1k+1x = \frac{kx_2 + x_1}{k + 1}, y = \frac{ky_2 + y_1}{k + 1}

💡Examples

Problem 1:

Find the coordinates of the point PP which divides the line segment joining the points A(4,−3)A(4, -3) and B(8,5)B(8, 5) in the ratio 3:13:1 internally.

Solution:

  1. Identify the given values: (x1,y1)=(4,−3)(x_1, y_1) = (4, -3), (x2,y2)=(8,5)(x_2, y_2) = (8, 5), m1=3m_1 = 3, and m2=1m_2 = 1.
  2. Apply the x-coordinate formula: x=m1x2+m2x1m1+m2=3(8)+1(4)3+1=24+44=284=7x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7.
  3. Apply the y-coordinate formula: y=m1y2+m2y1m1+m2=3(5)+1(−3)3+1=15−34=124=3y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3.
  4. The coordinates of point PP are (7,3)(7, 3).

Explanation:

We use the Section Formula because the point divides the line in a specific ratio. By substituting the coordinates and the ratio values into the formula, we solve for the specific xx and yy values of the dividing point.

Problem 2:

If the mid-point of the segment joining A(a,b+1)A(a, b+1) and B(a+2,b−3)B(a+2, b-3) is M(5,−2)M(5, -2), find the values of aa and bb.

Solution:

  1. Identify the given values: (x1,y1)=(a,b+1)(x_1, y_1) = (a, b+1), (x2,y2)=(a+2,b−3)(x_2, y_2) = (a+2, b-3), and the mid-point M(x,y)=(5,−2)M(x, y) = (5, -2).
  2. Use the x-coordinate of the mid-point: x=x1+x22⇒5=a+(a+2)2⇒10=2a+2⇒2a=8⇒a=4x = \frac{x_1 + x_2}{2} \Rightarrow 5 = \frac{a + (a + 2)}{2} \Rightarrow 10 = 2a + 2 \Rightarrow 2a = 8 \Rightarrow a = 4.
  3. Use the y-coordinate of the mid-point: y=y1+y22⇒−2=(b+1)+(b−3)2⇒−4=2b−2⇒2b=−2⇒b=−1y = \frac{y_1 + y_2}{2} \Rightarrow -2 = \frac{(b + 1) + (b - 3)}{2} \Rightarrow -4 = 2b - 2 \Rightarrow 2b = -2 \Rightarrow b = -1.
  4. The values are a=4a = 4 and b=−1b = -1.

Explanation:

Since MM is the mid-point, we set up two separate equations (one for xx and one for yy) using the mid-point formula and solve for the unknown variables aa and bb.