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Coordinate Geometry - Use section formula for internal division to find coordinates of partition points

Grade 10CBSE

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

🔑Concepts

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The Section Formula allows us to find the coordinates of a point P(x,y)P(x, y) that divides a line segment joining two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) into a specific ratio m1:m2m_1 : m_2 internally.

A line segment AB divided by point P in the ratio m1:m2.
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When the ratio is not known, it is often easier to assume the ratio as k:1k:1. By substituting m1=km_1 = k and m2=1m_2 = 1 in the section formula, we solve for kk using either the xx or yy coordinate of the dividing point.

Line segment AB divided by P in ratio k:1
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A special case of the section formula occurs when m1=m2=1m_1 = m_2 = 1. This defines the midpoint MM, which is the average of the coordinates: M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).

Line segment showing the midpoint dividing it into two equal parts.
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Trisection of a line segment means dividing it into three equal parts. This requires finding two points, PP and QQ. Point PP divides the segment in ratio 1:21:2, and point QQ divides it in ratio 2:12:1.

📐Formulae

Section Formula (Internal): P(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)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)

Midpoint Formula: M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

Ratio k:1k:1 Formula: P(x,y)=(kx2+x1k+1,ky2+y1k+1)P(x, y) = \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1} \right)

Centroid of a Triangle: G=(x1+x2+x33,y1+y2+y33)G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

💡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 Section Formula for the x-coordinate: 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 Section Formula for the y-coordinate: 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 internal section formula by substituting the endpoints and the given ratio to find the specific coordinates of the point PP located on segment ABAB.

Problem 2:

In what ratio does the point P(−4,6)P(-4, 6) divide the line segment joining the points A(−6,10)A(-6, 10) and B(3,−8)B(3, -8)?

Solution:

  1. Let the ratio be k:1k:1.
  2. Use the x-coordinate formula: x=kx2+x1k+1x = \frac{kx_2 + x_1}{k + 1}.
  3. Substitute the values −4=k(3)+(−6)k+1-4 = \frac{k(3) + (-6)}{k + 1}.
  4. Cross-multiply: −4(k+1)=3k−6  ⟹  −4k−4=3k−6-4(k + 1) = 3k - 6 \implies -4k - 4 = 3k - 6.
  5. Rearrange terms: −4k−3k=−6+4  ⟹  −7k=−2-4k - 3k = -6 + 4 \implies -7k = -2.
  6. Solve for kk: k=27k = \frac{2}{7}.
  7. The ratio is 2:72:7.

Explanation:

To find an unknown ratio, we assume it is k:1k:1, set up an equation using one of the coordinates (either x or y), and solve for kk. Since kk is positive, the division is internal.

Problem 3:

Find the coordinates of the points of trisection of the line segment joining the points A(2,−2)A(2, -2) and B(−7,4)B(-7, 4).

Graph showing points A and B with trisection points P and Q.

Solution:

Let PP and QQ be the points of trisection. PP divides ABAB in ratio 1:21:2. Using Section Formula for P(x,y)P(x, y): x=1(−7)+2(2)1+2=−7+43=−1x = \frac{1(-7) + 2(2)}{1+2} = \frac{-7+4}{3} = -1 y=1(4)+2(−2)1+2=4−43=0y = \frac{1(4) + 2(-2)}{1+2} = \frac{4-4}{3} = 0 So, PP is (−1,0)(-1, 0). QQ is the midpoint of PBPB or divides ABAB in 2:12:1. Using 2:12:1: x=2(−7)+1(2)2+1=−14+23=−4x = \frac{2(-7) + 1(2)}{2+1} = \frac{-14+2}{3} = -4 y=2(4)+1(−2)2+1=8−23=2y = \frac{2(4) + 1(-2)}{2+1} = \frac{8-2}{3} = 2 So, QQ is (−4,2)(-4, 2).

Explanation:

Trisection divides the segment into three equal lengths. We calculate the coordinates of the first point using ratio 1:21:2 and the second using 2:12:1.

Problem 4:

Find the ratio in which the y-axis divides the line segment joining the points (5,−6)(5, -6) and (−1,−4)(-1, -4). Also, find the point of intersection.

Line segment crossing the y-axis at point P.

Solution:

Let the y-axis divide the segment ABAB at point P(0,y)P(0, y) in the ratio k:1k:1. Since PP lies on the y-axis, its x-coordinate is 00. Using Section Formula for x: 0=k(−1)+1(5)k+10 = \frac{k(-1) + 1(5)}{k+1} 0=−k+5  ⟹  k=50 = -k + 5 \implies k = 5 So the ratio is 5:15:1. Now, find yy using k=5k=5: y=5(−4)+1(−6)5+1=−20−66=−266=−133y = \frac{5(-4) + 1(-6)}{5+1} = \frac{-20 - 6}{6} = \frac{-26}{6} = -\frac{13}{3} The point of intersection is (0,−133)(0, -\frac{13}{3}).

Explanation:

On the y-axis, the x-coordinate is always zero. We use this property to find the unknown ratio kk first, then use kk to find the y-coordinate.