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Coordinate Geometry - Section Formula

Grade 10CBSE

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

🔑Concepts

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The Section Formula allows finding the coordinates of a point P(x,y)P(x, y) that divides a line segment joining A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) in a given ratio m1:m2m_1:m_2. The coordinates are given by x=m1x2+m2x1m1+m2x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} and y=m1y2+m2y1m1+m2y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}.

A line segment AB divided by point P in ratio m1:m2
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When the ratio m1:m2m_1:m_2 is 1:11:1, the point PP becomes the midpoint of the segment ABAB. The formula simplifies to x=x1+x22x = \frac{x_1 + x_2}{2} and y=y1+y22y = \frac{y_1 + y_2}{2}.

Midpoint M dividing line segment AB into two equal parts
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Points of Trisection: To divide a line segment into three equal parts, we need two points PP and QQ. PP divides ABAB in ratio 1:21:2, and QQ divides ABAB in ratio 2:12:1.

Line segment divided into three equal segments by points P and Q
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The Centroid of a triangle is the point of intersection of its medians. It divides each median in the ratio 2:12:1 from the vertex. If vertices are (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3), the centroid GG is (x1+x2+x33,y1+y2+y33)(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}).

📐Formulae

x=m1x2+m2x1m1+m2,y=m1y2+m2y1m1+m2x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}

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

Centroid G=(x1+x2+x33,y1+y2+y33)\text{Centroid } 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 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:

Given: (x1,y1)=(4,−3)(x_1, y_1) = (4, -3), (x2,y2)=(8,5)(x_2, y_2) = (8, 5), m1=3m_1 = 3, m2=1m_2 = 1. Using the section formula: x=3(8)+1(4)3+1=24+44=284=7x = \frac{3(8) + 1(4)}{3+1} = \frac{24 + 4}{4} = \frac{28}{4} = 7 y=3(5)+1(−3)3+1=15−34=124=3y = \frac{3(5) + 1(-3)}{3+1} = \frac{15 - 3}{4} = \frac{12}{4} = 3 The coordinates are (7,3)(7, 3).

Explanation:

We substitute the given coordinates and the ratio into the internal section formula to find the xx and yy coordinates of the required point.

Problem 2:

In what ratio does the yy-axis divide the line segment joining the points A(5,−6)A(5, -6) and B(−1,−4)B(-1, -4)?

Solution:

Let the yy-axis divide ABAB in the ratio k:1k:1 at point P(0,y)P(0, y). Using the xx-coordinate formula: 0=k(−1)+1(5)k+10 = \frac{k(-1) + 1(5)}{k+1} 0=−k+50 = -k + 5 k=5k = 5 The ratio is 5:15:1.

Explanation:

Any point on the yy-axis has an xx-coordinate of 00. By setting the xx-coordinate section formula equal to 00, we can solve for the ratio kk.

Problem 3:

Find the coordinates of the centroid of a triangle whose vertices are A(0,6)A(0, 6), B(8,12)B(8, 12), and C(1,0)C(1, 0).

Solution:

Using the centroid formula: x=0+8+13=93=3x = \frac{0 + 8 + 1}{3} = \frac{9}{3} = 3 y=6+12+03=183=6y = \frac{6 + 12 + 0}{3} = \frac{18}{3} = 6 The centroid is G(3,6)G(3, 6).

Explanation:

The centroid coordinates are the average of the xx-coordinates and yy-coordinates of the three vertices.

Problem 4:

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).

Plot showing points A, P, Q, and B on a coordinate plane

Solution:

Let PP and QQ be the points of trisection. PP divides ABAB in ratio 1:21:2. Using section formula for PP: 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 ratio 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 means dividing the segment into three equal parts. We calculate the first point using the ratio 1:21:2 and the second point using the ratio 2:12:1.

Problem 5:

Find the ratio in which the point P(x,2)P(x, 2) divides the line segment joining the points A(12,5)A(12, 5) and B(4,−3)B(4, -3). Also, find the value of xx.

Line segment AB with point P lying on it

Solution:

Let the ratio be k:1k:1. Using the yy-coordinate of the section formula: 2=k(−3)+1(5)k+12 = \frac{k(-3) + 1(5)}{k + 1} 2(k+1)=−3k+52(k + 1) = -3k + 5 2k+2=−3k+52k + 2 = -3k + 5 5k=3  ⟹  k=3/55k = 3 \implies k = 3/5 The ratio is 3:53:5. Now, find xx using the xx-coordinate formula with ratio 3:53:5: x=3(4)+5(12)3+5x = \frac{3(4) + 5(12)}{3 + 5} x=12+608=728=9x = \frac{12 + 60}{8} = \frac{72}{8} = 9

Explanation:

When the ratio is unknown, assume it is k:1k:1. Use the known coordinate (here, y=2y=2) to solve for kk, then use kk to find the unknown coordinate xx.