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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 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) internally in a given ratio m:nm:n. This is derived using the properties of similar triangles.

A line segment AB divided by point P in the ratio m:n
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The Mid-point Formula is a special case of the Section Formula where the ratio is 1:11:1. It provides the average of the xx-coordinates and the yy-coordinates of the endpoints.

Line segment with M exactly in the center
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The Centroid of a triangle is the point where its three medians intersect. It divides each median in the ratio 2:12:1 from the vertex to the midpoint of the opposite side.

Triangle showing the intersection of medians at the centroid G
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Trisection of a segment involves finding two points, PP and QQ, that divide the segment into three equal parts. PP divides ABAB in 1:21:2, and QQ divides ABAB in 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.

Problem 3:

Find the ratio in which the yy-axis divides the line segment joining the points A(−4,2)A(-4, 2) and B(3,6)B(3, 6). Also, find the coordinates of the point of intersection.

Line segment AB crossing the y-axis at point P

Solution:

Let the yy-axis divide the segment ABAB at point P(0,y)P(0, y) in the ratio k:1k:1. Using the section formula for the x-coordinate: x=kx2+x1k+1x = \frac{kx_2 + x_1}{k + 1} 0=k(3)+(−4)k+10 = \frac{k(3) + (-4)}{k + 1} 0=3k−40 = 3k - 4 3k=4  ⟹  k=433k = 4 \implies k = \frac{4}{3} So, the ratio is 4:34:3. Now find the y-coordinate using m1=4,m2=3m_1=4, m_2=3: y=4(6)+3(2)4+3y = \frac{4(6) + 3(2)}{4 + 3} y=24+67=307y = \frac{24 + 6}{7} = \frac{30}{7} The point of intersection is (0,307)(0, \frac{30}{7}).

Explanation:

Points on the y-axis always have an x-coordinate of 0. By setting the x-coordinate formula to zero, we can solve for the unknown ratio kk.

Problem 4:

Find the coordinates of the centroid of △ABC\triangle ABC whose vertices are A(−1,0)A(-1, 0), B(5,−2)B(5, -2), and C(8,2)C(8, 2).

Triangle with vertices A, B, C and its centroid G at (4,0)

Solution:

The coordinates of the centroid G(x,y)G(x, y) are given by: x=x1+x2+x33,y=y1+y2+y33x = \frac{x_1 + x_2 + x_3}{3}, y = \frac{y_1 + y_2 + y_3}{3} Substitute the values: x=−1+5+83=123=4x = \frac{-1 + 5 + 8}{3} = \frac{12}{3} = 4 y=0+(−2)+23=03=0y = \frac{0 + (-2) + 2}{3} = \frac{0}{3} = 0 The coordinates of the centroid are G(4,0)G(4, 0).

Explanation:

The centroid is calculated by taking the arithmetic mean of the x-coordinates and the y-coordinates of the three vertices.