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Finding Ratios And Mid-Points

Use the k:1 technique to find unknown ratios, and apply the mid-point formula for 1:1 division.

11 micro-lessons

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Finding Unknown Ratios

Concept introduction for solving for ratios instead of points.

The $k:1$ Assumption

When you know the coordinates of the point that divides a line segment, but you need to find the ratio of division, it is mathematically simpler to assume the ratio is $k : 1$ instead of $m_1 : m_2$ .

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Finding Ratios and Mid-Points Visual

Trisection (1:2 and 2:1) benefits from visual segmenting.

coordinate-graph: A horizontal line segment AB with point A at (2, -2) and point B at (-7, 4). Two intermediate points P…

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Why Use k:1?

Explanation of why k:1 simplifies the section formula for unknown ratios.

Key Takeaway

Replacing $m_{1} : m_{2}$ with $k : 1$ reduces the problem from two variables to just one ( $k$ ). If $k$ evaluates to a positive number, the division is internal.

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Example 7: Solving for the Ratio

Worked example finding the ratio given the division point.

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

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Example 9: Division by an Axis

Worked example finding the ratio when the y-axis cuts a line segment.

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

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Faded Example: Ratio by a Point

Guided practice finding the ratio k:1.

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To find the ratio in which the point $(-1, 6)$ divides the line segment joining $(-3, 10)$ and $(6, -8)$ , we assume the ratio is $k : 1$ . Using the section formula for the x-coordinate, we set up the equation: $-1 = \frac{6k - 3}{$ $}$ . Multiplying both sides by the denominator gives $-k - 1 = 6k - 3$ . Rearranging the terms, we get $7k =$ , which means $k = \frac{2}{7}$ . Therefore, the required ratio is .

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Division by x-axis

MCQ testing condition for x-axis division.

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

← previousThe Section Formula next →Review: Coordinate Geometry

Finding Unknown Ratios

Concept introduction for solving for ratios instead of points.

The $k:1$ Assumption

1 / 3

Example 7: Solving for the Ratio

Worked example finding the ratio given the division point.

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

1 / 5

Example 9: Division by an Axis

Worked example finding the ratio when the y-axis cuts a line segment.

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

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Faded Example: Ratio by a Point

Guided practice finding the ratio k:1.

0 of 3 blanks filled

Division by x-axis

MCQ testing condition for x-axis division.

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

Finding Ratios And Mid-Points

Finding Unknown Ratios

The k:1k:1k:1 Assumption

Finding Ratios and Mid-Points Visual

Why Use k:1?

Key Takeaway

Example 7: Solving for the Ratio

Example 9: Division by an Axis

Faded Example: Ratio by a Point

Division by x-axis

Finding Unknown Ratios

The k:1k:1k:1 Assumption

Finding Ratios and Mid-Points Visual

Why Use k:1?

Key Takeaway

Example 7: Solving for the Ratio

Example 9: Division by an Axis

Faded Example: Ratio by a Point

Division by x-axis

The $k:1$ Assumption

The $k:1$ Assumption