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Algebra - Coordinate Geometry (Reflection, Section Formula, Equation of a Line)

Grade 10ICSE

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

🔑Concepts

Reflection in Axes and Origin: Reflection is a transformation where a point is 'flipped' over a line. In the xx-axis, the point (x,y)(x, y) becomes (x,y)(x, -y), appearing as a vertical mirror image across the horizontal axis. In the yy-axis, (x,y)(x, y) becomes (x,y)(-x, y), representing a horizontal flip across the vertical axis. Reflection in the origin (0,0)(0, 0) transforms (x,y)(x, y) into (x,y)(-x, -y), which visually looks like a 180180^{\circ} rotation around the center.

Invariant Points: A point is considered invariant with respect to a line (or point) of reflection if its coordinates remain unchanged after the transformation. Visually, these are points that lie exactly on the mirror line. For example, any point on the xx-axis like (5,0)(5, 0) is invariant when reflected in the xx-axis.

Section Formula (Internal Division): This formula determines the coordinates of a point PP that divides a line segment ABAB into two parts with a specific ratio m1:m2m_1:m_2. Visually, if PP is closer to AA, m1m_1 is smaller than m2m_2. The formula uses a weighted average of the coordinates of the endpoints A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2).

Midpoint and Centroid: The midpoint is a special case of the section formula where the ratio is 1:11:1, representing the exact center of a line segment. The centroid of a triangle is the point where its three medians intersect; visually, it is the 'balance point' of the triangle, calculated by averaging the xx and yy coordinates of the three vertices.

Slope (Gradient) of a Line: The slope mm represents the steepness and direction of a line. It is defined as the tangent of the angle of inclination θ\theta (m=tanθm = \tan \theta). On a graph, a positive slope indicates the line rises from left to right, while a negative slope indicates it falls. A horizontal line has a slope of 00, and a vertical line has an undefined slope.

Forms of the Equation of a Line: A straight line can be expressed in different forms depending on the given data. The Slope-Intercept form y=mx+cy = mx + c is most common, where cc is the yy-intercept (the point where the line crosses the vertical axis). The Point-Slope form is used when one point and the slope are known, describing how the line radiates from that specific point.

Parallel and Perpendicular Lines: Geometrically, parallel lines have the same steepness and never meet, meaning their slopes are equal (m1=m2m_1 = m_2). Perpendicular lines intersect at a 9090^{\circ} angle; algebraically, the product of their slopes is 1-1 (m1m2=1m_1 \cdot m_2 = -1), which means one slope is the negative reciprocal of the other.

📐Formulae

Reflection in xx-axis: P(x,y)P(x,y)P(x, y) \rightarrow P'(x, -y)

Reflection in yy-axis: P(x,y)P(x,y)P(x, y) \rightarrow P'(-x, y)

Reflection in Origin: P(x,y)P(x,y)P(x, y) \rightarrow P'(-x, -y)

Distance Formula: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Section Formula: 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)

Centroid Formula: 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)

Slope (mm): m=y2y1x2x1=tanθm = \frac{y_2 - y_1}{x_2 - x_1} = \tan \theta

Slope-Intercept Form: y=mx+cy = mx + c

Point-Slope Form: yy1=m(xx1)y - y_1 = m(x - x_1)

💡Examples

Problem 1:

Point A(4,2)A(-4, 2) is reflected in the xx-axis to AA'. Point BB is the reflection of AA' in the origin. Find the coordinates of AA' and BB. Also, find the equation of the line ABAB.

Solution:

  1. Reflection of A(4,2)A(-4, 2) in the xx-axis: yy changes sign A(4,2)\rightarrow A'(-4, -2). \n2. Reflection of A(4,2)A'(-4, -2) in the origin: both xx and yy change signs B(4,2)\rightarrow B(4, 2). \n3. To find the equation of line ABAB, first find the slope mm using A(4,2)A(-4, 2) and B(4,2)B(4, 2): \nm=224(4)=08=0m = \frac{2 - 2}{4 - (-4)} = \frac{0}{8} = 0. \n4. Since the slope is 00, the line is horizontal. Using yy1=m(xx1)y - y_1 = m(x - x_1): \ny2=0(x+4)y=2y - 2 = 0(x + 4) \rightarrow y = 2.

Explanation:

We first applied reflection rules for the xx-axis and the origin. Since the yy-coordinates of AA and BB are the same, the line is horizontal, resulting in a slope of 00 and an equation of the form y=ky = k.

Problem 2:

Find the ratio in which the line segment joining P(3,10)P(-3, 10) and Q(6,8)Q(6, -8) is divided by the point R(1,y)R(-1, y). Also, find the value of yy.

Solution:

  1. Let the ratio be k:1k:1. Using the xx-coordinate of the section formula: \nx=m1x2+m2x1m1+m21=k(6)+1(3)k+1x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \rightarrow -1 = \frac{k(6) + 1(-3)}{k + 1}. \n2. Multiply by (k+1)(k+1): k1=6k3-k - 1 = 6k - 3. \n3. Solve for kk: 2=7kk=272 = 7k \rightarrow k = \frac{2}{7}. So the ratio is 2:72:7. \n4. Now find yy using the ratio 2:72:7: \ny=2(8)+7(10)2+7=16+709=549=6y = \frac{2(-8) + 7(10)}{2 + 7} = \frac{-16 + 70}{9} = \frac{54}{9} = 6.

Explanation:

By assuming the ratio k:1k:1, we use the known xx-coordinate of the division point to solve for kk. Once the ratio is established, the yy-coordinate is found by substituting the ratio back into the section formula.