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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

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Reflection across axes and the origin transforms a point P(x,y)P(x, y) by changing the signs of its coordinates. Reflection in the xx-axis produces P′(x,−y)P'(x, -y), reflection in the yy-axis produces P′(−x,y)P'(-x, y), and reflection in the origin produces P′(−x,−y)P'(-x, -y). Points lying on the line of reflection are called invariant points as their coordinates remain unchanged.

Coordinate plane showing reflection of point (3,2) across x-axis, y-axis, and origin.
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The Section Formula determines 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. If PP lies between AA and BB, it is internal division.

Line segment AB divided by point P in ratio m1:m2.
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The slope of a line, mm, represents its inclination. It is calculated as the tangent of the angle θ\theta that the line makes with the positive xx-axis (m=tan⁡θm = \tan \theta) or using two points on the line: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. Parallel lines have equal slopes (m1=m2m_1 = m_2), while perpendicular lines have slopes whose product is −1-1 (m1×m2=−1m_1 \times m_2 = -1).

A line intersecting the x-axis at an angle theta.
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The equation of a line can be expressed in different forms depending on the available data: Slope-Intercept form (y=mx+cy = mx + c) where cc is the yy-intercept, or Point-Slope form (y−y1=m(x−x1)y - y_1 = m(x - x_1)) when a specific point on the line and its slope are known.

📐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=(x2−x1)2+(y2−y1)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=y2−y1x2−x1=tan⁡θm = \frac{y_2 - y_1}{x_2 - x_1} = \tan \theta

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

Point-Slope Form: y−y1=m(x−x1)y - y_1 = m(x - x_1)

💡Examples

Problem 1:

Point A(−4,2)A(-4, 2) is reflected in the xx-axis to A′A'. Point BB is the reflection of A′A' in the origin. Find the coordinates of A′A' 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).
  2. Reflection of A′(−4,−2)A'(-4, -2) in the origin: both xx and yy change signs →B(4,2)\rightarrow B(4, 2).
  3. 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): m=2−24−(−4)=08=0m = \frac{2 - 2}{4 - (-4)} = \frac{0}{8} = 0.
  4. Since the slope is 00, the line is horizontal. Using y−y1=m(x−x1)y - y_1 = m(x - x_1): y−2=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: x=m1x2+m2x1m1+m2→−1=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}.
  2. Multiply by (k+1)(k+1): −k−1=6k−3-k - 1 = 6k - 3.
  3. Solve for kk: 2=7k→k=272 = 7k \rightarrow k = \frac{2}{7}. So the ratio is 2:72:7.
  4. Now find yy using the ratio 2:72:7: y=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.

Problem 3:

Find the equation of the perpendicular bisector of the line segment joining A(2,3)A(2, 3) and B(6,−1)B(6, -1).

Line segment AB and its perpendicular bisector passing through midpoint M.

Solution:

  1. Find the midpoint MM of ABAB: M=(2+62,3−12)=(4,1)M = \left( \frac{2+6}{2}, \frac{3-1}{2} \right) = (4, 1).
  2. Find the slope of ABAB (mABm_{AB}): mAB=−1−36−2=−44=−1m_{AB} = \frac{-1 - 3}{6 - 2} = \frac{-4}{4} = -1.
  3. Slope of the perpendicular bisector (m⊥m_{\perp}): Since m1×m2=−1m_1 \times m_2 = -1, m⊥=−1−1=1m_{\perp} = \frac{-1}{-1} = 1.
  4. Use point-slope form with M(4,1)M(4, 1) and m=1m = 1: y−1=1(x−4)⇒y−1=x−4⇒x−y−3=0y - 1 = 1(x - 4) \Rightarrow y - 1 = x - 4 \Rightarrow x - y - 3 = 0.

Explanation:

A perpendicular bisector passes through the midpoint of a segment at a 90∘90^{\circ} angle. We calculate the midpoint to find a point on the line and use the negative reciprocal of the segment's slope to find the bisector's slope.

Problem 4:

The point P(2,−3)P(2, -3) is reflected in the yy-axis to P′P'. P′P' is then reflected in the line y=1y = 1 to P′′P''. Find the coordinates of P′P' and P′′P''.

Points P, P' (y-axis reflection), and P'' (reflection across y=1).

Solution:

  1. Reflection of P(2,−3)P(2, -3) in the yy-axis: P′(x,y)→P′(−x,y)P'(x, y) \rightarrow P'(-x, y). Thus, P′=(−2,−3)P' = (-2, -3).
  2. Reflection of P′(−2,−3)P'(-2, -3) in the line y=1y = 1: The xx-coordinate remains −2-2. The distance of y=−3y = -3 from y=1y = 1 is 1−(−3)=41 - (-3) = 4 units. The image P′′P'' will be 44 units above the line y=1y = 1. So, yP′′=1+4=5y_{P''} = 1 + 4 = 5. Thus, P′′=(−2,5)P'' = (-2, 5).

Explanation:

Reflecting in the y-axis flips the sign of the x-coordinate. Reflecting across a horizontal line y=ky = k keeps the x-coordinate the same and moves the y-coordinate such that the line y=ky = k is the midpoint between the original and reflected y-values.