Sigma Percentile
JEE Main 2021 (24 February Shift 2)
LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: Let . If the mirror image of the point with respect to the line is , then is equal to:

Select Answer:

Visualized Solution

Visualize the Mirror Image

  • Let be the given point.
  • Let be its mirror image.
  • The mirror is the line .

Properties of Mirror Image

  • The line segment is perpendicular to the line .
  • The midpoint of lies exactly on the line .

Coordinates of Midpoint

  • Midpoint formula:
  • Substitute and :

Simplify Midpoint

  • -coordinate:
  • -coordinate:
  • -coordinate:

Midpoint Lies on Line

  • Since is on the line , its coordinates must satisfy the line's equation.
  • Line :

Substitute into Line Equation

  • Substitute , , into :

Simplify the First Ratio

  • First ratio:
  • Take LCM in numerator:
  • Simplify:

Simplify the Second Ratio

  • Second ratio:
  • Take LCM in numerator:
  • Simplify:

Simplify the Third Ratio

  • Third ratio:
  • Take LCM in numerator:
  • Simplify:

Equate Ratios to Find

  • The simplified equation is:
  • To find , equate the first and third ratios:

Solve for

  • Cross-multiply:
  • Divide by 2:
  • Expand:
  • Rearrange:

Substitute to Find

  • Now use the first and second ratios:
  • Substitute :

Solve for

  • Simplify numerator:
  • Divide:
  • Multiply by 10:

Calculate Final Answer

  • We need to find the value of .
  • Substitute and :

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

In the realm of 3D geometry, reflecting a point across a line defined by
results in an image point . This process relies on the principle of symmetry, where the line acts as the perpendicular bisector of the segment .

The Midpoint

Our Anchor in the Void
The midpoint of the segment must lie on the line . We calculate the coordinates of by averaging the coordinates of and :
Simplifying the components, we obtain:
Since lies on the line , it must satisfy the given symmetric equations of the line.

The Algebraic Bridge

Substituting the coordinates of into the line equation, we get:
Simplifying the numerators, the expression becomes:

Solving the Mystery

To determine the value of , we equate the first and third ratios:
Cross-multiplying and simplifying, we have , which reduces to . Expanding this yields , leading to , or .
Next, we solve for by equating the first and second ratios:
This results in , which gives .

The Final Calculation

With the values of and determined, we calculate the final required value:
Through the application of geometric symmetry and algebraic substitution, we have successfully resolved the coordinates and reached the final result of 88.

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