Q.35 The distance between the two points of intersection of x² + y = 7 and x + y = 7 (rounded off to two decimal places) is ____________.
Distance Between Points of Intersection of x² + y = 7 and x + y = 7
The two curves given are:
- Parabola: x² + y = 7
- Line: x + y = 7
To find their points of intersection, substitute from the line into the parabola.
Step 1: Substitute y = 7 − x
From the line: y = 7 - x.
Substitute into x² + y = 7:
x² + (7 - x) = 7
x² - x = 0
x(x - 1) = 0
So, x = 0 or x = 1.
Step 2: Find Corresponding y-values
- If x = 0 → y = 7 → point A(0, 7)
- If x = 1 → y = 6 → point B(1, 6)
Step 3: Apply Distance Formula
Distance between A(0, 7) and B(1, 6):
d = √[(1 − 0)² + (6 − 7)²]
= √(1 + 1)
= √2 ≈ 1.4142
Rounded to two decimal places: 1.41
Final Answer
Distance = 1.41 (rounded to two decimal places)
Introduction
The distance between intersection points of x² + y = 7 and x + y = 7 is a classic coordinate geometry problem. It tests substitution into curves and application of the distance formula, a common topic in IIT JAM mathematics.
Step-by-Step Explanation
Rewrite the line as y = 7 − x, substitute into the parabola, solve x² − x = 0, giving x = 0 and x = 1. Corresponding y-values are 7 and 6. Thus, intersection points are A(0, 7) and B(1, 6).
Distance Formula Application
d = √[(1 − 0)² + (6 − 7)²] = √2 ≈ 1.41Common Mistakes
- Solving the wrong substituted equation (e.g., x² + x = 7)
- Trying to use parallel line distance formula incorrectly
- Forgetting rounding to two decimals
The correct distance is: 1.41
