NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 Question 10
Welcome! In this post, we will solve Class 10 Maths Chapter 7 Exercise 7.1 Question 10 step by step using the ncert solutions approach — in simple, teacher-to-student language. No prior knowledge of coordinate geometry is assumed. Let’s dive in! 🎯
🎓 Our Expert’s Approach
To solve this problem, our experts first recognize that “equidistant” means the distances from point &&(x, y)&& to both given points are equal. The strategy is to apply the Distance Formula to both distances, set them equal, square both sides to remove the square roots, and then simplify algebraically. This method is chosen for its clarity and reliability, reflecting years of teaching experience in CBSE Coordinate Geometry. The result will be a clean linear equation in &&x&& and &&y&&.
Understanding the Question 🧐
Q10. Find a relation between &&x&& and &&y&& such that the point &&(x, y)&& is equidistant from the points &&(3, 6)&& and &&(-3, 4)&&.
💡 What does “equidistant” mean? It means the distance from &&(x, y)&& to &&(3, 6)&& is exactly the same as the distance from &&(x, y)&& to &&(-3, 4)&&.
⚠️ Caution: Do NOT confuse equidistant with midpoint. Equidistant means equal distances — the point &&(x, y)&& can be anywhere on a line, not just the midpoint.
📝 Solution: Step-by-Step
Let us name our three points:
- &&A = (x, y)&& → the unknown point
- &&B = (3, 6)&& → first given point
- &&C = (-3, 4)&& → second given point
Since &&A&& is equidistant from &&B&& and &&C&&, we have: &&AB = AC&&
The distance formula between two points &&(x_1, y_1)&& and &&(x_2, y_2)&& is:
&&d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}&&
This formula comes from the Pythagorean theorem applied on a coordinate plane. Always remember it! 📐
Here, &&x_1 = x,\ y_1 = y,\ x_2 = 3,\ y_2 = 6&&
&&AB = \sqrt{(3 – x)^2 + (6 – y)^2}&&
Here, &&x_1 = x,\ y_1 = y,\ x_2 = -3,\ y_2 = 4&&
&&AC = \sqrt{(-3 – x)^2 + (4 – y)^2}&&
Since the point is equidistant:
&&\sqrt{(3 – x)^2 + (6 – y)^2} = \sqrt{(-3 – x)^2 + (4 – y)^2}&&
Squaring both sides removes the square roots safely (both sides are non-negative distances).
&&(3 – x)^2 + (6 – y)^2 = (-3 – x)^2 + (4 – y)^2&&
Left Side (LHS):
- &&(3 – x)^2 = 9 – 6x + x^2&&
- &&(6 – y)^2 = 36 – 12y + y^2&&
- LHS &&= 9 – 6x + x^2 + 36 – 12y + y^2&&
- LHS &&= x^2 + y^2 – 6x – 12y + 45&&
Right Side (RHS):
- &&(-3 – x)^2 = (3 + x)^2 = 9 + 6x + x^2&&
- &&(4 – y)^2 = 16 – 8y + y^2&&
- RHS &&= 9 + 6x + x^2 + 16 – 8y + y^2&&
- RHS &&= x^2 + y^2 + 6x – 8y + 25&&
⚠️ Key Caution: &&(-3-x)^2 = (3+x)^2&&, NOT &&(-3)^2 – 2(3)(x) + x^2&&. Always expand carefully!
&&x^2 + y^2 – 6x – 12y + 45 = x^2 + y^2 + 6x – 8y + 25&&
Cancel &&x^2&& and &&y^2&& from both sides (they appear on both sides):
&&-6x – 12y + 45 = 6x – 8y + 25&&
Bring all terms to the left side:
&&-6x – 6x – 12y + 8y + 45 – 25 = 0&&
&&-12x – 4y + 20 = 0&&
Divide the entire equation by &&-4&&:
&&3x + y – 5 = 0&&
🔍 What does this mean geometrically? The equation &&3x + y = 5&& represents a straight line. Every point on this line is equidistant from &&(3, 6)&& and &&(-3, 4)&&. This line is actually the perpendicular bisector of the segment joining &&(3, 6)&& and &&(-3, 4)&&!
Conclusion and Key Points ✅
In this ncert solutions problem, we used the Distance Formula to find the relation between &&x&& and &&y&& for a point equidistant from two given points. The key steps were:
- Set up &&AB = AC&& using the Distance Formula.
- Square both sides to eliminate square roots.
- Expand using algebraic identities &&(a-b)^2 = a^2 – 2ab + b^2&&.
- Cancel common terms (&&x^2&&, &&y^2&&) and simplify.
- Divide by the common factor to get the simplest form.
The final relation is: &&3x + y = 5&&
💡 Smart Trick / Shortcut
The locus of all points equidistant from two points is always the perpendicular bisector of the line segment joining those two points. You can verify: Midpoint of &&(3,6)&& and &&(-3,4)&& is &&\left(\frac{3+(-3)}{2}, \frac{6+4}{2}\right) = (0, 5)&&. Check: &&3(0) + 5 = 5&& ✅ — the midpoint lies on our answer line!
📌 Points to Remember
- Distance Formula: &&d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}&&
- Equidistant → Set the two distances equal.
- Always square both sides to remove square roots.
- &&x^2&& and &&y^2&& cancel out, giving a linear equation.
- &&(-a – x)^2 = (a + x)^2&& — expand carefully!
- The result is the equation of the perpendicular bisector of the two points.
❓ Frequently Asked Questions (FAQ)
Q1. What is the distance formula used in this question?
The distance formula between two points &&(x_1, y_1)&& and &&(x_2, y_2)&& is: &&d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}&&. It is derived from the Pythagorean theorem and is the foundation of Chapter 7 Coordinate Geometry.
Q2. What does “equidistant” mean in coordinate geometry?
A point &&P(x, y)&& is said to be equidistant from two points &&A&& and &&B&& if &&PA = PB&&, i.e., the distance from &&P&& to &&A&& equals the distance from &&P&& to &&B&&. In this question, &&(x, y)&& is equidistant from &&(3, 6)&& and &&(-3, 4)&&.
Q3. Why do we square both sides in this problem?
We square both sides to eliminate the square root sign from the distance formula. Since both sides represent distances (non-negative values), squaring is valid and does not introduce extra solutions. This simplifies the equation from &&\sqrt{\ldots} = \sqrt{\ldots}&& to a polynomial equation.
Q4. Why do &&x^2&& and &&y^2&& cancel out in the solution?
After expanding both sides, the terms &&x^2&& and &&y^2&& appear on both the LHS and RHS with the same coefficient (&&+1&&). When we subtract RHS from LHS, these terms cancel, leaving a linear equation in &&x&& and &&y&&. This is why the answer is always a straight line (linear relation).
Q5. What is the geometric meaning of the relation &&3x + y = 5&&?
The equation &&3x + y = 5&& represents a straight line in the coordinate plane. Geometrically, this line is the perpendicular bisector of the segment joining &&(3, 6)&& and &&(-3, 4)&&. Every point on this line is exactly equidistant from both given points.
Q6. How can I verify the answer &&3x + y = 5&&?
Take any point on the line &&3x + y = 5&&, for example the midpoint &&(0, 5)&&. Distance from &&(0,5)&& to &&(3,6)&&: &&\sqrt{(3-0)^2 + (6-5)^2} = \sqrt{9+1} = \sqrt{10}&&. Distance from &&(0,5)&& to &&(-3,4)&&: &&\sqrt{(-3-0)^2 + (4-5)^2} = \sqrt{9+1} = \sqrt{10}&&. Both distances are equal ✅ — the answer is verified!
Q7. Is this question important for CBSE Board Exams?
Yes! Questions based on the equidistant condition using the Distance Formula are frequently asked in CBSE Class 10 board exams. This type of question tests your understanding of the Distance Formula and algebraic simplification. It is a must-practice question from Exercise 7.1.
Q8. What is the difference between equidistant point and midpoint?
The midpoint is one specific point that is equidistant from both endpoints of a segment. An equidistant point &&(x, y)&& can be any point satisfying the condition &&PA = PB&&. The set of all such points forms the perpendicular bisector line, of which the midpoint is just one particular point.
📚 Further Reading
For the official NCERT textbook and additional resources, visit the NCERT Official Website.
