Maharashtra Board Class 11 Maths Part 1 Chapter 5 Locus and Straight Line 5.1 Solutions

Question 1. If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Answer:
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3) and B(2, 1).
PA = PB
.. PA² = PB²
.. (x - 1)² + (y - 3)² = (x - 2)² + (y - 1)²
.. x² - 2x + 1 + y² - 6y + 9 = x² - 4x + 4 + y² - 2y + 1
.. -2x - 6y + 10 = -4x - 2y + 5
.. 2x - 4y + 5 = 0
.. The required equation of locus is 2x - 4y + 5 = 0.
In simple words: The problem asks us to find the set of all points P that are equidistant from two given points A and B. By equating the squares of the distances PA and PB, we use the distance formula to set up an algebraic equation, which simplifies to a linear equation representing the locus.

🎯 Exam Tip: Remember to square both sides (PA² = PB²) to eliminate square roots early on, simplifying calculations significantly. Always check for common terms that can be cancelled from both sides of the equation.

Question 2. A(-5, 2) and B(4, 1). Find the equation of the locus of point P, which is equidistant from A and B.
Answer:
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(-5, 2) and B(4, 1).
.. PA = PB
... PA² = PB²
.. (x + 5)² + (y - 2)² = (x - 4)² + (y - 1)²
.. x² + 10x + 25 + y² - 4y + 4 = x² - 8x + 16 + y² - 2y + 1
.. 10x - 4y + 29 = -8x - 2y + 17
.. 18x - 2y + 12 = 0
.. 9x - y + 6 = 0
.. The required equation of locus is 9x - y - 6 = 0
In simple words: This question is similar to the first, requiring us to find the locus of points equidistant from two given points A and B. The method involves setting the squared distances PA² and PB² equal, expanding the binomials, and simplifying to obtain a linear equation.

🎯 Exam Tip: Be careful with signs when expanding squared terms, especially `(x + 5)²` or `(x - 4)²`. A common mistake is mismanaging the negative signs, which can lead to incorrect final equations.

Question 3. If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Answer:
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
.:. AP² = 4BP²
.. (x - 2)² + (y - 0)² = 4[(x - 0)² + (y - 3)²]
.. x² - 4x + 4 + y² = 4(x² + y² y² - 6y + 9)
.. x² - 4x + 4 + y² = 4x² + 4y² - 24y + 36
.. 3x² + 3y² + 4x - 24y + 32 = 0
.. The required equation of locus is 3x² + 3y² + 4x - 24y + 32 = 0
In simple words: Here, the locus of point P is defined by a ratio of distances, AP = 2BP. Squaring both sides gives AP² = 4BP², allowing us to use the distance formula. Expanding and collecting terms results in a quadratic equation, which represents a circle if the coefficients of x² and y² are equal.

🎯 Exam Tip: When dealing with ratios like AP = nBP, always square both sides to get AP² = n²BP². This eliminates square roots and sets up for an easier algebraic expansion. Pay close attention to the distributive property when multiplying by `n²`.

Question 4. If A(4, 1) and B(5, 4), find the equation of the locus of point P if PA² = 3PB².
Answer:
Solution:
Let P(x, y) be any point on the required locus.
Given, A(4, 1), B(5, 4) and PA² = 3PB²
.. (x - 4)² + (y - 1)² = 3[(x - 5)² + (y - 4)²]
.. x² - 8x + 16 + y² - 2y + 1 = 3(x² - 10x + 25 + y² - 8y + 16)
.. x² - 8x + y² - 2y + 17 = 3x² - 30x + 75 + 3y² - 24y + 48
.. 2x² + 2y² - 22x - 22y + 106 = 0
.. x² + y² - 11x - 11y + 53 = 0
.. The required equation of locus is x² + y² - 11x - y² - 11x - 11y + 53 = 0.
In simple words: This problem involves a direct relationship between the squared distances from point P to two fixed points A and B (PA² = 3PB²). Expanding the distance formulas and simplifying yields a quadratic equation in x and y, which describes the desired locus.

🎯 Exam Tip: Distribute the constant multiplier (in this case, 3) carefully to all terms within the brackets on the right side of the equation. Any arithmetic errors during distribution will lead to an incorrect final equation.

Question 5. A(2, 4) and B(5, 8), find the equation of the locus of point P such that PA² - PB² = 13.
Answer:
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 4), B(5, 8) and PA² - PB² = 13
.. [(x - 2)² + (y - 4)²] - [(x - 5)² + (y - 8)²] = 13
.. (x² - 4x + 4 + y² - 8y + 16) - (x² - 10x + 25 + y² - 16y + 64) = 13
.. 6x + 8y - 69 = 13
.. 6x + 8y - 82 = 0
.. 3x + 4y - 41 = 0
.. The required equation of locus is 3x + 4y - 41 = 0
In simple words: This question asks for the locus of points P where the difference of the squares of its distances from two fixed points A and B is constant (PA² - PB² = 13). Expanding the distance formulas and simplifying results in a linear equation, indicating a straight line as the locus.

🎯 Exam Tip: When subtracting entire expressions, ensure you distribute the negative sign to all terms within the second bracket. This is a common source of error. Carefully combine like terms to arrive at the simplest form of the equation.

Question 6. A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends a right angle at P. (∠APB = 90°)
Answer:
Solution:
Let P(x. y) be any point on the required locus.
Given, A(1, 6) and B(3, 5), ∠APB = 90°
.. ΔΑΡΒ is a right-angled triangle.

ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक समकोण त्रिभुज APB दिखाया गया है, जहाँ कोण P 90° का है। बिंदु A के निर्देशांक (1, 6) हैं, बिंदु B के निर्देशांक (3, 5) हैं, और बिंदु P के निर्देशांक (x, y) हैं। यह आरेख पाइथागोरस प्रमेय के अनुप्रयोग को दर्शाता है।
By Pythagoras theorem,
AP² + PB² = AB²
.. [(x - 1)² + (y - 6)²] + [(x - 3)² + (y - 5)²] = (1 - 3)² + (6 - 5)²
.. x² - 2x + 1 + y² - 12y + 36 + x² - 6x + 9 + y² - 10y + 25 = 4 + 1
.. 2x² + 2y² - 8x - 22y + 66 = 0
.. x² + y² - 4x - 11y + 33 = 0
.. The required equation of locus is x² + y² - 4x - 11y + 33 = 0
In simple words: The problem states that the line segment AB subtends a right angle at point P, meaning triangle APB is a right-angled triangle with the right angle at P. Applying the Pythagorean theorem (AP² + PB² = AB²) and using the distance formula, we can derive a quadratic equation that represents a circle, which is the locus of P.

🎯 Exam Tip: Recognize that "subtends a right angle" implies the use of the Pythagorean theorem. Calculate the squared distances AP², PB², and AB² accurately. Simplify the equation by combining like terms and dividing by common factors to get the simplest form.

Question 7. If the origin is shifted to the point O'(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points (a) A(1, 3) (b) B(2, 5)
Answer:
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
.. x = X + h and y = Y + k
.. x = X + 2 and y = Y + 3 .....(i)
(a) Given, A(x, y) = A(1, 3)
x = X + 2 and y = Y + 3 .....[From (i)]
.. 1 = X + 2 and 3 = Y + 3
.. X = -1 and Y = 0
.. the new co-ordinates of point A are (-1, 0).
(b) Given, B(x, y) = B(2, 5)
x = X + 2 andy = Y + 3 ......[From (i)]
.. 2 = X + 2 and 5 = Y +3
.. X = 0 and Y = 2
.. the new co-ordinates of point B are (0, 2).
In simple words: When the origin is shifted, the new coordinates (X, Y) are related to the old coordinates (x, y) by the transformation x = X + h and y = Y + k, where (h, k) is the new origin. We use this relationship to find the new coordinates of given points.

🎯 Exam Tip: Clearly identify the old coordinates (x, y) and the new origin (h, k). Be precise when substituting values into the transformation equations `x = X + h` and `y = Y + k` and solving for X and Y.

Question 8. If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points (a) C(5, 4) (b) D(3, 3)
Answer:
Solution:
Origin is shifted to (1, 3) = (h, k)
Let the new co-ordinates be (X, Y)
x = X + h and y = Y + k
.: x = X + 1 and 7 = Y + 3 .....(i)
(a) Given, C(X, Y) = C(5, 4)
.. x = X + 1 andy = Y + 3 .....[From(i)]
∴ x = 5 + 1 = 6 and y = 4 + 3 = 7
.. the old co-ordinates of point C are (6, 7).
(b) Given, D(X, Y) = D(3, 3)
.. x = X + 1 and y = Y + 3 .....[From (i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
.. the old co-ordinates of point D are (4, 6).
In simple words: This question is the reverse of the previous one. Given the new origin (h, k) and the new coordinates (X, Y) of points, we use the same transformation formulas x = X + h and y = Y + k to calculate the original coordinates (x, y) of those points.

🎯 Exam Tip: Distinguish carefully between old and new coordinates. Remember that `(h, k)` is the *shift* of the origin. It's crucial to correctly identify which variables are known and which need to be solved for. Double-check your arithmetic, especially additions.

Question 9. If the co-ordinates (5, 14) change to (8, 3) by the shift of origin, find the co-ordinates of the point, where the origin is shifted.
Answer:
Solution:
Let the origin be shifted to (h, k).
Given, (x,y) = (5, 14), (X, Y) = (8, 3)
Since, x = X + h and y = Y + k
.. 5 = 8 + h and 14 = 3 + k
.. h = -3 and k = 11
.. the co-ordinates of the point, where the origin is shifted are (-3, 11).
In simple words: In this problem, we are given the original coordinates (x, y) and the new coordinates (X, Y) of a point after an origin shift. Using the transformation equations x = X + h and y = Y + k, we solve for h and k to find the coordinates of the shifted origin.

🎯 Exam Tip: Set up two separate equations for x and y based on the transformation formulas. Solve these simple linear equations for h and k. Ensure clear distinction between original coordinates, new coordinates, and the shift values.

Question 10. Obtain the new equations of the following loci if the origin is shifted to the point O'(2, 2), the direction of axes remaining the same:
(a) 3x - y + 2 = 0
(b) x² + y² - 3x = 7
(c) xy - 2x - 2y + 4 = 0
Answer:
Solution:
Given, (h, k) = (2, 2)
Let (X, Y) be the new co-ordinates of the point (x, y).
.. x = X + h and y = Y + k
.. x = X + 2 and y = Y + 2
(a) Substituting the values of x and y in the equation 3x - y + 2 = 0, we get
3(X + 2) - (Y + 2) + 2 = 0
.. 3X + 6 - Y - 2 + 2 = 0
.. 3X - Y + 6 = 0, which is the new equation of locus.
(b) Substituting the values of x and y in the equation x² + y² - 3x = 7, we get
(X + 2)² + (Y + 2)² - 3(X + 2) = 7
.. X² + 4X + 4 + Y² + 4Y + 4 - 3X - 6 = 7
.. X² + Y² + X + 4Y - 5 = 0, which is the new equation of locus.
(c) Substituting the values of x and y in the equation xy - 2x - 2y + 4 = 0, we get
(X + 2) (Y + 2) - 2(X + 2) - 2(Y + 2) + 4 = 0
.. XY + 2X + 2Y + 4 - 2X - 4 - 2Y - 4 + 4 = 0
.. XY = 0, which is the new equation of locus.
In simple words: To find the new equation of a locus after shifting the origin, substitute x = X + h and y = Y + k (where (h, k) is the new origin) into the original equation. Then simplify the resulting expression to get the equation in terms of X and Y.

🎯 Exam Tip: Remember to substitute `(X+h)` for every `x` and `(Y+k)` for every `y` in the original equation. Pay meticulous attention to expanding binomials and distributing terms, especially for quadratic expressions, to avoid algebraic errors.