Samacheer Kalvi Class 11 Business Maths Solutions Chapter 3 Analytical Geometry Exercise 3.1

Question 1. Find the locus of a point which is equidistant from (1, 3) and x-axis.
Answer: Let P \( (x_1, y_1) \) be any point on the locus. Let A be the fixed point \( (1, 3) \). The distance of point P from the x-axis is \( y_1 \). We are given that the distance from P to A is equal to the distance from P to the x-axis, so \( PA = y_1 \). Squaring both sides, we get \( PA^2 = y_1^2 \). Using the distance formula for \( PA^2 \), we have \( (x_1 - 1)^2 + (y_1 - 3)^2 = y_1^2 \). Expanding this equation gives \( x_1^2 - 2x_1 + 1 + y_1^2 - 6y_1 + 9 = y_1^2 \). Subtracting \( y_1^2 \) from both sides and simplifying, we find the locus of the point \( (x_1, y_1) \) is \( x^2 - 2x - 6y + 10 = 0 \). This equation shows all points that meet the given condition.
In simple words: We are looking for all points that are the same distance away from point (1, 3) and the x-axis. Using the distance formula and simplifying, we get an equation that describes the path of these points.

🎯 Exam Tip: When finding a locus, remember to represent the moving point as \( (x, y) \) and use the distance formula or other geometric relations to set up an equation, then simplify it.

Question 2. A point moves so that it is always at a distance of 4 units from the point (3, -2).
Answer: Let P \( (x_1, y_1) \) be any point on the locus. Let A be the fixed point \( (3, -2) \). We are given that the distance between P and A is always 4 units, so \( PA = 4 \). Squaring both sides, we get \( PA^2 = 16 \). Using the distance formula for \( PA^2 \), we have \( (x_1 - 3)^2 + (y_1 - (-2))^2 = 16 \). This simplifies to \( (x_1 - 3)^2 + (y_1 + 2)^2 = 16 \). Expanding this gives \( x_1^2 - 6x_1 + 9 + y_1^2 + 4y_1 + 4 = 16 \). Combining like terms and moving 16 to the left side, we get \( x_1^2 + y_1^2 - 6x_1 + 4y_1 + 13 - 16 = 0 \). Therefore, the locus of the point \( (x_1, y_1) \) is \( x^2 + y^2 - 6x + 4y - 3 = 0 \). This represents a circle centered at (3, -2) with a radius of 4.
In simple words: We need to find the path of a point that stays 4 units away from the point (3, -2). This path is a circle, and its equation is found by using the distance formula and simplifying.

🎯 Exam Tip: A point always at a fixed distance from another point describes a circle. Be careful with signs when substituting coordinates into the distance formula, especially with negative values.

Question 3. If the distance of a point from the points (2, 1) and (1, 2) are in the ratio 2 : 1, then find the locus of the point.
Answer: Let P \( (x_1, y_1) \) be any point on the locus. Let A be the point \( (2, 1) \) and B be the point \( (1, 2) \). We are given that the ratio of the distance from P to A and P to B is \( 2:1 \), meaning \( PA : PB = 2 : 1 \).
\( \implies \frac{PA}{PB} = \frac{2}{1} \)
\( \implies PA = 2PB \)
Squaring both sides to remove the square root from the distance formula:
\( \implies PA^2 = (2PB)^2 \)
\( \implies PA^2 = 4PB^2 \)
Using the distance formula:
\( \implies (x_1 - 2)^2 + (y_1 - 1)^2 = 4[(x_1 - 1)^2 + (y_1 - 2)^2] \)
Expanding both sides:
\( \implies x_1^2 - 4x_1 + 4 + y_1^2 - 2y_1 + 1 = 4[x_1^2 - 2x_1 + 1 + y_1^2 - 4y_1 + 4] \)
\( \implies x_1^2 + y_1^2 - 4x_1 - 2y_1 + 5 = 4x_1^2 - 8x_1 + 4y_1^2 - 16y_1 + 20 \)
Move all terms to one side to set the equation to zero:
\( \implies 0 = 4x_1^2 - x_1^2 + 4y_1^2 - y_1^2 - 8x_1 + 4x_1 - 16y_1 + 2y_1 + 20 - 5 \)
\( \implies 0 = 3x_1^2 + 3y_1^2 - 4x_1 - 14y_1 + 15 \)
Thus, the locus of the point \( (x_1, y_1) \) is \( 3x^2 + 3y^2 - 4x - 14y + 15 = 0 \). This final equation represents the path of all such points.
In simple words: A point moves so that its distance from point (2,1) is twice its distance from point (1,2). We use the distance formula, set up the ratio, and then simplify the equation to find the exact path the point takes.

🎯 Exam Tip: When dealing with ratios of distances, always square both sides of the equation \( PA = k \cdot PB \) early to eliminate square roots and make algebraic manipulation easier.

Question 4. Find a point on the x-axis which is equidistant from the points (7, -6) and (3, 4).
Answer: Let P \( (x_1, 0) \) be any point on the x-axis. The y-coordinate is 0 because the point lies on the x-axis. Let A be the point \( (7, -6) \) and B be the point \( (3, 4) \). We are given that P is equidistant from A and B, so \( PA = PB \).
Squaring both sides:
\( \implies PA^2 = PB^2 \)
Using the distance formula:
\( \implies (x_1 - 7)^2 + (0 - (-6))^2 = (x_1 - 3)^2 + (0 - 4)^2 \)
\( \implies (x_1 - 7)^2 + (6)^2 = (x_1 - 3)^2 + (-4)^2 \)
Expanding the squares:
\( \implies x_1^2 - 14x_1 + 49 + 36 = x_1^2 - 6x_1 + 9 + 16 \)
\( \implies x_1^2 - 14x_1 + 85 = x_1^2 - 6x_1 + 25 \)
Subtract \( x_1^2 \) from both sides and rearrange to solve for \( x_1 \):
\( \implies -14x_1 + 85 = -6x_1 + 25 \)
\( \implies 85 - 25 = -6x_1 + 14x_1 \)
\( \implies 60 = 8x_1 \)
\( \implies x_1 = \frac{60}{8} \)
Simplify the fraction:
\( \implies x_1 = \frac{15}{2} \)
Therefore, the required point on the x-axis is \( (\frac{15}{2}, 0) \). This point is exactly in the middle of the distance between A and B, but specifically along the x-axis.
In simple words: We need to find a point on the x-axis that is the same distance from two other given points. We use the distance formula and the fact that any point on the x-axis has a y-coordinate of zero to solve for the x-coordinate.

🎯 Exam Tip: Remember that a point on the x-axis always has coordinates \( (x, 0) \), and a point on the y-axis always has coordinates \( (0, y) \). This helps simplify the distance formula significantly.

Question 5. If A(-1, 1) and B(2, 3) are two fixed points, then find the locus of a point P so that the area of triangle APB = 8 sq.units.
Answer: Let the point P be \( (x_1, y_1) \). The fixed points are A \( (-1, 1) \) and B \( (2, 3) \). We are given that the area of triangle APB is 8 square units. The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is \( \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \).
Given Area \( = 8 \).
So, \( \frac{1}{2} |x_1(1 - 3) + (-1)(3 - y_1) + 2(y_1 - 1)| = 8 \)
Multiplying by 2:
\( \implies |x_1(-2) + (-1)(3 - y_1) + 2(y_1 - 1)| = 16 \)
Expanding the terms inside the absolute value:
\( \implies |-2x_1 - 3 + y_1 + 2y_1 - 2| = 16 \)
Combine like terms:
\( \implies |-2x_1 + 3y_1 - 5| = 16 \)
This means there are two possibilities for the expression inside the absolute value:
Either \( -2x_1 + 3y_1 - 5 = 16 \) or \( -2x_1 + 3y_1 - 5 = -16 \).

Case 1: \( -2x_1 + 3y_1 - 5 = 16 \)
\( \implies -2x_1 + 3y_1 - 21 = 0 \)
Multiplying by -1 to make the first term positive:
\( \implies 2x_1 - 3y_1 + 21 = 0 \)

Case 2: \( -2x_1 + 3y_1 - 5 = -16 \)
\( \implies -2x_1 + 3y_1 + 11 = 0 \)
Multiplying by -1 to make the first term positive:
\( \implies 2x_1 - 3y_1 - 11 = 0 \)
The locus of the point P \( (x_1, y_1) \) is therefore given by the two parallel lines: \( 2x - 3y + 21 = 0 \) and \( 2x - 3y - 11 = 0 \). These lines represent all points where a triangle formed with A and B would have an area of 8 square units.
In simple words: We are looking for points that, when connected with two other fixed points, form a triangle with a specific area of 8. We use the area formula for a triangle in coordinate geometry, substitute the given points, and solve for the path of our moving point. This gives two possible straight lines.

🎯 Exam Tip: When using the area of a triangle formula with coordinates, remember the absolute value sign. This often leads to two possible equations, representing two parallel lines equidistant from the base AB.