OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Exercise 16 (I)

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(i)

Question 1. Its distance from the x-axis is \( c^2 \).
Answer: Let \( (h, k) \) be any point on the locus. The distance of this point from the x-axis is given by \( |k| \). According to the given condition, this distance is \( c^2 \).
Therefore, we have \( |k| = c^2 \).
The locus of point \( (h, k) \) is found by replacing \( h \) with \( x \) and \( k \) with \( y \).
So, the locus of the point is \( |y| = c^2 \). This means the point always stays at a fixed distance from the x-axis, either \( c^2 \) units above or \( c^2 \) units below.
In simple words: We are looking for all points that are a specific distance \( c^2 \) away from the x-axis. This means the y-coordinate of these points will always be \( c^2 \) or \( -c^2 \).

🎯 Exam Tip: Remember that the distance of a point \( (x, y) \) from the x-axis is \( |y| \) and from the y-axis is \( |x| \). Always use absolute values for distances.

Question 2. Its distance from the origin is 5.
Answer: Let \( (h, k) \) be any point on the locus. The given point is the origin \( (0, 0) \).
The distance of \( (h, k) \) from the origin \( (0, 0) \) is found using the distance formula:
\( \sqrt{(h-0)^2 + (k-0)^2} = 5 \)
\( \implies \sqrt{h^2 + k^2} = 5 \)
To remove the square root, we square both sides of the equation:
\( ( \sqrt{h^2 + k^2} )^2 = 5^2 \)
\( \implies h^2 + k^2 = 25 \)
The locus of point \( (h, k) \) is given by replacing \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the equation of the locus is \( x^2 + y^2 = 25 \). This equation describes a circle centered at the origin with a radius of 5 units.
In simple words: We need to find all points that are 5 units away from the center point (0,0). Using the distance formula and squaring both sides gives us the equation of a circle.

🎯 Exam Tip: The standard equation of a circle centered at the origin with radius \( r \) is \( x^2 + y^2 = r^2 \). Squaring both sides is a common step to remove square roots in locus problems.

Question 3. The sum of the squares of its distances from the points (2, 4) and (- 3, - 1) is 30.
Answer: Let \( P(h, k) \) be any point on the locus. The two given points are \( A(2, 4) \) and \( B(-3, -1) \).
According to the given condition, the sum of the squares of the distances from P to A and P to B is 30.
\( PA^2 + PB^2 = 30 \)
Using the distance formula, \( PA^2 = (h-2)^2 + (k-4)^2 \) and \( PB^2 = (h-(-3))^2 + (k-(-1))^2 = (h+3)^2 + (k+1)^2 \).
So, we have:
\( (h-2)^2 + (k-4)^2 + (h+3)^2 + (k+1)^2 = 30 \)
\( \implies (h^2 - 4h + 4) + (k^2 - 8k + 16) + (h^2 + 6h + 9) + (k^2 + 2k + 1) = 30 \)
\( \implies 2h^2 + 2k^2 + 2h - 6k + 30 = 30 \)
\( \implies 2h^2 + 2k^2 + 2h - 6k = 0 \)
Divide the entire equation by 2:
\( \implies h^2 + k^2 + h - 3k = 0 \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the locus of point \( (h, k) \) is given by:
\( x^2 + y^2 + x - 3y = 0 \). This is the equation of a circle.
In simple words: We set up an equation where the squared distances from a moving point \( P \) to two fixed points \( A \) and \( B \) add up to 30. After expanding and simplifying the equation, we get the final locus.

🎯 Exam Tip: When dealing with sums of squares of distances, carefully expand all binomial terms and combine like terms. This often leads to a simpler quadratic equation.

Question 4. Its distance from the x-axis is half its distance from the y-axis.
Answer: Let \( P(h, k) \) be any point on the locus.
The equation of the x-axis is \( y = 0 \). The distance of \( P(h, k) \) from the x-axis is \( |k| \).
The equation of the y-axis is \( x = 0 \). The distance of \( P(h, k) \) from the y-axis is \( |h| \).
According to the given condition:
\( |k| = \frac{1}{2} |h| \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the locus of point \( P(h, k) \) is given by:
\( |y| = \frac{1}{2} |x| \)
To remove the absolute values, we can square both sides:
\( (|y|)^2 = \left(\frac{1}{2} |x|\right)^2 \)
\( \implies y^2 = \frac{1}{4} x^2 \)
Multiply both sides by 4:
\( \implies 4y^2 = x^2 \)
\( \implies x^2 - 4y^2 = 0 \). This represents a pair of straight lines \( x = \pm 2y \).
In simple words: We want points where the distance to the x-axis is half the distance to the y-axis. Setting up this rule using absolute values for distances and then squaring both sides gives the equation for the path of these points.

🎯 Exam Tip: Remember that distance from an axis is always positive, hence the use of absolute values. Squaring both sides is a common technique to eliminate absolute values when solving locus problems.

Question 5. Its distance from the y-axis is equal to its distance from the point (1, 1).
Answer: Let \( P(h, k) \) be any point on the locus. The given point is \( A(1, 1) \).
The equation of the y-axis is \( x = 0 \). The distance of \( P(h, k) \) from the y-axis is \( |h| \).
The distance of \( P(h, k) \) from \( A(1, 1) \) is given by the distance formula:
\( \sqrt{(h-1)^2 + (k-1)^2} \)
According to the given condition, these two distances are equal:
\( |h| = \sqrt{(h-1)^2 + (k-1)^2} \)
To eliminate the absolute value and the square root, we square both sides:
\( h^2 = (h-1)^2 + (k-1)^2 \)
\( \implies h^2 = (h^2 - 2h + 1) + (k^2 - 2k + 1) \)
\( \implies h^2 = h^2 - 2h + 1 + k^2 - 2k + 1 \)
Subtract \( h^2 \) from both sides and simplify the constants:
\( \implies 0 = -2h + k^2 - 2k + 2 \)
Rearrange the terms to get the standard form:
\( \implies k^2 - 2k - 2h + 2 = 0 \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the required locus of \( P(h, k) \) is given by:
\( y^2 - 2y - 2x + 2 = 0 \). This is the equation of a parabola.
In simple words: We set the distance from the y-axis (which is \( |h| \)) equal to the distance from a fixed point \( (1,1) \). Squaring both sides helps remove the absolute value and square root, and then we simplify the equation to find the path of the point.

🎯 Exam Tip: When solving locus problems, always clearly define the moving point \( (h,k) \) and set up the equation based on the given conditions. Be careful when expanding squared terms like \( (h-1)^2 \).

Question 6. Find the locus of a point which is equidistant from the points (1, 0) and (- 1, 0).
Answer: Let \( P(h, k) \) be any point on the locus. The given points are \( A(1, 0) \) and \( B(-1, 0) \).
According to the given condition, point P is equidistant from A and B.
\( |PA| = |PB| \)
To simplify, we can square both sides to remove square roots from the distance formula:
\( PA^2 = PB^2 \)
Using the distance formula:
\( (h-1)^2 + (k-0)^2 = (h-(-1))^2 + (k-0)^2 \)
\( \implies (h-1)^2 + k^2 = (h+1)^2 + k^2 \)
Subtract \( k^2 \) from both sides:
\( \implies (h-1)^2 = (h+1)^2 \)
Expand both sides:
\( \implies h^2 - 2h + 1 = h^2 + 2h + 1 \)
Subtract \( h^2 \) and 1 from both sides:
\( \implies -2h = 2h \)
Add \( 2h \) to both sides:
\( \implies 0 = 4h \)
Divide by 4:
\( \implies h = 0 \)
Replacing \( h \) with \( x \), the required locus of \( P(h, k) \) is given by:
\( x = 0 \). This is the equation of the y-axis.
In simple words: We are looking for all points that are the same distance from two fixed points. When we set up the distance equation and simplify, we find that all such points must lie on the y-axis.

🎯 Exam Tip: A point equidistant from two fixed points lies on the perpendicular bisector of the segment joining those points. In this case, the y-axis is the perpendicular bisector of the segment connecting (1,0) and (-1,0).

Question 7. A(2, 0) and B(4, 0) are two given points. A point P moves so that \( PA^2 + PB^2 = 10 \). Find the locus of P.
Answer: Let \( P(h, k) \) be the required locus. The given points are \( A(2, 0) \) and \( B(4, 0) \).
According to the given condition, the sum of the squares of the distances from P to A and P to B is 10.
\( PA^2 + PB^2 = 10 \)
Using the distance formula, \( PA^2 = (h-2)^2 + (k-0)^2 = (h-2)^2 + k^2 \) and \( PB^2 = (h-4)^2 + (k-0)^2 = (h-4)^2 + k^2 \).
So, we have:
\( (h-2)^2 + k^2 + (h-4)^2 + k^2 = 10 \)
\( \implies (h^2 - 4h + 4) + k^2 + (h^2 - 8h + 16) + k^2 = 10 \)
\( \implies 2h^2 + 2k^2 - 12h + 20 = 10 \)
Subtract 10 from both sides:
\( \implies 2h^2 + 2k^2 - 12h + 10 = 0 \)
Divide the entire equation by 2:
\( \implies h^2 + k^2 - 6h + 5 = 0 \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the required locus of \( P(h, k) \) is given by:
\( x^2 + y^2 - 6x + 5 = 0 \). This is the equation of a circle.
In simple words: We find the path of a point P such that the squares of its distances from two fixed points, A and B, add up to 10. By expanding and simplifying the distance formulas, we get the equation of the locus.

🎯 Exam Tip: Remember to use the correct distance formula for \( PA^2 \) and \( PB^2 \). Careful expansion of squared terms and combining like terms are crucial for arriving at the correct locus equation.

Question 8. Find the locus of a point such that the sum of its distances from the points (0, 2) and (0, - 2) is 6.
Answer: Let \( P(h, k) \) be any point on the locus. The given points are \( A(0, 2) \) and \( B(0, -2) \).
According to the given condition, the sum of the distances from P to A and P to B is 6. This is the definition of an ellipse.
\( PA + PB = 6 \)
Let \( PA = l \) and \( PB = m \), so \( l + m = 6 \).
Using the distance formula:
\( l = \sqrt{(h-0)^2 + (k-2)^2} = \sqrt{h^2 + (k-2)^2} \)
\( m = \sqrt{(h-0)^2 + (k+2)^2} = \sqrt{h^2 + (k+2)^2} \)
Now, we subtract the squares of the distances:
\( l^2 - m^2 = (h^2 + (k-2)^2) - (h^2 + (k+2)^2) \)
\( \implies l^2 - m^2 = (k^2 - 4k + 4) - (k^2 + 4k + 4) \)
\( \implies l^2 - m^2 = -8k \)
We know that \( l^2 - m^2 = (l-m)(l+m) \). Since \( l+m = 6 \):
\( (l-m)(6) = -8k \)
\( \implies l-m = \frac{-8k}{6} = \frac{-4k}{3} \)
Now we have a system of two linear equations for \( l \) and \( m \):
1. \( l + m = 6 \)
2. \( l - m = -\frac{4k}{3} \)
Adding (1) and (2):
\( 2l = 6 - \frac{4k}{3} \)
\( \implies l = 3 - \frac{2k}{3} \)
Now substitute the expression for \( l \) back into its original distance formula:
\( \sqrt{h^2 + (k-2)^2} = 3 - \frac{2k}{3} \)
Square both sides:
\( h^2 + (k-2)^2 = \left(3 - \frac{2k}{3}\right)^2 \)
\( \implies h^2 + k^2 - 4k + 4 = 9 - 2(3)\left(\frac{2k}{3}\right) + \left(\frac{2k}{3}\right)^2 \)
\( \implies h^2 + k^2 - 4k + 4 = 9 - 4k + \frac{4k^2}{9} \)
Notice that the \( -4k \) terms cancel out on both sides:
\( \implies h^2 + k^2 + 4 = 9 + \frac{4k^2}{9} \)
Multiply the entire equation by 9 to clear the fraction:
\( \implies 9h^2 + 9k^2 + 36 = 81 + 4k^2 \)
Rearrange the terms:
\( \implies 9h^2 + 9k^2 - 4k^2 + 36 - 81 = 0 \)
\( \implies 9h^2 + 5k^2 - 45 = 0 \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the required locus of \( P(h, k) \) is given by:
\( 9x^2 + 5y^2 - 45 = 0 \). This is the standard equation of an ellipse.
In simple words: The problem describes an ellipse because the sum of distances from a point to two fixed points is constant. We set up equations for distances, combine them to simplify, and then remove square roots by squaring to find the equation of the ellipse.

🎯 Exam Tip: This type of problem often leads to an ellipse. The trick is to isolate one square root term before squaring, or to use the difference of squares identity \( (l-m)(l+m) \), which simplifies the algebra significantly.

Question 9. Find the locus of a point, so that the join of points (- 5, 1) and (3, 2) subtends a right angle at the moving point.
Answer: Let \( P(h, k) \) be any point on the locus. The given points are \( A(-5, 1) \) and \( B(3, 2) \).
The condition that the line joining points A and B subtends a right angle at P means that the angle \( \angle APB \) is \( 90^\circ \).
In a right-angled triangle, by Pythagoras theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, AB is the hypotenuse.
\( PA^2 + PB^2 = AB^2 \)

Using the distance formula for \( PA^2 \), \( PB^2 \), and \( AB^2 \):
\( PA^2 = (h - (-5))^2 + (k - 1)^2 = (h+5)^2 + (k-1)^2 \)
\( PB^2 = (h - 3)^2 + (k - 2)^2 \)
\( AB^2 = (3 - (-5))^2 + (2 - 1)^2 = (3+5)^2 + (1)^2 = 8^2 + 1^2 = 64 + 1 = 65 \)
Substitute these into the Pythagorean equation:
\( (h+5)^2 + (k-1)^2 + (h-3)^2 + (k-2)^2 = 65 \)
Expand the squared terms:
\( (h^2 + 10h + 25) + (k^2 - 2k + 1) + (h^2 - 6h + 9) + (k^2 - 4k + 4) = 65 \)
Combine like terms:
\( \implies 2h^2 + 2k^2 + (10h - 6h) + (-2k - 4k) + (25 + 1 + 9 + 4) = 65 \)
\( \implies 2h^2 + 2k^2 + 4h - 6k + 39 = 65 \)
Subtract 65 from both sides:
\( \implies 2h^2 + 2k^2 + 4h - 6k + 39 - 65 = 0 \)
\( \implies 2h^2 + 2k^2 + 4h - 6k - 26 = 0 \)
Divide the entire equation by 2:
\( \implies h^2 + k^2 + 2h - 3k - 13 = 0 \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the required locus of \( P(h, k) \) is given by:
\( x^2 + y^2 + 2x - 3y - 13 = 0 \). This is the equation of a circle, which is known as the circle with AB as diameter.
In simple words: If a line segment forms a right angle at a moving point, that point traces a circle with the line segment as its diameter. We use the distance formula and the Pythagorean theorem to set up an equation, then simplify it to find the circle's equation.

🎯 Exam Tip: The condition that a line segment subtends a right angle at a point implies that the point lies on a circle whose diameter is the given line segment. Use the distance formula and the Pythagorean theorem carefully.

Question 10. Two points A and B with co-ordinates (5, 3) and (3, - 2) are given. A point P moves so that the area of \( \triangle PAB \) is constant and equal to 9 square units. Find the equation to the locus of the point P.
Answer: Let \( P(h, k) \) be any point on the locus. The given points are \( A(5, 3) \) and \( B(3, -2) \).
The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\( \text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)
Here, \( (h, k) \) is \( (x_1, y_1) \), \( (5, 3) \) is \( (x_2, y_2) \), and \( (3, -2) \) is \( (x_3, y_3) \).
Area of \( \triangle PAB = \frac{1}{2} | h(3 - (-2)) + 5(-2 - k) + 3(k - 3) | \)
\( = \frac{1}{2} | h(5) - 10 - 5k + 3k - 9 | \)
\( = \frac{1}{2} | 5h - 2k - 19 | \)
Given that the area of \( \triangle PAB \) is 9 square units:
\( \frac{1}{2} | 5h - 2k - 19 | = 9 \)
Multiply by 2:
\( \implies | 5h - 2k - 19 | = 18 \)
This absolute value equation means there are two possibilities:
\( 5h - 2k - 19 = 18 \) OR \( 5h - 2k - 19 = -18 \)
Case 1: \( 5h - 2k - 19 = 18 \)
\( \implies 5h - 2k - 37 = 0 \)
Case 2: \( 5h - 2k - 19 = -18 \)
\( \implies 5h - 2k - 1 = 0 \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the locus of \( P(h, k) \) consists of two straight lines:
\( 5x - 2y - 1 = 0 \) and \( 5x - 2y - 37 = 0 \). These are two parallel lines, which is expected as the base AB is fixed and the height (distance from P to AB) must be constant for the area to be constant.
In simple words: The problem asks for the path of a point P such that the triangle formed by P and two fixed points A and B always has the same area. Using the triangle area formula with coordinates, we set it equal to 9. This gives two possible linear equations, representing two parallel lines.

🎯 Exam Tip: Remember the coordinate geometry formula for the area of a triangle. The absolute value in the area formula leads to two separate equations, representing two parallel lines equidistant from the base formed by A and B.

Question 11. Show that (1, 2) lies on the locus \( x^2 + y^2 - 4x - 6y + 11 = 0 \).
Answer: The equation of the given locus is:
\( x^2 + y^2 - 4x - 6y + 11 = 0 \) ...(1)
To show that the point \( (1, 2) \) lies on this locus, we substitute \( x = 1 \) and \( y = 2 \) into the left-hand side (L.H.S.) of the equation.
L.H.S. \( = (1)^2 + (2)^2 - 4(1) - 6(2) + 11 \)
\( = 1 + 4 - 4 - 12 + 11 \)
\( = 5 - 4 - 12 + 11 \)
\( = 1 - 12 + 11 \)
\( = -11 + 11 \)
\( = 0 \)
Since L.H.S. \( = 0 \), which is equal to the right-hand side (R.H.S.) of the equation, the point \( (1, 2) \) satisfies the equation. This demonstrates that the point \( (1, 2) \) lies on the given locus.
In simple words: To check if a point is on a line or curve, put its x and y values into the equation. If both sides of the equation become equal, then the point is on the line/curve.

🎯 Exam Tip: Always show your substitution steps clearly. A point lies on a curve if and only if its coordinates satisfy the equation of the curve.

Question 12. Does the point (3, 0) lie on the curve \( 3x^2 + y^2 - 4x + 7 = 0 \)?
Answer: The equation of the given curve is:
\( 3x^2 + y^2 - 4x + 7 = 0 \) ...(1)
To check if the point \( (3, 0) \) lies on this curve, we substitute \( x = 3 \) and \( y = 0 \) into the left-hand side (L.H.S.) of the equation.
L.H.S. \( = 3(3)^2 + (0)^2 - 4(3) + 7 \)
\( = 3(9) + 0 - 12 + 7 \)
\( = 27 - 12 + 7 \)
\( = 15 + 7 \)
\( = 22 \)
Since L.H.S. \( = 22 \), and the R.H.S. of the equation is \( 0 \), L.H.S. \( \neq \) R.H.S.
Therefore, the point \( (3, 0) \) does not satisfy the equation. This means the point \( (3, 0) \) does not lie on the given curve.
In simple words: We put the x and y values from the point (3,0) into the curve's equation. If the calculation does not result in 0 (which is the right side of the equation), then the point is not on the curve.

🎯 Exam Tip: Be careful with calculations involving squares and signs. A clear demonstration that L.H.S. \( \neq \) R.H.S. is necessary to prove that a point does not lie on a curve.

Question 13. Find the condition that the point (h, k) may lie on the curve \( x^2 + y^2 + 5x + 11y - 2 = 0 \).
Answer: The equation of the given curve is:
\( x^2 + y^2 + 5x + 11y - 2 = 0 \) ...(1)
For the point \( (h, k) \) to lie on this curve, its coordinates must satisfy the equation of the curve.
This means we simply replace \( x \) with \( h \) and \( y \) with \( k \) in the equation.
Substituting \( x = h \) and \( y = k \) into equation (1), we get:
\( h^2 + k^2 + 5h + 11k - 2 = 0 \)
This is the required condition for the point \( (h, k) \) to lie on the given curve.
In simple words: If a point is on a curve, its coordinates must work in the curve's equation. So, we just replace 'x' with 'h' and 'y' with 'k' in the equation.

🎯 Exam Tip: The condition for a point \( (h,k) \) to lie on a curve is simply its equation with \( x \) replaced by \( h \) and \( y \) replaced by \( k \). No further calculations are needed unless asked to prove something specific.

Question 14. If the line \( (2 + k) x - (2 - k) y + (4k + 14) = 0 \) passes through the point (- 1, 21), find k.
Answer: The equation of the given line is:
\( (2 + k) x - (2 - k) y + (4k + 14) = 0 \) ...(1)
Given that the line passes through the point \( (-1, 21) \), the coordinates of this point must satisfy the equation of the line.
Substitute \( x = -1 \) and \( y = 21 \) into equation (1):
\( (2 + k)(-1) - (2 - k)(21) + (4k + 14) = 0 \)
Expand the terms:
\( \implies -2 - k - (42 - 21k) + 4k + 14 = 0 \)
\( \implies -2 - k - 42 + 21k + 4k + 14 = 0 \)
Combine the constant terms: \( -2 - 42 + 14 = -44 + 14 = -30 \)
Combine the terms with \( k \): \( -k + 21k + 4k = 24k \)
So the equation becomes:
\( 24k - 30 = 0 \)
Add 30 to both sides:
\( \implies 24k = 30 \)
Divide by 24 to solve for \( k \):
\( \implies k = \frac{30}{24} \)
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, 6:
\( \implies k = \frac{5}{4} \)
In simple words: If a point is on a line, its coordinates can be plugged into the line's equation. We substitute the given point's x and y values into the equation and then solve for 'k'.

🎯 Exam Tip: When a point lies on a line, its coordinates will satisfy the line's equation. Be careful with signs when expanding and combining terms, especially with negative numbers.

Question 15. A is the point (- 1, 0) and B is the point (1, 1). Find a point on the line \( 4x + 5y = 4 \), which is equidistant from A and B.
Answer: Let \( P(h, k) \) be the required point on the line \( 4x + 5y = 4 \).
Since P lies on the line, its coordinates must satisfy the line's equation:
\( 4h + 5k = 4 \) ...(1)
The point P is also equidistant from points \( A(-1, 0) \) and \( B(1, 1) \).
\( |PA| = |PB| \)
Square both sides to eliminate square roots:
\( PA^2 = PB^2 \)
Using the distance formula:
\( (h - (-1))^2 + (k - 0)^2 = (h - 1)^2 + (k - 1)^2 \)
\( \implies (h+1)^2 + k^2 = (h-1)^2 + (k-1)^2 \)
Expand the squared terms:
\( \implies (h^2 + 2h + 1) + k^2 = (h^2 - 2h + 1) + (k^2 - 2k + 1) \)
Cancel \( h^2 \) and \( k^2 \) from both sides:
\( \implies 2h + 1 = -2h + 1 - 2k + 1 \)
\( \implies 2h + 1 = -2h - 2k + 2 \)
Move all terms to one side:
\( \implies 2h + 2h + 2k + 1 - 2 = 0 \)
\( \implies 4h + 2k - 1 = 0 \) ...(2)
Now we have a system of two linear equations (1) and (2) for \( h \) and \( k \):
1. \( 4h + 5k = 4 \)
2. \( 4h + 2k = 1 \) (rearranging (2))
Subtract equation (2) from equation (1):
\( (4h + 5k) - (4h + 2k) = 4 - 1 \)
\( \implies 3k = 3 \)
\( \implies k = 1 \)
Substitute \( k = 1 \) into equation (1):
\( 4h + 5(1) = 4 \)
\( \implies 4h + 5 = 4 \)
\( \implies 4h = 4 - 5 \)
\( \implies 4h = -1 \)
\( \implies h = -\frac{1}{4} \)
Therefore, the coordinates of the required point are \( \left(-\frac{1}{4}, 1\right) \). This point is the intersection of the given line and the perpendicular bisector of the segment AB.
In simple words: We need to find a point that lies on a specific line AND is the same distance from two other points. We use the line's equation and the "equidistant" rule to create two equations, then solve them together to find the point's coordinates.

🎯 Exam Tip: A point equidistant from two fixed points lies on the perpendicular bisector of the line segment joining them. Combine this condition with the line equation to form a system of simultaneous equations.

Question 16. The co-ordinates of the point S are (4, 0) and a point P has coordinates (x, y). Express \( PS^2 \) in terms of x and y. Given that M is the foot of the perpendicular from P to the y-axis and that the point P moves so that the lengths PS and PM are equal, prove that the locus of P is \( y^2 - 8x + 16 = 0 \). Find the co-ordinates of one of the two points on the curve whose distance from S is 20 units.
Answer: Given coordinates of point S are \( (4, 0) \) and point P are \( (x, y) \).
First, express \( PS^2 \) in terms of \( x \) and \( y \) using the distance formula:
\( PS^2 = (x - 4)^2 + (y - 0)^2 \)
\( PS^2 = (x - 4)^2 + y^2 \)
Next, M is the foot of the perpendicular from P to the y-axis. The equation of the y-axis is \( x = 0 \).
So, if \( P = (x, y) \), the coordinates of M are \( (0, y) \).
The distance \( PM \) is the distance between \( P(x, y) \) and \( M(0, y) \):
\( PM = \sqrt{(x-0)^2 + (y-y)^2} = \sqrt{x^2 + 0^2} = \sqrt{x^2} = |x| \)
So, \( PM^2 = x^2 \).

Given that P moves so that \( PS = PM \):
\( PS^2 = PM^2 \)
\( (x - 4)^2 + y^2 = x^2 \)
Expand \( (x-4)^2 \):
\( x^2 - 8x + 16 + y^2 = x^2 \)
Subtract \( x^2 \) from both sides:
\( \implies -8x + 16 + y^2 = 0 \)
Rearrange the terms:
\( \implies y^2 - 8x + 16 = 0 \). This is the required locus, which is a parabola.
Now, find the coordinates of points on this curve whose distance from S is 20 units.
Let \( Q(h, k) \) be a point on the curve. So, \( k^2 - 8h + 16 = 0 \) ...(1)
The distance from S\( (4, 0) \) to Q\( (h, k) \) is 20 units.
\( QS = 20 \)
\( QS^2 = 20^2 = 400 \)
Using the distance formula for \( QS^2 \):
\( (h - 4)^2 + (k - 0)^2 = 400 \)
\( \implies (h - 4)^2 + k^2 = 400 \) ...(2)
From equation (1), we can express \( k^2 \) as \( k^2 = 8h - 16 \).
Substitute this expression for \( k^2 \) into equation (2):
\( (h - 4)^2 + (8h - 16) = 400 \)
Expand \( (h-4)^2 \):
\( h^2 - 8h + 16 + 8h - 16 = 400 \)
The \( -8h \) and \( +8h \) terms cancel, and the \( +16 \) and \( -16 \) terms cancel:
\( \implies h^2 = 400 \)
Take the square root of both sides:
\( \implies h = \pm \sqrt{400} \)
\( \implies h = \pm 20 \)
Now find the corresponding values of \( k \) using \( k^2 = 8h - 16 \).
Case 1: When \( h = 20 \)
\( k^2 = 8(20) - 16 \)
\( \implies k^2 = 160 - 16 \)
\( \implies k^2 = 144 \)
\( \implies k = \pm 12 \)
So, two points are \( (20, 12) \) and \( (20, -12) \).
Case 2: When \( h = -20 \)
\( k^2 = 8(-20) - 16 \)
\( \implies k^2 = -160 - 16 \)
\( \implies k^2 = -176 \)
Since \( k^2 \) cannot be negative for real values of \( k \), there are no real points on the curve with \( h = -20 \).
Thus, the required points on the curve whose distance from S is 20 units are \( (20, 12) \) and \( (20, -12) \).
In simple words: First, we find the path of point P by setting its distance from point S equal to its distance from the y-axis. This gives us the equation of the locus. Then, we find specific points on this path that are 20 units away from S by solving a system of equations, making sure to only consider real solutions.

🎯 Exam Tip: The condition \( PS = PM \) (distance from a fixed point equals distance from a fixed line) defines a parabola. Remember to check for real solutions when taking square roots; a negative value under a square root for a real coordinate means no such real point exists.

Question 17. Find the ratio in which the line joining the points (6, 12) and (4, 9) is divided by the curve \( x^2 + y^2 = 4 \).
Answer: Let the point \( P(h, k) \) divide the line segment joining points \( A(6, 12) \) and \( B(4, 9) \) in the ratio \( \lambda : 1 \).

Using the section formula, the coordinates of P are:
\( P\left(\frac{\lambda x_2 + 1 x_1}{\lambda + 1}, \frac{\lambda y_2 + 1 y_1}{\lambda + 1}\right) = P\left(\frac{4\lambda + 6}{\lambda + 1}, \frac{9\lambda + 12}{\lambda + 1}\right) \)
Since point P lies on the curve \( x^2 + y^2 = 4 \), its coordinates must satisfy this equation:
\( \left(\frac{4\lambda + 6}{\lambda + 1}\right)^2 + \left(\frac{9\lambda + 12}{\lambda + 1}\right)^2 = 4 \)
\( \implies \frac{(4\lambda + 6)^2 + (9\lambda + 12)^2}{(\lambda + 1)^2} = 4 \)
\( \implies (4\lambda + 6)^2 + (9\lambda + 12)^2 = 4(\lambda + 1)^2 \)
Expand the terms:
\( (16\lambda^2 + 48\lambda + 36) + (81\lambda^2 + 216\lambda + 144) = 4(\lambda^2 + 2\lambda + 1) \)
Combine like terms on the left side:
\( \implies (16\lambda^2 + 81\lambda^2) + (48\lambda + 216\lambda) + (36 + 144) = 4\lambda^2 + 8\lambda + 4 \)
\( \implies 97\lambda^2 + 264\lambda + 180 = 4\lambda^2 + 8\lambda + 4 \)
Move all terms to one side to form a quadratic equation:
\( \implies 97\lambda^2 - 4\lambda^2 + 264\lambda - 8\lambda + 180 - 4 = 0 \)
\( \implies 93\lambda^2 + 256\lambda + 176 = 0 \)
Factor this quadratic equation. We can find factors \( (3\lambda+4) \) and \( (31\lambda+44) \).
\( (3\lambda + 4)(31\lambda + 44) = 0 \)
This gives two possible values for \( \lambda \):
\( 3\lambda + 4 = 0 \implies 3\lambda = -4 \implies \lambda = -\frac{4}{3} \)
\( 31\lambda + 44 = 0 \implies 31\lambda = -44 \implies \lambda = -\frac{44}{31} \)
Since both values of \( \lambda \) are negative, the line segment is divided externally by the curve.
The required ratios are \( 4:3 \) and \( 44:31 \) externally.
In simple words: We find the coordinates of any point P on the line segment AB using the section formula, assuming a ratio \( \lambda : 1 \). Then we plug these coordinates into the equation of the curve \( x^2 + y^2 = 4 \). Solving the resulting quadratic equation for \( \lambda \) gives us the ratio(s) in which the curve divides the line segment. A negative \( \lambda \) means external division.

🎯 Exam Tip: When using the section formula to find a ratio, if the value of \( \lambda \) is negative, it indicates external division. Remember to carefully expand the squared binomials to avoid algebraic errors.

Question 18. AB is a line of fixed length, 6 units, joining the points A(t, 0) and B which lies on the positive y-axis. P is a point on AB distant 2 units from A. Express the co-ordinates of B and P in terms of t. Find the locus of P as t varies.
Answer: Let the coordinates of B be \( (0, k) \) since B lies on the positive y-axis. The coordinates of A are \( (t, 0) \).
The length of AB is 6 units, so \( AB^2 = 6^2 = 36 \).
Using the distance formula for AB:
\( (t - 0)^2 + (0 - k)^2 = 36 \)
\( \implies t^2 + k^2 = 36 \)
Since B lies on the positive y-axis, \( k \) must be positive:
\( k^2 = 36 - t^2 \implies k = \sqrt{36 - t^2} \)
So, the coordinates of B are \( (0, \sqrt{36 - t^2}) \).
Point P is on AB and is distant 2 units from A. This means \( AP = 2 \). Since \( AB = 6 \), then \( PB = AB - AP = 6 - 2 = 4 \).
Therefore, P divides the line segment AB in the ratio \( AP : PB = 2 : 4 = 1 : 2 \). Let \( P = (\alpha, \beta) \).
Using the section formula for P:

\( \alpha = \frac{1 \cdot 0 + 2 \cdot t}{1+2} = \frac{2t}{3} \)
\( \beta = \frac{1 \cdot \sqrt{36 - t^2} + 2 \cdot 0}{1+2} = \frac{\sqrt{36 - t^2}}{3} \)
Now we need to find the locus of P by eliminating \( t \).
From the expression for \( \alpha \):
\( 3\alpha = 2t \implies t = \frac{3\alpha}{2} \)
Substitute this \( t \) into the expression for \( \beta \):
\( \beta = \frac{\sqrt{36 - \left(\frac{3\alpha}{2}\right)^2}}{3} \)
\( 3\beta = \sqrt{36 - \frac{9\alpha^2}{4}} \)
Square both sides:
\( (3\beta)^2 = 36 - \frac{9\alpha^2}{4} \)
\( 9\beta^2 = 36 - \frac{9\alpha^2}{4} \)
Multiply the entire equation by 4 to clear the fraction:
\( 4(9\beta^2) = 4(36) - 4\left(\frac{9\alpha^2}{4}\right) \)
\( \implies 36\beta^2 = 144 - 9\alpha^2 \)
Rearrange the terms:
\( \implies 9\alpha^2 + 36\beta^2 = 144 \)
Divide the entire equation by 9:
\( \implies \alpha^2 + 4\beta^2 = 16 \)
Replacing \( \alpha \) with \( x \) and \( \beta \) with \( y \), the required locus of \( P(\alpha, \beta) \) is given by:
\( x^2 + 4y^2 = 16 \). This is the equation of an ellipse.
In simple words: We use the fixed length of the rod to find the y-coordinate of point B in terms of 't'. Then, we use the ratio in which P divides AB to find P's coordinates in terms of 't'. Finally, we eliminate 't' to get the equation of the path that P follows.

🎯 Exam Tip: Carefully determine the ratio in which P divides AB. Remember that if P is 2 units from A on a 6-unit line, it is 4 units from B, leading to a 1:2 ratio. Use substitution to eliminate the parameter \( t \) and find the locus.

Question 19. A rod of length l slides with its ends on two perpendicular lines. Find the locus of its mid-point.
Answer: Let the two perpendicular lines be the x-axis and the y-axis.
Let the rod AB have length \( l \). Suppose it intersects the x-axis at point \( A(a, 0) \) and the y-axis at point \( B(0, b) \).
Let \( P(h, k) \) be the midpoint of the rod AB, whose locus we need to find.
Using the midpoint formula:
\( h = \frac{a + 0}{2} \implies a = 2h \)
\( k = \frac{0 + b}{2} \implies b = 2k \)

Consider the right-angled triangle formed by the origin O, A, and B, i.e., \( \triangle AOB \).
By the Pythagorean theorem:
\( OA^2 + OB^2 = AB^2 \)
The length of OA is \( |a| \) and OB is \( |b| \). The length of AB is \( l \).
\( a^2 + b^2 = l^2 \)
Now substitute \( a = 2h \) and \( b = 2k \) into this equation:
\( (2h)^2 + (2k)^2 = l^2 \)
\( \implies 4h^2 + 4k^2 = l^2 \)
Replacing \( h \) with \( x \) and \( k \) with \( y \), the locus of \( P(h, k) \) is given by:
\( 4x^2 + 4y^2 = l^2 \)
This can be written as \( x^2 + y^2 = \frac{l^2}{4} \). This is the equation of a circle centered at the origin with radius \( \frac{l}{2} \).
In simple words: When a ladder slides against a wall and floor, its midpoint traces a circle. We call the ends of the rod (a,0) and (0,b), and its midpoint (h,k). Using the Pythagorean theorem and midpoint formula, we find the equation of the circle.

🎯 Exam Tip: When a rod slides between two perpendicular lines, its midpoint always traces a circle. Remembering this property helps you anticipate the final form of the locus equation. Applying the midpoint and distance formulas are key steps.

Question 20. If O is the origin and Q is a variable point on \( x^2 = 4y \), find the locus of the mid-point of OQ.
Answer: Let \( O \) be the origin \( (0, 0) \).
Let \( Q(h, k) \) be a variable point on the curve \( x^2 = 4y \).
Since Q lies on the curve, its coordinates must satisfy the equation:
\( h^2 = 4k \) ...(1)
Let \( P(\alpha, \beta) \) be the midpoint of the line segment OQ.
Using the midpoint formula for P:
\( \alpha = \frac{0 + h}{2} \implies h = 2\alpha \)
\( \beta = \frac{0 + k}{2} \implies k = 2\beta \)
Now, substitute the expressions for \( h \) and \( k \) (in terms of \( \alpha \) and \( \beta \)) into equation (1):
\( (2\alpha)^2 = 4(2\beta) \)
\( \implies 4\alpha^2 = 8\beta \)
Divide both sides by 4:
\( \implies \alpha^2 = 2\beta \)
Replacing \( \alpha \) with \( x \) and \( \beta \) with \( y \), the locus of the midpoint \( P(\alpha, \beta) \) is given by:
\( x^2 = 2y \). This is also the equation of a parabola.
In simple words: We take a moving point Q on a parabola and find the midpoint P between Q and the origin. By using the midpoint formula and substituting back into the original parabola's equation, we find the path that P traces, which is another parabola.

🎯 Exam Tip: This type of problem involves a parameter \( (h,k) \) for the variable point Q. Express the midpoint's coordinates in terms of this parameter, then eliminate the parameter using the equation of the original curve to find the locus.