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

Question 1. The lines x – 2y + 6 = 0 and 2x – y – 10 = 0 intersect at P. Without finding the co-ordinates of P prove that the equation of the line through P and the origin of co-ordinates is perpendicular to 39x + 33y – 580 = 0.
Answer: Let the given equations of the lines be:
\( x - 2y + 6 = 0 \) ...(1)
\( 2x - y - 10 = 0 \) ...(2)
Let the coordinates of the point of intersection be \( P(h, k) \). Since \( P(h, k) \) lies on both lines, we can write:
\( h - 2k = -6 \) ...(3)
\( 2h - k = 10 \) ...(4)
To find the ratio \( \frac{k}{h} \), divide equation (3) by equation (4):
\( \frac{h - 2k}{2h - k} = \frac{-6}{10} = \frac{-3}{5} \)
Now, cross-multiply:
\( 5(h - 2k) = -3(2h - k) \)
\( 5h - 10k = -6h + 3k \)
Gather the h and k terms:
\( 5h + 6h = 3k + 10k \)
\( 11h = 13k \)

\( \implies \frac{k}{h} = \frac{11}{13} \)
The slope of the line OP (joining P(h, k) and the origin (0, 0)) is \( m_1 = \frac{k - 0}{h - 0} = \frac{k}{h} \).
So, \( m_1 = \frac{11}{13} \).
Now, find the slope of the given line \( 39x + 33y - 580 = 0 \). The slope of a line \( Ax + By + C = 0 \) is \( -\frac{A}{B} \).
Let \( m_2 \) be this slope:
\( m_2 = \frac{-39}{33} = \frac{-13}{11} \)
To check if the lines are perpendicular, we multiply their slopes:
\( m_1 m_2 = \left(\frac{11}{13}\right) \times \left(\frac{-13}{11}\right) = -1 \)
Since the product of the slopes is -1, the line joining P to the origin is perpendicular to the line \( 39x + 33y - 580 = 0 \). This method saves time by avoiding the calculation of exact coordinates for P.
In simple words: First, we find the relationship between the x and y coordinates of point P where the two lines cross. Then, we find the slope of the line connecting P to the origin (0,0). After that, we find the slope of the other line given in the problem. If we multiply these two slopes and get -1, it means the lines are perpendicular.

🎯 Exam Tip: To prove lines are perpendicular, always show that the product of their slopes is -1. This often avoids needing to find exact intersection points, saving time.

Question 2. A point P moves so that its distance from the line given x = − 3 is equal to its distance from the point (3, 0). Show that the locus of P is y² = 12x.
Answer: Let the coordinates of point P be \( (h, k) \).
The given line is \( x = -3 \), which can be written as \( x + 3 = 0 \).
The distance of point \( (h, k) \) from the line \( Ax + By + C = 0 \) is given by \( \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \).
So, the distance of \( P(h, k) \) from \( x + 3 = 0 \) is \( \frac{|h + 3|}{\sqrt{1^2 + 0^2}} = |h + 3| \).
The distance of point \( P(h, k) \) from the fixed point \( (3, 0) \) is found using the distance formula:
\( \sqrt{(h - 3)^2 + (k - 0)^2} = \sqrt{(h - 3)^2 + k^2} \)
According to the problem statement, these two distances are equal:
\( |h + 3| = \sqrt{(h - 3)^2 + k^2} \)
To remove the square root and absolute value, square both sides:
\( (h + 3)^2 = (h - 3)^2 + k^2 \)
Expand both sides:
\( h^2 + 6h + 9 = h^2 - 6h + 9 + k^2 \)
Subtract \( h^2 \) and 9 from both sides:
\( 6h = -6h + k^2 \)
Move \( -6h \) to the left side:
\( 6h + 6h = k^2 \)
\( 12h = k^2 \)
Finally, replace \( (h, k) \) with \( (x, y) \) to get the locus:
The locus of P is \( y^2 = 12x \). This equation represents a parabola, a common type of locus.
In simple words: We imagine a point P moving so that it is always the same distance from a line and a fixed point. When we use the distance formulas and set them equal, then simplify the equation, we get \( y^2 = 12x \). This equation describes the path that point P traces.

🎯 Exam Tip: When finding a locus, always start by assuming the moving point has coordinates \( (h, k) \), apply the given conditions using distance formulas, and then replace \( (h, k) \) with \( (x, y) \) at the end.

Question 3. A (2, 5), B(4,- 11) are two fixed points and C is a point which moves on the line 3x + 4y + 5 = 0. Find the locus of the centroid of the triangle ABC.
Answer: Let the coordinates of the moving point C be \( (h, k) \).
Since C lies on the line \( 3x + 4y + 5 = 0 \), we can write:
\( 3h + 4k + 5 = 0 \) ...(1)
Let the centroid of triangle ABC be \( G(\alpha, \beta) \). The formula for the centroid of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is \( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \).
Using the given points A(2, 5), B(4, -11) and C(h, k):
\( \alpha = \frac{2 + 4 + h}{3} \)
\( \alpha = \frac{6 + h}{3} \)

\( \implies 3\alpha = 6 + h \)
\( \implies h = 3\alpha - 6 \) ...(2)
And for the y-coordinate:
\( \beta = \frac{5 - 11 + k}{3} \)
\( \beta = \frac{-6 + k}{3} \)

\( \implies 3\beta = -6 + k \)
\( \implies k = 3\beta + 6 \) ...(3)
Now, substitute the expressions for \( h \) and \( k \) from (2) and (3) into equation (1) since \( (h, k) \) satisfies that equation:
\( 3(3\alpha - 6) + 4(3\beta + 6) + 5 = 0 \)
Expand the terms:
\( 9\alpha - 18 + 12\beta + 24 + 5 = 0 \)
Combine the constant terms:
\( 9\alpha + 12\beta + 11 = 0 \)
Finally, replace \( (\alpha, \beta) \) with \( (x, y) \) to find the locus of the centroid:
The locus of the centroid is \( 9x + 12y + 11 = 0 \). This shows that the centroid also moves along a straight line.
In simple words: We have two fixed points and one point that moves along a line. We want to find the path (locus) of the center (centroid) of the triangle formed by these three points. By using the formulas for centroid and the line equation, we find that the centroid also follows a straight line.

🎯 Exam Tip: For locus problems involving a centroid, express the coordinates of the moving point (h, k) in terms of the centroid's coordinates (α, β), then substitute these into the equation of the line on which the point (h, k) moves.

Question 4. Find the cartesian equation of the curve whose parametric equations are :
(i) x = t, y = 3 t + 5;
(ii) x = t, y = t²;
(iii) x = 4 cos θ, y = 4 sin θ;
(iv) x = 4 cos θ; y = 3 sin θ.
Answer:
(i) Given parametric equations are:
\( x = t \) ...(1)
\( y = 3t + 5 \) ...(2)
To find the Cartesian equation, we eliminate the parameter \( t \). Substitute (1) into (2):
\( y = 3x + 5 \)
This is the required locus, which represents a straight line.

(ii) Given parametric equations are:
\( x = t \) ...(1)
\( y = t^2 \) ...(2)
To find the Cartesian equation, we eliminate the parameter \( t \). Substitute (1) into (2):
\( y = x^2 \)
This is the required locus, which represents a parabola.

(iii) Given parametric equations are:
\( x = 4 \cos \theta \)
\( \implies \frac{x}{4} = \cos \theta \) ...(1)
\( y = 4 \sin \theta \)
\( \implies \frac{y}{4} = \sin \theta \) ...(2)
To eliminate \( \theta \), we use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \). Square and add equations (1) and (2):
\( \left(\frac{x}{4}\right)^2 + \left(\frac{y}{4}\right)^2 = \cos^2 \theta + \sin^2 \theta \)
\( \frac{x^2}{16} + \frac{y^2}{16} = 1 \)
Multiply by 16:
\( x^2 + y^2 = 16 \)
This is the required locus, which represents a circle centered at the origin with radius 4.

(iv) Given parametric equations are:
\( x = 4 \cos \theta \)
\( \implies \frac{x}{4} = \cos \theta \) ...(1)
\( y = 3 \sin \theta \)
\( \implies \frac{y}{3} = \sin \theta \) ...(2)
To eliminate \( \theta \), we use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \). Square and add equations (1) and (2):
\( \left(\frac{x}{4}\right)^2 + \left(\frac{y}{3}\right)^2 = \cos^2 \theta + \sin^2 \theta \)
\( \frac{x^2}{16} + \frac{y^2}{9} = 1 \)
This is the required locus, which represents an ellipse. The process of eliminating the parameter allows us to see the shape of the curve.
In simple words: For each pair of equations, we want to get rid of the "t" or "θ" to find an equation that only has "x" and "y". This new equation shows the actual path or shape of the curve. For parts with sine and cosine, we use the fact that \( \sin^2 \theta + \cos^2 \theta = 1 \) to combine them.

🎯 Exam Tip: The key to converting parametric equations to Cartesian form is to eliminate the parameter. Look for direct substitution opportunities or trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 5. Find the locus of the point of intersection of the lines x = \frac{a}{m^2} and y = \frac{2a}{m}, where m is a parameter.
Answer: Given equations of lines are:
\( x = \frac{a}{m^2} \) ...(1)
\( y = \frac{2a}{m} \) ...(2)
To find the locus of the point of intersection, we need to eliminate the parameter \( m \) from equations (1) and (2).
From equation (2), we can express \( m \) in terms of \( y \):
\( m = \frac{2a}{y} \)
Now, substitute this expression for \( m \) into equation (1):
\( x = \frac{a}{\left(\frac{2a}{y}\right)^2} \)
Simplify the denominator:
\( x = \frac{a}{\frac{4a^2}{y^2}} \)
Invert and multiply:
\( x = a \times \frac{y^2}{4a^2} \)
\( x = \frac{y^2}{4a} \)
Rearrange the terms to get the standard form of the locus:
\( 4ax = y^2 \)
\( y^2 = 4ax \)
This is the required locus, which represents a parabola. The elimination of the parameter reveals the underlying Cartesian equation of the curve.
In simple words: We have two equations for x and y that depend on a changing value 'm'. To find the actual path (locus) these points create, we need to get rid of 'm'. We solve one equation for 'm' and then put that into the other equation. After simplifying, we get the final equation relating only x and y.

🎯 Exam Tip: When eliminating a parameter from two equations, it's often easiest to solve one equation for the parameter and substitute that expression into the other equation. Simplify carefully to arrive at the Cartesian form.

Question 6. Find the intersection S of the lines x – ty + t² = 0, tx + y – t³ – 2t = 0 Show that S lies on the curve whose equation is y² = 4x. Sketch this curve.
Answer: Given equations of the lines are:
\( x - ty + t^2 = 0 \) ...(1)
\( tx + y - t^3 - 2t = 0 \) ...(2)
To find the point of intersection \( S(x, y) \), we need to solve these two equations simultaneously for \( x \) and \( y \).
From (1), \( y = \frac{x + t^2}{t} \). Substitute this into (2):
\( tx + \frac{x + t^2}{t} - t^3 - 2t = 0 \)
Multiply the entire equation by \( t \) to clear the denominator:
\( t^2x + x + t^2 - t^4 - 2t^2 = 0 \)
Group the \( x \) terms and constant terms:
\( (t^2 + 1)x + t^2 - t^4 - 2t^2 = 0 \)
\( (t^2 + 1)x - t^4 - t^2 = 0 \)
\( (t^2 + 1)x = t^4 + t^2 \)
Factor out \( t^2 \) on the right side:
\( (t^2 + 1)x = t^2(t^2 + 1) \)
Since \( t^2 + 1 \) cannot be zero, we can divide both sides by \( t^2 + 1 \):
\( x = t^2 \)
Now substitute \( x = t^2 \) back into equation (1) to find \( y \):
\( t^2 - ty + t^2 = 0 \)
\( 2t^2 - ty = 0 \)
\( ty = 2t^2 \)
If \( t \ne 0 \), we can divide by \( t \):
\( y = 2t \)
So, the point of intersection \( S \) is \( (t^2, 2t) \).
Let \( S \) be a point \( (h, k) \) on the locus. Then we have:
\( h = t^2 \) ...(3)
\( k = 2t \) ...(4)
To find the locus, we eliminate the parameter \( t \). From (4), \( t = \frac{k}{2} \). Substitute this into (3):
\( h = \left(\frac{k}{2}\right)^2 \)
\( h = \frac{k^2}{4} \)

\( \implies k^2 = 4h \)
Replace \( (h, k) \) with \( (x, y) \):
The locus of S is \( y^2 = 4x \). This equation confirms that S lies on the specified curve, which is a parabola opening to the right, with its vertex at the origin. Parabolic shapes are common in physics, such as the trajectory of a projectile.
In simple words: First, we solve the two given line equations to find the exact point where they meet. This point's x and y coordinates will depend on 't'. Next, we remove 't' from these coordinates to get a single equation relating x and y. This equation, \( y^2 = 4x \), tells us the path that the intersection point follows. Finally, we draw this path, which is a parabola.

🎯 Exam Tip: When sketching a locus, identify the type of curve (e.g., parabola, circle, ellipse). Plot a few key points, such as the vertex or intercepts, to ensure accuracy, and always label the axes and the equation of the curve.

Question 7. Find the locus of the middle point of the portion of the line x cos a + y sin a = p, where p is a constant, intercepted between the axes.
Answer: The equation of the given line is:
\( x \cos \alpha + y \sin \alpha = p \) ...(1)
To find the points where the line intercepts the axes:
When the line meets the x-axis, \( y = 0 \). Substitute \( y = 0 \) into (1):
\( x \cos \alpha = p \)
\( x = \frac{p}{\cos \alpha} \)
So, the x-intercept is \( A\left(\frac{p}{\cos \alpha}, 0\right) \).
When the line meets the y-axis, \( x = 0 \). Substitute \( x = 0 \) into (1):
\( y \sin \alpha = p \)
\( y = \frac{p}{\sin \alpha} \)
So, the y-intercept is \( B\left(0, \frac{p}{\sin \alpha}\right) \).
Let \( P(h, k) \) be the middle point of the portion of the line AB intercepted between the axes. Using the midpoint formula:
\( h = \frac{\frac{p}{\cos \alpha} + 0}{2} = \frac{p}{2 \cos \alpha} \)

\( \implies \cos \alpha = \frac{p}{2h} \) ...(2)
And
\( k = \frac{0 + \frac{p}{\sin \alpha}}{2} = \frac{p}{2 \sin \alpha} \)

\( \implies \sin \alpha = \frac{p}{2k} \) ...(3)
To find the locus, we eliminate \( \alpha \) using the trigonometric identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \). Square equations (2) and (3) and add them:
\( \left(\frac{p}{2h}\right)^2 + \left(\frac{p}{2k}\right)^2 = \cos^2 \alpha + \sin^2 \alpha \)
\( \frac{p^2}{4h^2} + \frac{p^2}{4k^2} = 1 \)
Factor out \( p^2 \) from the left side:
\( p^2 \left(\frac{1}{4h^2} + \frac{1}{4k^2}\right) = 1 \)
\( p^2 \left(\frac{k^2 + h^2}{4h^2k^2}\right) = 1 \)
Multiply both sides by \( 4h^2k^2 \):
\( p^2(h^2 + k^2) = 4h^2k^2 \)
Replace \( (h, k) \) with \( (x, y) \) to find the locus:
The required locus of P is \( p^2(x^2 + y^2) = 4x^2y^2 \). This equation describes the path of the midpoint as the line changes its angle but keeps its distance from the origin constant.
In simple words: Imagine a line that cuts through the x and y axes. This line is always the same distance 'p' from the center point (origin). We want to find the path of the exact middle point of the line segment that is between the axes. By using geometry and a bit of algebra to remove the angle 'α', we find a special equation relating x and y that shows the path this midpoint takes.

🎯 Exam Tip: For locus problems involving midpoints, first find the coordinates of the endpoints of the segment, then use the midpoint formula. Remember the relationship between intercepts and the general form of a line.