Chapter 9 — Straight Lines

Theory

The slope–intercept form of a straight line is given by \(y = mx + c\), where \(m\) represents the slope of the line and \(c\) represents the y-intercept (the point where the line cuts the y-axis). To convert any linear equation into this form, we isolate \(y\) on one side. The coefficient of \(x\) gives the slope, while the constant term gives the intercept.

Solution Roadmap

• Rearrange the equation to isolate \(y\).
• Express the equation in the form \(y = mx + c\).
• Identify slope \(m\) and y-intercept \(c\).
• Interpret the nature of the line (increasing, decreasing, horizontal).

Solution

For the first equation, start with \(x+7y=0\). To express it in slope–intercept form \(y=mx+c\), isolate \(y\). \[ \begin{aligned} x+7y=0\\ 7y=-x\\ y=-\frac{1}{7}x+0\\ m=-\frac{1}{7},\quad c=0 \end{aligned} \] The slope is negative, so the line is decreasing and passes through the origin.

For the second equation, consider \(6x+3y-5=0\). \[ \begin{aligned} 6x+3y-5=0\\ 3y=-6x+5\\ y=-2x+\frac{5}{3}\\ m=-2,\quad c=\frac{5}{3} \end{aligned} \] The slope is steep and negative, and the line intersects the y-axis above the origin.

For the third equation, \(y=0\) already represents a horizontal line. \[ \begin{aligned} y=0\cdot x+0\\ m=0,\quad c=0 \end{aligned} \] This is the x-axis itself.

Significance

This question builds the foundational skill of converting general linear equations into slope–intercept form, which is crucial for graphical interpretation and analytical geometry. In board examinations, it directly tests conceptual clarity and algebraic manipulation. For competitive exams like JEE and NDA, this concept is frequently used in coordinate geometry problems involving angle between lines, parallelism, perpendicularity, and graph-based reasoning.

Theory

The intercept form of a straight line is given by \( \frac{x}{a} + \frac{y}{b} = 1 \), where \(a\) and \(b\) are the intercepts on the \(x\)-axis and \(y\)-axis respectively. The value of \(a\) is obtained by putting \(y=0\), and \(b\) is obtained by putting \(x=0\). This form is particularly useful for quickly sketching graphs and understanding how a line interacts with coordinate axes.

Solution Roadmap

• Rearrange the equation so that constant term is on the right side.
• Divide the entire equation by the constant to make RHS equal to \(1\).
• Convert into the form \( \frac{x}{a} + \frac{y}{b} = 1 \).
• Identify intercepts \(a\) and \(b\).
• Check special cases like horizontal or vertical lines.

Solution

For the first equation, start with \(3x+2y-12=0\). \[ \begin{aligned} 3x+2y-12=0\\ 3x+2y=12\\ \frac{3x}{12}+\frac{2y}{12}=1\\ \frac{x}{4}+\frac{y}{6}=1\\ a=4,\quad b=6 \end{aligned} \] Thus, the intercepts are \( (4,0) \) and \( (0,6) \).

For the second equation, consider \(4x-3y=6\). \[ \begin{aligned} 4x-3y=6\\ \frac{4x}{6}-\frac{3y}{6}=1\\ \frac{2}{3}x-\frac{1}{2}y=1\\ \frac{x}{3/2}+\frac{y}{-2}=1\\ a=\frac{3}{2},\quad b=-2 \end{aligned} \] Hence, the intercepts are \( \left(\frac{3}{2},0\right) \) and \( (0,-2) \).

For the third equation, take \(3y+2=0\). \[ \begin{aligned} 3y+2=0\\ y=-\frac{2}{3} \end{aligned} \] This represents a horizontal line parallel to the \(x\)-axis. It cuts the \(y\)-axis at \( (0,-\frac{2}{3}) \) and does not have a finite \(x\)-intercept, hence it cannot be expressed in intercept form.

Significance

This problem strengthens the understanding of intercept form, which is essential for graphical interpretation of straight lines. In board exams, it helps in quick plotting and conceptual clarity. In competitive exams like JEE, intercept concepts are frequently used in problems involving area of triangles, region analysis, and coordinate geometry transformations. Recognizing special cases such as horizontal lines also prevents common mistakes.

Theory

The perpendicular distance of a point \((x_1, y_1)\) from a line \(Ax + By + C = 0\) is given by \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] This formula represents the shortest distance from the point to the line, measured along the perpendicular direction. It is essential to first convert the given equation into standard form.

Solution Roadmap

• Convert the given equation into standard form \(Ax + By + C = 0\).
• Identify coefficients \(A\), \(B\), and \(C\).
• Substitute the point coordinates into the distance formula.
• Simplify carefully to obtain the final distance.

Solution

Begin by rewriting the given line in the standard form: \[ \begin{aligned} 12(x+6)&=5(y-2)\\ 12x+72&=5y-10\\ 12x-5y+82&=0 \end{aligned} \] Thus, \(A=12\), \(B=-5\), and \(C=82\).

Using the distance formula for the point \((-1,1)\): \[ \begin{aligned} d&=\frac{|12(-1)+(-5)(1)+82|}{\sqrt{12^2+(-5)^2}}\\ &=\frac{|-12-5+82|}{\sqrt{144+25}}\\ &=\frac{65}{\sqrt{169}}\\ &=\frac{65}{13}\\ &=5 \end{aligned} \]

Therefore, the distance of the point \((-1,1)\) from the given line is \(5\) units.

Significance

This concept is fundamental in coordinate geometry as it provides the shortest distance between a point and a line. In board examinations, it is a direct application-based question with high scoring potential. For competitive exams like JEE, it forms the basis for advanced topics such as distance between parallel lines, shortest distance problems, and optimization in geometry. Accuracy in algebraic simplification is critical to avoid calculation errors.

Theory

Any point on the \(x\)-axis has coordinates of the form \((a,0)\). The perpendicular distance of a point from a line \(Ax+By+C=0\) is given by \[ d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \] Since distance is always non-negative, solving such equations involves considering both positive and negative cases after removing the modulus.

Solution Roadmap

• Assume a general point on the \(x\)-axis as \((a,0)\).
• Convert the given line into standard form.
• Apply the distance formula.
• Solve the resulting modulus equation.
• Obtain all possible points.

Solution

Let the required point on the \(x\)-axis be \((a,0)\). Convert the given line into standard form: \[ \begin{aligned} \frac{x}{3}+\frac{y}{4}=1\\ 4x+3y-12=0 \end{aligned} \]

Using the distance formula: \[ \begin{aligned} 4=\frac{|4a+3(0)-12|}{\sqrt{4^2+3^2}}\\ 4=\frac{|4a-12|}{5} \end{aligned} \]

Multiply both sides: \[ |4a-12|=20 \] This gives two cases: \[ \begin{aligned} 4a-12=20 \Rightarrow a=8\\ 4a-12=-20 \Rightarrow a=-2 \end{aligned} \]

Hence, the required points are \((8,0)\) and \((-2,0)\).

Significance

This problem integrates the concept of intercept form with the distance formula, making it highly important for both board exams and competitive exams. It develops the ability to handle modulus equations and interpret geometric constraints algebraically. In exams like JEE, such problems often appear in transformed forms involving loci and distance conditions, making this a foundational application.

Theory

The distance between two parallel lines of the form \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\) is given by \[ d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} \] This formula is valid because parallel lines have identical normal vectors \((a, b)\), ensuring the shortest distance between them is constant everywhere.

Solution Roadmap

• Verify that both lines are parallel (same coefficients of \(x\) and \(y\)).
• Identify \(a\), \(b\), \(c_1\), and \(c_2\).
• Apply the distance formula directly.
• Simplify carefully.

Solution

In the first case, the lines are: \[ 15x + 8y - 34 = 0 \quad \text{and} \quad 15x + 8y + 31 = 0 \] Here, \(a = 15\), \(b = 8\), \(c_1 = -34\), \(c_2 = 31\). \[ \begin{aligned} d &= \frac{|31 - (-34)|}{\sqrt{15^2 + 8^2}}\\ &= \frac{65}{\sqrt{225 + 64}}\\ &= \frac{65}{\sqrt{289}}\\ &= \frac{65}{17} \end{aligned} \]

In the second case, the lines are: \[ l(x+y)+p=0 \quad \text{and} \quad l(x+y)-r=0 \] Rewriting: \[ lx + ly + p = 0 \quad \text{and} \quad lx + ly - r = 0 \] Here, \(a = l\), \(b = l\), \(c_1 = p\), \(c_2 = -r\). \[ \begin{aligned} d &= \frac{|-r - p|}{\sqrt{l^2 + l^2}}\\ &= \frac{|p + r|}{\sqrt{2l^2}}\\ &= \frac{|p + r|}{l\sqrt{2}} \end{aligned} \]

Significance

This problem is a direct application of the distance between parallel lines, a high-frequency concept in both board and competitive examinations. It is particularly important for solving problems involving strips, regions between lines, and shortest distance calculations. In JEE-level problems, this concept is often extended to 3D geometry and optimization problems, making it a critical foundational tool.

Theory

Two lines are parallel if they have the same slope. For a line in standard form \(Ax + By + C = 0\), the slope is given by \(m = -\frac{A}{B}\). Alternatively, we can convert the equation into slope–intercept form \(y = mx + c\) to directly identify the slope. A line parallel to a given line will have the same slope but a different constant term.

Solution Roadmap

• Convert the given line into slope–intercept form to find slope.
• Use point–slope form with the given point.
• Simplify the equation into standard form.

Solution

First rewrite the given line: \[ \begin{aligned} 3x - 4y + 2 = 0\\ 4y = 3x + 2\\ y = \frac{3}{4}x + \frac{1}{2}\\ m = \frac{3}{4} \end{aligned} \]

Since the required line is parallel, it has the same slope \(m = \frac{3}{4}\). Using point–slope form with point \((-2, 3)\): \[ \begin{aligned} y - 3 = \frac{3}{4}(x + 2) \end{aligned} \]

Simplifying: \[ \begin{aligned} 4y - 12 = 3x + 6\\ 4y = 3x + 18\\ 3x - 4y + 18 = 0 \end{aligned} \]

Hence, the required equation is \(3x - 4y + 18 = 0\).

Significance

This problem tests the concept of parallel lines and point–slope form, both of which are fundamental in coordinate geometry. In board exams, it is a direct and scoring question. In competitive exams like JEE, this concept is frequently used in problems involving families of lines, distance between lines, and geometric constraints. Strong understanding ensures faster problem solving and fewer algebraic errors.

Theory

For a line in the form \(Ax + By + C = 0\), the slope is \(m = -\frac{A}{B}\). If two lines are perpendicular, their slopes satisfy the relation \(m_1 \cdot m_2 = -1\). Also, a line with \(x\)-intercept \(a\) passes through the point \((a,0)\).

Solution Roadmap

• Convert the given line into slope–intercept form to find its slope.
• Use perpendicular condition to find the required slope.
• Use the given \(x\)-intercept to get a point.
• Apply point–slope form and simplify.

Solution

First rewrite the given line: \[ \begin{aligned} x - 7y + 5 = 0\\ -7y = -x - 5\\ y = \frac{1}{7}x + \frac{5}{7}\\ m_1 = \frac{1}{7} \end{aligned} \]

For a perpendicular line: \[ m_1 \cdot m_2 = -1 \Rightarrow \frac{1}{7} \cdot m_2 = -1 \Rightarrow m_2 = -7 \]

The required line has \(x\)-intercept \(3\), so it passes through \((3,0)\). Using point–slope form: \[ \begin{aligned} y - 0 = -7(x - 3)\\ y = -7x + 21\\ 7x + y - 21 = 0 \end{aligned} \]

Hence, the required equation is \(7x + y - 21 = 0\).

Significance

This problem combines the concepts of slope, perpendicularity, and intercepts. It is a standard question in board exams and forms the base for more advanced coordinate geometry problems in competitive exams like JEE. Mastery of such problems is essential for tackling questions involving orthogonal lines, shortest distance, and geometric constructions.

Theory

The angle \(\theta\) between two lines having slopes \(m_1\) and \(m_2\) is given by \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| \] The acute angle is usually required unless specified otherwise. For perpendicular lines, \(m_1 m_2 = -1\), and for parallel lines, \(m_1 = m_2\).

Solution Roadmap

• Convert both equations into slope–intercept form.
• Identify slopes \(m_1\) and \(m_2\).
• Substitute into the angle formula.
• Simplify carefully and determine the angle.

Solution

Convert the first equation: \[ \begin{aligned} \sqrt{3}x+y=1\\ y=-\sqrt{3}x+1\\ m_1=-\sqrt{3} \end{aligned} \]

Convert the second equation: \[ \begin{aligned} x+\sqrt{3}y=1\\ \sqrt{3}y=1-x\\ y=-\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}\\ m_2=-\frac{1}{\sqrt{3}} \end{aligned} \]

Using the formula: \[ \begin{aligned} \tan\theta&=\left|\frac{-\sqrt{3}-\left(-\frac{1}{\sqrt{3}}\right)}{1+\left(-\sqrt{3}\right)\left(-\frac{1}{\sqrt{3}}\right)}\right|\\ &=\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{2}\right|\\ &=\left|\frac{-\frac{2}{\sqrt{3}}}{2}\right|\\ &=\frac{1}{\sqrt{3}} \end{aligned} \]

Therefore, \(\theta = 30^\circ\). The other angle between the lines is \(180^\circ - 30^\circ = 150^\circ\).

Significance

This problem tests the concept of angle between two lines, a core topic in coordinate geometry. It is frequently asked in board exams and is highly relevant for competitive exams like JEE, where it is used in problems involving perpendicularity, angle bisectors, and pair of lines. Strong command over slope interpretation and algebraic simplification is essential for speed and accuracy.

Theory

The slope of a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2-y_1}{x_2-x_1}\). If two lines are perpendicular, their slopes satisfy the condition \(m_1 \cdot m_2 = -1\). This condition is central to solving problems involving right angles in coordinate geometry.

Solution Roadmap

• Find the slope of the line passing through \((h,3)\) and \((4,1)\).
• Find the slope of the given line.
• Apply perpendicular condition \(m_1 m_2 = -1\).
• Solve for \(h\).

Solution

The slope of the line through \((h,3)\) and \((4,1)\) is: \[ \begin{aligned} m_1=\frac{1-3}{4-h}=\frac{-2}{4-h} \end{aligned} \]

Rewrite the given line: \[ \begin{aligned} 7x-9y-19=0\\ -9y=-7x+19\\ y=\frac{7}{9}x-\frac{19}{9}\\ m_2=\frac{7}{9} \end{aligned} \]

Since the lines are perpendicular: \[ \begin{aligned} m_1 m_2=-1\\ \frac{-2}{4-h}\cdot\frac{7}{9}=-1\\ \frac{-14}{9(4-h)}=-1 \end{aligned} \]

Solving: \[ \begin{aligned} -14=-9(4-h)\\ -14=-36+9h\\ 9h=22\\ h=\frac{22}{9} \end{aligned} \]

Therefore, the required value of \(h\) is \(\frac{22}{9}\).

Significance

This problem integrates slope calculation and perpendicularity conditions, both of which are fundamental in coordinate geometry. It is commonly tested in board exams and forms a base for advanced problems in competitive exams like JEE, especially in topics involving orthogonal lines, projections, and geometric constraints. Careful algebraic handling is essential to avoid sign errors.

Theory

A line in the form \(Ax + By + C = 0\) has slope \(m = -\frac{A}{B}\). Any line parallel to it must have the same slope. The point–slope form of a line passing through \((x_1, y_1)\) with slope \(m\) is \(y - y_1 = m(x - x_1)\). Using these two ideas, we can derive the required equation.

Solution Roadmap

• Find the slope of the given line.
• Use the fact that parallel lines have equal slopes.
• Apply point–slope form using the given point.
• Rearrange to obtain the required form.

Solution

Consider the given line: \[ \begin{aligned} Ax + By + C = 0\\ By = -Ax - C\\ y = -\frac{A}{B}x - \frac{C}{B} \end{aligned} \] Hence, the slope of the given line is \(m = -\frac{A}{B}\).

A line parallel to this line will have the same slope. Let the required line pass through \((x_1, y_1)\). Using point–slope form: \[ \begin{aligned} y - y_1 = -\frac{A}{B}(x - x_1) \end{aligned} \]

Multiply both sides by \(B\): \[ \begin{aligned} B(y - y_1) = -A(x - x_1) \end{aligned} \]

Rearranging: \[ \begin{aligned} A(x - x_1) + B(y - y_1) = 0 \end{aligned} \]

Hence proved.

Significance

This is a fundamental derivation in coordinate geometry that is frequently used in both board exams and competitive exams. It provides a direct formula for writing equations of parallel lines, eliminating the need for repeated slope calculations. In advanced problems, this concept is used in families of lines, distance problems, and geometric proofs, making it an essential result to remember.

Theory

The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| \] For a given angle, this relation may yield two possible slopes, corresponding to the two possible orientations of lines making that angle.

Solution Roadmap

• Use angle formula with \(m_1=2\) and \(\theta=60^\circ\).
• Solve for slope \(m\) (consider both cases).
• Use point–slope form with point \((2,3)\).
• Write final equation(s).

Solution

Given \(m_1 = 2\) and \(\theta = 60^\circ\), we use \[ \tan 60^\circ = \sqrt{3} = \left|\frac{2 - m}{1 + 2m}\right| \]

Removing modulus gives two cases:

Case 1: \[ \begin{aligned} \sqrt{3} = \frac{2 - m}{1 + 2m}\\ \sqrt{3}(1 + 2m) = 2 - m\\ \sqrt{3} + 2\sqrt{3}m = 2 - m\\ m(1 + 2\sqrt{3}) = 2 - \sqrt{3}\\ m = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \end{aligned} \]

Case 2: \[ \begin{aligned} -\sqrt{3} = \frac{2 - m}{1 + 2m}\\ -\sqrt{3}(1 + 2m) = 2 - m\\ -\sqrt{3} - 2\sqrt{3}m = 2 - m\\ m(1 - 2\sqrt{3}) = 2 + \sqrt{3}\\ m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \end{aligned} \]

Thus, two possible slopes exist.

Using point–slope form for point \((2,3)\):

First line: \[ y - 3 = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}(x - 2) \]

Second line: \[ y - 3 = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}(x - 2) \]

These are the required equations of the lines making an angle of \(60^\circ\) with the given line.

Significance

This problem is highly important as it introduces the concept of multiple possible solutions when angle between lines is given. It strengthens algebraic manipulation and conceptual clarity. In board exams, it tests understanding of angle formula, while in competitive exams like JEE, such problems appear in advanced coordinate geometry involving pair of lines, angle bisectors, and locus problems.

Theory

The right (perpendicular) bisector of a line segment is a line that passes through the midpoint of the segment and is perpendicular to it. The midpoint of two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\), and the slope of a perpendicular line is the negative reciprocal of the given slope.

Solution Roadmap

• Find the midpoint of the given line segment.
• Calculate the slope of the line joining the points.
• Find the perpendicular slope.
• Use point–slope form to obtain the equation.

Solution

Midpoint of the segment: \[ \begin{aligned} x_1=\frac{3+(-1)}{2}=1,\\ y_1=\frac{4+2}{2}=3 \end{aligned} \] So, midpoint is \((1,3)\).

Slope of the line joining the points: \[ \begin{aligned} m_1=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2} \end{aligned} \]

Slope of the perpendicular bisector: \[ m_2=-\frac{1}{m_1}=-2 \]

Equation using point–slope form: \[ \begin{aligned} y-3=-2(x-1)\\ y-3=-2x+2\\ y+2x-5=0 \end{aligned} \]

Hence, the required equation is \(y+2x-5=0\).

Significance

This problem is fundamental in coordinate geometry as it combines midpoint, slope, and perpendicular concepts. It is frequently asked in board exams and is highly relevant for competitive exams like JEE, where perpendicular bisectors are used in locus problems, triangle geometry, and coordinate proofs. Mastery of this concept is essential for solving advanced geometric constructions analytically.

Theory

The foot of the perpendicular from a point to a line is the point where the perpendicular from the given point meets the line. To find it, we form the equation of the perpendicular line through the given point and then solve it simultaneously with the given line. The slope of a perpendicular line is the negative reciprocal of the given slope.

Solution Roadmap

• Find slope of the given line.
• Determine slope of perpendicular line.
• Form equation of perpendicular through given point.
• Solve both equations to get intersection point.

Solution

Rewrite the given line: \[ \begin{aligned} 3x-4y-16=0\\ -4y=-3x+16\\ y=\frac{3}{4}x-4\\ m_1=\frac{3}{4} \end{aligned} \]

Slope of perpendicular: \[ m_2=-\frac{4}{3} \]

Equation of perpendicular through \((-1,3)\): \[ \begin{aligned} y-3=-\frac{4}{3}(x+1)\\ 3y-9=-4x-4\\ 4x+3y-5=0 \end{aligned} \]

Solve with given line: \[ \begin{aligned} 3x-4y-16=0\\ 4x+3y-5=0 \end{aligned} \]

Eliminate \(y\): \[ \begin{aligned} 9x-12y-48=0\\ 16x+12y-20=0\\ 25x-68=0\\ x=\frac{68}{25} \end{aligned} \]

Substitute back: \[ \begin{aligned} 4\left(\frac{68}{25}\right)+3y-5=0\\ \frac{272}{25}+3y=\frac{125}{25}\\ 3y=-\frac{147}{25}\\ y=-\frac{49}{25} \end{aligned} \]

Hence, the foot of the perpendicular is \(\left(\frac{68}{25},-\frac{49}{25}\right)\).

Significance

This problem is fundamental for understanding projections and shortest distance in coordinate geometry. It is frequently asked in board exams and forms a base for advanced problems in competitive exams like JEE, especially in topics like distance of a point from a line, reflection, and optimization. Accuracy in solving simultaneous equations is crucial.

Theory

If a line is perpendicular to another, the product of their slopes is \(-1\). Also, the slope of a line joining two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{y_2-y_1}{x_2-x_1}\). Since the foot of the perpendicular from the origin lies on the given line, that point satisfies the equation of the line.

Solution Roadmap

• Find slope of perpendicular from origin using given point.
• Use perpendicular condition to find slope \(m\).
• Substitute given point into line equation to find \(c\).

Solution

The perpendicular is drawn from the origin \((0,0)\) to the point \((-1,2)\). So, slope of the perpendicular: \[ \begin{aligned} m_2=\frac{2-0}{-1-0}=-2 \end{aligned} \]

If the given line has slope \(m\), then: \[ \begin{aligned} m \cdot m_2=-1\\ m(-2)=-1\\ m=\frac{1}{2} \end{aligned} \]

So the equation becomes: \[ y=\frac{1}{2}x+c \]

Since \((-1,2)\) lies on the line: \[ \begin{aligned} 2=\frac{1}{2}(-1)+c\\ 2=-\frac{1}{2}+c\\ c=\frac{5}{2} \end{aligned} \]

Hence, \(m=\frac{1}{2}\) and \(c=\frac{5}{2}\).

Significance

This problem connects perpendicularity with coordinate geometry and reinforces the concept of slope relations. It is commonly asked in board exams and is important for competitive exams like JEE, where such ideas are extended to projections, shortest distance problems, and coordinate transformations. Understanding geometric meaning behind algebraic steps is key here.

Theory

The perpendicular distance from the origin to a line \(ax + by + c = 0\) is given by \[ d=\frac{|c|}{\sqrt{a^2+b^2}} \] Also, the trigonometric identities \(\sin^2\theta+\cos^2\theta=1\) and \(\sin 2\theta = 2\sin\theta\cos\theta\) are essential for simplification in such problems.

Solution Roadmap

• Convert both equations into standard form.
• Apply distance formula from origin.
• Simplify using trigonometric identities.
• Combine expressions to prove the result.

Solution

First line: \[ x \cos \theta - y \sin \theta - k \cos 2\theta = 0 \] Distance from origin: \[ \begin{aligned} p&=\frac{| -k \cos 2\theta |}{\sqrt{\cos^2\theta+\sin^2\theta}}\\ &=|k\cos 2\theta| \end{aligned} \] Hence, \[ p^2 = k^2 \cos^2 2\theta \]

Second line: \[ x \sec \theta + y \;\text{cosec} \theta - k = 0 \] Distance from origin: \[ \begin{aligned} q&=\frac{| -k |}{\sqrt{\sec^2\theta+\text{cosec}^2\theta}} \end{aligned} \]

Simplify denominator: \[ \begin{aligned} \sec^2\theta+\text{cosec}^2\theta &=\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}\\ &=\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}\\ &=\frac{1}{\sin^2\theta\cos^2\theta} \end{aligned} \]

Thus, \[ \begin{aligned} q&=|k|\sin\theta\cos\theta\\ &=\frac{|k|}{2}\sin 2\theta \end{aligned} \] Hence, \[ q^2=\frac{k^2}{4}\sin^2 2\theta \]

Now, \[ \begin{aligned} p^2+4q^2 &=k^2\cos^2 2\theta+4\cdot\frac{k^2}{4}\sin^2 2\theta\\ &=k^2(\cos^2 2\theta+\sin^2 2\theta)\\ &=k^2 \end{aligned} \]

Hence proved that \(p^2 + 4q^2 = k^2\).

Significance

This is a high-quality conceptual problem combining coordinate geometry with trigonometry. It is important for board exams due to its proof-based nature and for competitive exams like JEE because it tests algebraic manipulation along with identity application. Such problems build strong problem-solving intuition and are frequently used in advanced geometry and transformation-based questions.

Theory

An altitude of a triangle is a line drawn from a vertex perpendicular to the opposite side. To find its equation, we first determine the slope of the opposite side and then use the negative reciprocal to get the slope of the altitude. The length of the altitude is the perpendicular distance from the vertex to the opposite side.

Solution Roadmap

• Find slope of side \(BC\).
• Determine slope of altitude (perpendicular).
• Form equation using point–slope form through \(A\).
• Find equation of \(BC\).
• Use distance formula to find altitude length.

Solution

Slope of \(BC\): \[ \begin{aligned} m_{BC}=\frac{2-(-1)}{1-4}=\frac{3}{-3}=-1 \end{aligned} \]

Slope of altitude: \[ m = 1 \]

Equation of altitude through \(A(2,3)\): \[ \begin{aligned} y-3=1(x-2)\\ y=x+1 \end{aligned} \]

Equation of line \(BC\): \[ \begin{aligned} y+1=-1(x-4)\\ y=-x+3\\ x+y-3=0 \end{aligned} \]

Length of altitude (distance from \(A\) to line \(BC\)): \[ \begin{aligned} d=\frac{|2+3-3|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \end{aligned} \]

Hence, equation of altitude is \(y=x+1\) and its length is \(\sqrt{2}\).

Significance

This problem integrates multiple core concepts: slope, perpendicularity, equation of line, and distance formula. It is highly important for board exams and forms a foundation for advanced coordinate geometry in competitive exams like JEE, especially in problems involving triangle geometry, orthocenter, and distance-based reasoning.

Theory

A line cutting the axes at intercepts \(a\) and \(b\) has equation in intercept form \(\frac{x}{a}+\frac{y}{b}=1\). The perpendicular distance from a point \((x_1,y_1)\) to a line \(Ax+By+C=0\) is given by \[ d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \] This formula is used to compute the distance of the origin from the given line.

Solution Roadmap

• Write the equation in intercept form.
• Convert into standard form.
• Apply distance formula from origin.
• Simplify and take reciprocal.

Solution

Equation of the line in intercept form: \[ \frac{x}{a}+\frac{y}{b}=1 \]

Convert to standard form: \[ \begin{aligned} \frac{x}{a}+\frac{y}{b}-1=0\ bx+ay-ab=0 \end{aligned} \]

Distance from origin: \[ \begin{aligned} p=\frac{| -ab |}{\sqrt{b^2+a^2}}=\frac{ab}{\sqrt{a^2+b^2}} \end{aligned} \]

Squaring: \[ p^2=\frac{a^2b^2}{a^2+b^2} \]

Taking reciprocal: \[ \begin{aligned} \frac{1}{p^2} =\frac{a^2+b^2}{a^2b^2} =\frac{1}{a^2}+\frac{1}{b^2} \end{aligned} \]

Hence proved.

Significance

This is a standard proof-based result combining intercept form and distance formula. It is frequently asked in board exams and is highly relevant for competitive exams like JEE, where such identities are used in advanced coordinate geometry problems involving loci, transformations, and optimization. It strengthens algebraic manipulation and conceptual clarity.