Chapter 9 — Straight Lines

Theory

The general equation of a straight line is \(Ax + By + C = 0\), where:

• If \(B = 0\), the line becomes \(Ax + C = 0\), which is a vertical line parallel to the y-axis.
• If \(A = 0\), the line becomes \(By + C = 0\), which is a horizontal line parallel to the x-axis.
• A line passes through the origin if the constant term \(C = 0\).

Solution Roadmap

• Identify coefficients \(A\), \(B\), and \(C\) from the given equation.
• Apply conditions:
  • Parallel to x-axis → set coefficient of \(x = 0\).
  • Parallel to y-axis → set coefficient of \(y = 0\).
  • Passing through origin → set constant term \(= 0\).
• Solve resulting equations in \(k\).

Solution

The given equation of the line is \[ \begin{aligned} (k-3)x-(4-k^{2})y+k^{2}-7k+6=0 \end{aligned{ \] Comparing with \(Ax + By + C = 0\), we get: \[ \begin{aligned} A &= k-3,\\ B &= -(4-k^2)\\&=k^2-4,\\\ C &= k^2-7k+6 \end{aligned{ \]

(a) Parallel to the x-axis

For a line parallel to the x-axis, \(A = 0\): \[ k-3=0 \Rightarrow k=3 \]

(b) Parallel to the y-axis

For a line parallel to the y-axis, \(B = 0\): \[ k^2-4=0 \Rightarrow k=\pm 2 \]

(c) Passing through the origin

For the line to pass through the origin, \(C=0\): \[ \begin{aligned} k^2-7k+6&=0\\ (k-6)(k-1)&=0\\ k&=1 \\\text{ or } \\k&=6 \end{aligned} \]

Illustration

Final Answer

• (a) Parallel to x-axis: \(k = 3\)
• (b) Parallel to y-axis: \(k = \pm 2\)
• (c) Passing through origin: \(k = 1, 6\)

Significance

This problem tests conceptual clarity of the general form of a straight line, which is fundamental for both board examinations and competitive exams like JEE. Questions involving parameter conditions (like \(k\)) are very common, especially in determining orientation and positional constraints of lines. Mastery of these conditions helps in quickly solving MCQs and multi-step coordinate geometry problems.

Theory

The intercept form of a straight line is: \[ \frac{x}{a} + \frac{y}{b} = 1 \] where \(a\) and \(b\) are the intercepts on the x-axis and y-axis respectively.

• The sum of intercepts is given by \(a + b\).
• The product of intercepts is given by \(ab\).
• If sum and product are known, intercepts can be found using quadratic equations.

Solution Roadmap

• Let intercepts be \(a\) and \(b\).
• Form equations using given sum and product.
• Solve quadratic to find possible values of \(a\) and \(b\).
• Substitute in intercept form to get required equations.

Solution

Let the intercepts on the x-axis and y-axis be \(a\) and \(b\) respectively. Given: \[ a + b = 1, \quad ab = -6 \]

From \(ab = -6\), we write: \[ b = -\frac{6}{a} \] Substituting into \(a + b = 1\): \[ a - \frac{6}{a} = 1 \] \[ a^2 - 6 = a \] \[ a^2 - a - 6 = 0 \] \[ (a - 3)(a + 2) = 0 \]

Hence, \[ a = 3 \quad \text{or} \quad a = -2 \]

Using \(a + b = 1\):

• If \(a = 3\), then \(b = -2\)
• If \(a = -2\), then \(b = 3\)

Therefore, the intercept pairs are: \[ (3, -2) \quad \text{and} \quad (-2, 3) \]

Using intercept form:

For \((a, b) = (3, -2)\): \[ \frac{x}{3} + \frac{y}{-2} = 1 \Rightarrow \frac{x}{3} - \frac{y}{2} = 1 \]

For \((a, b) = (-2, 3)\): \[ \frac{x}{-2} + \frac{y}{3} = 1 \Rightarrow \frac{y}{3} - \frac{x}{2} = 1 \]

Illustration

Final Answer

The required equations of the lines are: \[ \frac{x}{3} - \frac{y}{2} = 1 \quad \text{and} \quad \frac{y}{3} - \frac{x}{2} = 1 \]

Significance

This problem is important for understanding intercept form and parameter-based geometry. Such questions frequently appear in board exams and competitive exams like JEE, especially in forms where intercepts are indirectly given through algebraic conditions. It strengthens the ability to convert geometric constraints into algebraic equations efficiently.

Theory

The perpendicular distance of a point \((x_1, y_1)\) from a line \(Ax + By + C = 0\) is given by: \[ \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

• Any point on the y-axis has coordinates \((0, b)\).
• The modulus sign leads to two cases, giving multiple possible points.

Solution Roadmap

• Assume a general point on the y-axis.
• Convert the given line into standard form.
• Apply distance formula.
• Solve resulting equation considering both modulus cases.

Solution

Let the required point on the y-axis be \((0, b)\).

The given line is: \[ \frac{x}{3} + \frac{y}{4} = 1 \] Multiplying by \(12\), we get: \[ 4x + 3y - 12 = 0 \]

Using the distance formula: \[ \text{Distance} = \frac{|4(0) + 3b - 12|}{\sqrt{4^2 + 3^2}} = 4 \] \[ \frac{|3b - 12|}{5} = 4 \] \[ |3b - 12| = 20 \]

Case 1: \[ 3b - 12 = 20 \Rightarrow 3b = 32 \Rightarrow b = \frac{32}{3} \]

Case 2: \[ 3b - 12 = -20 \Rightarrow 3b = -8 \Rightarrow b = \frac{-8}{3} \]

Hence, the required points are: \[ \left(0, \frac{32}{3}\right), \quad \left(0, \frac{-8}{3}\right) \]

Illustration

Final Answer

\[ \left(0, \frac{32}{3}\right) \quad \text{and} \quad \left(0, \frac{-8}{3}\right) \]

Significance

This is a standard application of the point-to-line distance formula, a high-frequency concept in board exams and competitive exams like JEE. Such problems often test the ability to translate geometric constraints into algebraic equations and handle modulus cases correctly. It also builds intuition about perpendicular distance and locus of points in coordinate geometry.

Theory

The perpendicular distance from a point \((0,0)\) to the line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is: \[ \text{Distance}=\frac{|x_1y_2 - y_1x_2|}{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}} \]

• Points of the form \((\cos\theta,\sin\theta)\) lie on the unit circle.
• Use trigonometric identities:
  • \(\sin A \cos B - \cos A \sin B = \sin(A-B)\)
  • \(\cos A \cos B + \sin A \sin B = \cos(A-B)\)
  • \(1 - \cos A = 2\sin^2 \frac{A}{2}\)

Solution Roadmap

• Substitute given coordinates into distance formula.
• Simplify numerator using sine identity.
• Simplify denominator using cosine identity.
• Reduce expression using half-angle identities.

Solution

Let the given points be \(A(\cos\theta,\sin\theta)\) and \(B(\cos\phi,\sin\phi)\).

Using the distance formula: \[ \text{Distance}=\frac{|\cos\theta\sin\phi - \sin\theta\cos\phi|}{\sqrt{(\cos\phi-\cos\theta)^2 + (\sin\phi-\sin\theta)^2}} \]

Simplifying numerator: \[ |\cos\theta\sin\phi - \sin\theta\cos\phi| = |\sin(\phi-\theta)| \]

Simplifying denominator: \[ (\cos\phi-\cos\theta)^2 + (\sin\phi-\sin\theta)^2 = 2 - 2\cos(\phi-\theta) \] \[ = 4\sin^2\frac{\phi-\theta}{2} \] Hence denominator becomes: \[ 2\left|\sin\frac{\phi-\theta}{2}\right| \]

Therefore, \[ \text{Distance} = \frac{|\sin(\phi-\theta)|}{2\left|\sin\frac{\phi-\theta}{2}\right|} = \left|\cos\frac{\phi-\theta}{2}\right| \]

Hence, the required perpendicular distance is \[ \left|\cos\frac{\phi-\theta}{2}\right| \]

Illustration

Final Answer

\[ \left|\cos\frac{\phi-\theta}{2}\right| \]

Significance

This problem is a strong application of coordinate geometry combined with trigonometry. It is highly relevant for board exams and competitive exams like JEE, where such mixed concept problems frequently appear. It builds efficiency in handling trigonometric identities within geometric frameworks, especially involving unit circle representations.

Theory

A line parallel to the y-axis is a vertical line and has the equation: \[ x = \text{constant} \]

• The point of intersection of two lines is obtained by solving them simultaneously.
• Any vertical line passing through \((x_0, y_0)\) is given by \(x = x_0\).

Solution Roadmap

• Solve the given pair of linear equations to find the intersection point.
• Use the x-coordinate of that point.
• Write equation of vertical line passing through that point.

Solution

The given lines are: \[ x - 7y + 5 = 0 \quad \text{and} \quad 3x + y = 0 \]

From the second equation: \[ y = -3x \] Substituting into the first equation: \[ x - 7(-3x) + 5 = 0 \] \[ x + 21x + 5 = 0 \] \[ 22x + 5 = 0 \Rightarrow x = -\frac{5}{22} \]

Substituting back: \[ y = -3\left(-\frac{5}{22}\right) = \frac{15}{22} \]

Hence, the point of intersection is: \[ \left(-\frac{5}{22}, \frac{15}{22}\right) \]

Since the required line is parallel to the y-axis, its equation is: \[ x = -\frac{5}{22} \]

Illustration

Final Answer

\[ x = -\frac{5}{22} \]

Significance

This problem reinforces solving simultaneous linear equations and understanding geometric meaning of vertical lines. Such questions are very common in board exams and form the basis for coordinate geometry problems in competitive exams like JEE, especially in identifying line orientation and constructing equations through given points.

Theory

The intercept form of a line is: \[ \frac{x}{a} + \frac{y}{b} = 1 \] where the line meets the y-axis at \((0, b)\).

• The slope of a line in the form \(Ax + By + C = 0\) is \(-\frac{A}{B}\).
• If two lines are perpendicular, then \(m_1 \cdot m_2 = -1\).
• Point–slope form: \(y - y_1 = m(x - x_1)\).

Solution Roadmap

• Find the y-intercept of the given line.
• Convert the line into slope-intercept form to get its slope.
• Find slope of perpendicular line.
• Use point-slope form to get the required equation.

Solution

The given line is: \[ \frac{x}{4} + \frac{y}{6} = 1 \]

To find the point where it meets the y-axis, put \(x = 0\): \[ \frac{y}{6} = 1 \Rightarrow y = 6 \] Hence, the point is \((0, 6)\).

Converting the equation into slope-intercept form: \[ 6x + 4y = 24 \] \[ 4y = -6x + 24 \] \[ y = -\frac{3}{2}x + 6 \] So, slope \(m_1 = -\frac{3}{2}\).

Let slope of required line be \(m_2\). Since the lines are perpendicular: \[ m_1 \cdot m_2 = -1 \] \[ -\frac{3}{2} \cdot m_2 = -1 \Rightarrow m_2 = \frac{2}{3} \]

Using point-slope form through \((0,6)\): \[ y - 6 = \frac{2}{3}(x - 0) \] \[ 3y - 18 = 2x \] \[ 3y - 2x - 18 = 0 \]

Illustration

Final Answer

\[ 3y - 2x - 18 = 0 \]

Significance

This problem integrates intercept form, slope concepts, and perpendicularity — a very common combination in board exams and JEE. It strengthens understanding of geometric interpretation of slopes and helps in quickly forming equations of lines under constraints, which is a key skill in coordinate geometry.

Theory

The triangle formed by three lines can be analyzed by finding their pairwise points of intersection. Once the vertices are known, the area can be calculated using the determinant (shoelace) formula: \[ \Delta = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

• The lines \(y=x\) and \(y=-x\) intersect at the origin.
• The line \(x=k\) is a vertical line.
• Symmetry plays an important role in simplifying calculations.

Solution Roadmap

• Find intersection points of each pair of lines.
• Determine coordinates of the triangle’s vertices.
• Apply the area formula.

Solution

The given lines are: \[ \begin{aligned} y-x&=0,\\ x+y&=0,\\ x-k&=0 \end{aligned} \]

Intersection of \(y-x=0\) and \(x+y=0\): \[ \begin{aligned} y&=x,\\ x+y&=0 \\\Rightarrow x+x&=0 \\\Rightarrow x&=0,\\ y&=0 \end{aligned} \] Vertex: \((0,0)\)

Intersection of \(x+y=0\) and \(x-k=0\): \[ \begin{aligned} x&=k,\\ k+y&=0 \\\Rightarrow y&=-k \end{aligned} \] Vertex: \((k,-k)\)

Intersection of \(y-x=0\) and \(x-k=0\): \[ x=k,\quad y=k \] Vertex: \((k,k)\)

Using the area formula: \[ \begin{aligned} \Delta &= \frac{1}{2}\left| 0(-k-k) + k(k-0) + k(0-(-k)) \right|\\ &= \frac{1}{2}\left|k^2 + k^2\right|\\ &= \left|k^2\right| \end{aligned} \]

Illustration

Final Answer

\[ |k^2| \]

Significance

This problem highlights symmetry in coordinate geometry and efficient use of intersection methods. It is important for board exams and highly relevant for competitive exams like JEE, where quick identification of geometric structure (like symmetry and vertical lines) can significantly reduce computation time.

Theory

Three lines are said to be concurrent if they pass through a common point. To ensure concurrency:

• First find the intersection point of any two lines.
• Then check whether this point satisfies the third line.

Solution Roadmap

• Solve first two equations to find their intersection point.
• Substitute this point into the third equation.
• Solve for \(p\).

Solution

The given lines are: \[ \begin{aligned} 3x+y-2&=0,\\ px+2y-3&=0,\\ 2x-y-3&=0 \end{aligned} \]

First, find the intersection of: \[ \begin{aligned} 3x+y-2&=0 \\ \text{and} \\ 2x-y-3&=0 \end{aligned} \]

Adding the two equations: \[ 5x - 5 = 0 \Rightarrow x = 1 \]

Substitute into \(2x - y - 3 = 0\): \[ \begin{aligned} 2(1) - y - 3 &= 0 \\\Rightarrow -y -1 &= 0 \\\Rightarrow y &= -1 \end{aligned} \]

So, the point of intersection is: \[ (1, -1) \]

For concurrency, this point must satisfy: \[ px + 2y - 3 = 0 \] Substituting: \[ p(1) + 2(-1) - 3 = 0 \] \[ p - 2 - 3 = 0 \Rightarrow p = 5 \]

Illustration

Final Answer

\[ p=5 \]

Significance

This problem is a standard application of concurrency of lines, frequently asked in both board exams and competitive exams like JEE. It builds the skill of reducing a geometric condition into an algebraic substitution problem, which is a common strategy in coordinate geometry.

Theory

Three lines are concurrent if they intersect at a single point \((x_0, y_0)\). Hence, this common point must satisfy all three equations.

• Equate pairs of equations to eliminate \(y_0\).
• Form relations in \(x_0\).
• Eliminate \(x_0\) to obtain a condition involving slopes and intercepts.

Solution Roadmap

• Assume common point \((x_0, y_0)\).
• Substitute into all three equations.
• Subtract equations pairwise.
• Eliminate \(x_0\) to derive required condition.

Solution

Since the three lines are concurrent, they pass through a common point \((x_0, y_0)\).

Therefore, \[ \begin{aligned} y_0 = m_1 x_0 + c_1,\\ y_0 = m_2 x_0 + c_2,\\ y_0 = m_3 x_0 + c_3 \end{aligned} \]

Subtracting the second from the first: \[ \begin{align} (m_1 - m_2)x_0 = c_2 - c_1 \tag{1} \end{align} \]

Subtracting the third from the second: \[ (m_2 - m_3)x_0 = c_3 - c_2 \tag{2} \]

Multiply (1) by \((m_3 - m_1)\) and (2) by \((m_1 - m_2)\): \[ (m_1 - m_2)(m_3 - m_1)x_0 = (c_2 - c_1)(m_3 - m_1) \] \[ (m_2 - m_3)(m_1 - m_2)x_0 = (c_3 - c_2)(m_1 - m_2) \]

Adding these, the terms involving \(x_0\) cancel, giving: \[ m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0 \]

Illustration

Final Result

\[ m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0 \]

Significance

This is a standard concurrency condition in coordinate geometry and is frequently used in competitive exams like JEE. It helps in quickly checking whether three given lines intersect at a single point without explicitly solving all equations. It also builds algebraic manipulation skills combined with geometric interpretation.

Theory

The angle between two lines with slopes \(m_1\) and \(m_2\) is given by: \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]

• If \(\theta = 45^\circ\), then \(\tan 45^\circ = 1\).
• This gives two cases due to modulus: \(+1\) and \(-1\).
• Use point–slope form: \(y - y_0 = m(x - x_0)\).

Solution Roadmap

• Find slope of the given line.
• Use angle formula to find possible slopes.
• Form equations using point–slope form.

Solution

The given line is: \[ x - 2y = 3 \] \[ y = \frac{1}{2}x - \frac{3}{2} \] Hence, slope \(m_2 = \frac{1}{2}\).

Let the slope of required line be \(m_1\). Using: \[ \frac{m_1 - \frac{1}{2}}{1 + \frac{m_1}{2}} = \pm 1 \]

Case 1: \[ \begin{aligned} \frac{m_1 - \frac{1}{2}}{1 + \frac{m_1}{2}} &= 1\\ m_1 - \frac{1}{2} &= 1 + \frac{m_1}{2}\\ \Rightarrow m_1 &= 3 \end{aligned} \]

Equation through \((3,2)\): \[ y - 2 = 3(x - 3) \Rightarrow y - 3x + 7 = 0 \]

Case 2: \[ \begin{aligned} \frac{m_1 - \frac{1}{2}}{1 + \frac{m_1}{2}} &= -1\\ m_1 - \frac{1}{2} &= -1 - \frac{m_1}{2}\\ \Rightarrow m_1 &= -\frac{1}{3} \end{aligned} \]

Equation through \((3,2)\): \[ \begin{aligned} y - 2 &= -\frac{1}{3}(x - 3) \\ \Rightarrow x + 3y - 9 &= 0 \end{aligned} \]

Illustration

Final Answer

\[ \begin{aligned} y - 3x + 7 &= 0 \\ \text{and} \\ x + 3y - 9 &= 0 \end{aligned} \]

Significance

This is a classic angle-between-lines problem, frequently asked in board exams and JEE. It strengthens understanding of slope relations and handling dual solutions arising from \(\tan\theta\). Such problems are often used in MCQs and multi-step coordinate geometry questions.

Theory

A line having equal intercepts on the coordinate axes has intercept form: \[ \begin{aligned} \frac{x}{a} + \frac{y}{a} &= 1 \\\Rightarrow x + y &= a \end{aligned} \]

• Intersection point of two lines is found by solving them simultaneously.
• Substitute the point into \(x + y = a\) to determine the constant.

Solution Roadmap

• Find intersection point of given lines.
• Use equal intercept form \(x+y=a\).
• Substitute the point to find \(a\).

Solution

The given lines are: \[ \begin{aligned} 4x+7y-3&=0,\\ 2x-3y+1&=0\\ 4x+7y&=3,\\ 2x-3y&=-1 \end{aligned} \]

Multiply second equation by 2: \[ 4x-6y=-2 \] Subtract from first: \[ 13y=5 \Rightarrow y=\frac{5}{13} \]

Substitute in \(2x-3y=-1\): \[ \begin{aligned} 2x-\frac{15}{13}&=-1\\ \Rightarrow 2x&=\frac{2}{13}\\ \Rightarrow x&=\frac{1}{13} \end{aligned} \]

Intersection point is: \[ \left(\frac{1}{13},\frac{5}{13}\right) \]

Required line: \(x+y=a\) \[ \frac{1}{13}+\frac{5}{13}=a=\frac{6}{13} \]

Hence, \[ x+y=\frac{6}{13} \Rightarrow 13x+13y-6=0 \]

Illustration

Final Answer

\[ 13x+13y-6=0 \]

Significance

This problem integrates concepts of intersection of lines and intercept form. It is important for board exams and competitive exams like JEE, where recognizing special forms like equal intercepts can simplify problems significantly and save time.

Theory

The angle between two lines with slopes \(m_1\) and \(m_2\) is given by: \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]

• A line passing through the origin has equation \(y = m_1 x\).
• Thus, \(\frac{y}{x} = m_1\).
• Both \(+\theta\) and \(-\theta\) cases must be considered.

Solution Roadmap

• Assume slope of required line \(= m_1\).
• Apply angle formula with given slope \(m\).
• Solve for \(m_1\).
• Substitute into \(\frac{y}{x}=m_1\).

Solution

The given line is \(y = mx + c\), so its slope is \(m\). Let the slope of the required line be \(m_1\).

Since the required line passes through the origin: \[ y = m_1 x \quad \Rightarrow \quad \frac{y}{x} = m_1 \]

Using the angle formula: \[ \tan\theta = \frac{m_1 - m}{1 + m_1 m} \]

Solving: \[ \tan\theta(1 + m_1 m) = m_1 - m \] \[ \tan\theta + m_1 m \tan\theta = m_1 - m \] \[ m_1(1 - m\tan\theta) = m + \tan\theta \] \[ m_1 = \frac{m + \tan\theta}{1 - m\tan\theta} \]

Similarly, taking the other case: \[ \tan\theta = \frac{m - m_1}{1 + m_1 m} \] we get: \[ m_1 = \frac{m - \tan\theta}{1 + m\tan\theta} \]

Hence, \[ m_1 = \frac{m \pm \tan\theta}{1 \mp m\tan\theta} \]

Therefore, \[ \frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta} \]

Illustration

Final Result

\[ \frac{y}{x}=\frac{m\pm \tan \theta}{1\mp m\tan \theta} \]

Significance

This is a standard derivation connecting slope and angle, widely used in coordinate geometry. It is very important for board exams and JEE, especially in problems involving family of lines, rotations, and angle transformations. Understanding this derivation helps in quickly forming equations without re-deriving each time.

Theory

If a line divides a line segment joining two points, the ratio can be found using:

• Intersection point of the given line with the segment.
• Section formula: \[ x = \frac{m x_2 + n x_1}{m+n}, \quad y = \frac{m y_2 + n y_1}{m+n} \]

Solution Roadmap

• Find equation of line joining given points.
• Find its intersection with \(x+y=4\).
• Use section formula to determine ratio.

Solution

Given points are \((-1,1)\) and \((5,7)\).

Slope of the line: \[ m = \frac{7-1}{5-(-1)} = \frac{6}{6} = 1 \]

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

Now find intersection with \(x + y = 4\): \[ y = x + 2 \] \[ x + (x+2) = 4 \Rightarrow 2x = 2 \Rightarrow x = 1 \] \[ y = 3 \] So intersection point is \((1,3)\).

Let ratio be \(m:n\). Using section formula: \[ 1 = \frac{5m + (-1)n}{m+n} \] \[ \begin{aligned} m+n &= 5m - n\\ \Rightarrow 2n &= 4m\\ \Rightarrow n &= 2m \end{aligned} \]

Hence, \[ m:n = 1:2 \]

Illustration

Final Answer

\[ 1:2 \]

Significance

This problem is a standard application of section formula and intersection of lines. It is frequently asked in board exams and JEE, especially in problems involving division of line segments and coordinate geometry relationships. It strengthens algebraic and geometric interpretation together.

Theory

Distance “along a line” means we measure distance between two points lying on that given direction.

• Find intersection of the given line with the direction line.
• Then compute distance between the given point and this intersection point.
• Use distance formula: \[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]

Solution Roadmap

• Find intersection of \(2x-y=0\) and \(4x+7y+5=0\).
• Compute distance from \((1,2)\) to that point.

Solution

Given direction line: \[ 2x-y=0 \Rightarrow y=2x \]

Substitute into \(4x+7y+5=0\): \[ 4x+7(2x)+5=0 \Rightarrow 18x+5=0 \Rightarrow x=-\frac{5}{18} \]

\[ y=2x=-\frac{5}{9} \] Intersection point: \[ \left(-\frac{5}{18},-\frac{5}{9}\right) \]

Distance from \((1,2)\): \[ d=\sqrt{\left(1+\frac{5}{18}\right)^2+\left(2+\frac{5}{9}\right)^2} \] \[ =\sqrt{\left(\frac{23}{18}\right)^2+\left(\frac{23}{9}\right)^2} \] \[ =\sqrt{\frac{23^2}{18^2}\left(1+4\right)} =\frac{23}{18}\sqrt{5} \]

Illustration

Final Answer

\[ \frac{23\sqrt{5}}{18} \]

Significance

This problem is important for understanding directional distance in coordinate geometry, a concept often tested in JEE and board exams. It strengthens the idea of combining intersection and distance formulas, especially when movement is constrained along a line.

Theory

If a line passes through a point and intersects another line, then:

• We can assume slope \(m\) and form its equation.
• Find intersection point using simultaneous equations.
• Apply distance formula between the given point and intersection point.
• Solve for slope \(m\).

Solution Roadmap

• Assume slope \(m\) and form equation through \((-1,2)\).
• Find intersection with \(x+y=4\).
• Apply distance condition \(=3\).
• Solve for \(m\) to determine direction.

Solution

Let slope of required line be \(m\). Equation through \((-1,2)\): \[ y-2 = m(x+1) \Rightarrow y = m(x+1)+2 \]

Substitute in \(x+y=4\): \[ \begin{aligned} x + m(x+1)+2 &= 4\\ \Rightarrow x(1+m) + m &= 2\\ \Rightarrow x &= \frac{2-m}{m+1} \end{aligned} \]

\[ y = 4 - x = \frac{5m+2}{m+1} \] Intersection point: \[ \left(\frac{2-m}{m+1}, \frac{5m+2}{m+1}\right) \]

Distance from \((-1,2)\) is 3: \[ \begin{aligned} 9 &= \left(\frac{-3}{m+1}\right)^2 + \left(\frac{-3m}{m+1}\right)^2\\ 9 &= \frac{9 + 9m^2}{(m+1)^2} \end{aligned} \]

\[ \begin{aligned} 9(m+1)^2 &= 9 + 9m^2\\ \Rightarrow 9m^2 + 18m + 9 &= 9 + 9m^2\\ \Rightarrow 18m &= 0\\ \Rightarrow m &= 0 \end{aligned} \]

Hence, the required line is parallel to the x-axis.

Illustration

Final Answer

Required direction: parallel to x-axis (i.e., slope \(m = 0\)).

Significance

This problem combines line equations, intersection, and distance constraints. It is important for board exams and competitive exams like JEE, especially for problems involving parametric slopes and geometric interpretation of distance.

Theory

If the legs of a right-angled triangle are parallel to the coordinate axes:

• One leg is parallel to x-axis → equation \(y = \text{constant}\).
• Other leg is parallel to y-axis → equation \(x = \text{constant}\).
• The right angle occurs at a point formed by combining x-coordinate of one point and y-coordinate of the other.

Solution Roadmap

• Identify endpoints of hypotenuse.
• Form possible right-angle vertices using coordinate combinations.
• Write equations of horizontal and vertical lines through those points.

Solution

Given endpoints of hypotenuse: \[ (1,3) \quad \text{and} \quad (-4,1) \]

Since the legs are parallel to the axes, the right angle must be at a point formed by combining coordinates.

Case 1: Take x from first point and y from second: \[ (1,1) \] The legs are: \[ x = 1 \quad \text{and} \quad y = 1 \]

Case 2: Take x from second point and y from first: \[ (-4,3) \] The legs are: \[ x = -4 \quad \text{and} \quad y = 3 \]

Hence, the required equations are: \[ x=1,\; y=1 \quad \text{or} \quad x=-4,\; y=3 \]

Illustration

Final Answer

\[ x=1,\; y=1 \quad \text{or} \quad x=-4,\; y=3 \]

Significance

This problem develops geometric visualization in coordinate geometry. It is frequently tested in board exams and JEE, especially for understanding perpendicularity, axis-parallel lines, and coordinate construction of triangles.

Theory

Reflection of a point across a line (mirror) follows:

• The image lies along the perpendicular from the point to the line.
• The mirror line bisects the segment joining the point and its image.
• The foot of perpendicular acts as the midpoint.

Solution Roadmap

• Find slope of given line and perpendicular slope.
• Find equation of perpendicular through given point.
• Find foot of perpendicular (intersection point).
• Use midpoint formula to get image.

Solution

Given line: \[ \begin{aligned} x + 3y &= 7 \\\Rightarrow y &= -\frac{1}{3}x + \frac{7}{3} \end{aligned} \] Slope = \(-\frac{1}{3}\), so perpendicular slope = \(3\).

Equation of perpendicular through \((3,8)\): \[ \begin{aligned} y - 8 &= 3(x - 3)\\ \Rightarrow y &= 3x - 1 \end{aligned} \]

Find intersection with given line: \[ \begin{aligned} x + 3(3x - 1) &= 7\\ \Rightarrow 10x &= 10\\ \Rightarrow x &= 1 \end{aligned} \] \[ y = 3(1) - 1 = 2 \] Foot of perpendicular = \((1,2)\)

Let image be \((x_1, y_1)\). Using midpoint: \[ \begin{aligned} 1 &= \frac{x_1 + 3}{2}, \\ 2 &= \frac{y_1 + 8}{2}\\ x_1 &= -1,\\ y_1 &= -4 \end{aligned} \]

Hence, image is: \[ (-1,-4) \]

Illustration

Final Answer

\[ (-1,-4) \]

Significance

This is a standard reflection problem, important for board exams and JEE. It combines perpendicular slope, intersection, and midpoint concepts. Mastery of this method helps in solving symmetry and reflection-based geometry problems efficiently.

Theory

If two lines are equally inclined to a third line, then the angles they make with the third line are equal in magnitude but opposite in sign.

• Angle between two lines: \[ \tan\theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
• Equal inclination ⇒ one angle is \(\theta\), the other is \(-\theta\).

Solution Roadmap

• Find slopes of given lines.
• Apply equal inclination condition.
• Solve resulting equation for \(m\).

Solution

Given lines: \[ \begin{aligned} y &= 3x + 1 \\\Rightarrow m_1 &= 3 \end{aligned} \] \[ \begin{aligned} 2y &= x + 3 \Rightarrow y \\&= \frac{1}{2}x + \frac{3}{2},\\ m_2 &= \frac{1}{2} \end{aligned} \]

Let slope of required line be \(m\). Using equal inclination: \[ \frac{3 - m}{1 + 3m} = -\frac{\frac{1}{2} - m}{1 + \frac{m}{2}} \]

Simplify RHS: \[ \frac{\frac{1}{2} - m}{1 + \frac{m}{2}} = \frac{1 - 2m}{2 + m} \]

Hence, \[ \frac{3 - m}{1 + 3m} = -\frac{1 - 2m}{2 + m} \]

Cross-multiplying: \[ (3 - m)(2 + m) = -(1 + 3m)(1 - 2m) \]

\[ \begin{aligned} 6 + m - m^2 &= - (1 + m - 6m^2)\\ 6 + m - m^2 &= -1 - m + 6m^2 \end{aligned} \]

\[ \begin{aligned} 7 + 2m - 7m^2 &= 0\\ \Rightarrow 7m^2 - 2m - 7 &= 0 \end{aligned} \]

Solving: \[ \begin{aligned} m &= \frac{2 \pm \sqrt{4 + 196}}{14}\\ &= \frac{2 \pm \sqrt{200}}{14}\\ &= \frac{2 \pm 10\sqrt{2}}{14}\\ &= \frac{1 \pm 5\sqrt{2}}{7} \end{aligned} \]

Illustration

Illustration

Final Answer

\[ m = \frac{1 \pm 5\sqrt{2}}{7} \]

Significance

This problem is important for understanding angle relationships between lines. It is frequently asked in board exams and JEE, especially in problems involving symmetry, angle bisectors, and equal inclination conditions.

Theory

The perpendicular distance of a point \((x,y)\) from a line \(Ax+By+C=0\) is: \[ \frac{|Ax+By+C|}{\sqrt{A^2+B^2}} \]

• If a linear combination of expressions in \(x\) and \(y\) equals a constant, it represents a straight line.
• When modulus signs are removed consistently (same region), the equation becomes linear.

Solution Roadmap

• Write expressions for perpendicular distances.
• Use given condition (sum = constant).
• Remove modulus with consistent sign.
• Show resulting equation is linear.

Solution

Let \(P(x,y)\) be the variable point.

Distance from \(x+y-5=0\): \[ \frac{|x+y-5|}{\sqrt{2}} \]

Distance from \(3x-2y+7=0\): \[ \frac{|3x-2y+7|}{\sqrt{13}} \]

Given: \[ \frac{|x+y-5|}{\sqrt{2}} + \frac{|3x-2y+7|}{\sqrt{13}} = 10 \]

For a fixed region, remove modulus: \[ \frac{x+y-5}{\sqrt{2}} + \frac{3x-2y+7}{\sqrt{13}} = 10 \]

Multiply by \(\sqrt{26}\): \[ \sqrt{13}(x+y-5) + \sqrt{2}(3x-2y+7) = 10\sqrt{26} \]

This is of the form: \[ ax + by + c = 0 \] which represents a straight line.

Hence, the locus of point \(P(x,y)\) is a straight line.

Final Result

The locus of point \(P\) is a straight line.

Significance

This problem introduces locus concepts using distance from lines, which is very important for board exams and JEE. It strengthens understanding of how geometric constraints translate into algebraic equations, especially in problems involving absolute values and regions.

Theory

The locus of points equidistant from two parallel lines is a line parallel to both, lying midway between them.

• Bring both lines to same form \(Ax + By + C = 0\).
• Use condition: \[ \frac{|Ax + By + C_1|}{\sqrt{A^2+B^2}} = \frac{|Ax + By + C_2|}{\sqrt{A^2+B^2}} \]
• This gives the equation of the required line.

Solution Roadmap

• Make coefficients of \(x\) and \(y\) same in both lines.
• Apply equidistance condition.
• Solve resulting equation.

Solution

Given lines: \[ 9x + 6y - 7 = 0 \quad \text{and} \quad 3x + 2y + 6 = 0 \]

Divide first equation by 3: \[ 3x + 2y - \frac{7}{3} = 0 \]

Let required line be equidistant from both. Then: \[ |3x + 2y - \frac{7}{3}| = |3x + 2y + 6| \]

Let \(3x + 2y = k\), then: \[ |k - \frac{7}{3}| = |k + 6| \]

Squaring: \[ \left(k - \frac{7}{3}\right)^2 = (k + 6)^2 \]

\[ k^2 - \frac{14}{3}k + \frac{49}{9} = k^2 + 12k + 36 \] \[ -\frac{50}{3}k = \frac{275}{9} \Rightarrow k = -\frac{11}{6} \]

Hence: \[ 3x + 2y = -\frac{11}{6} \]

Multiply by 6: \[ 18x + 12y + 11 = 0 \]

Illustration

Final Answer

\[ 18x + 12y + 11 = 0 \]

Significance

This problem is important for understanding loci and parallel line geometry. It is frequently used in board exams and JEE, especially in problems involving distance between lines and symmetry. Recognizing midpoint lines helps solve such questions quickly.

Theory

For reflection in the x-axis:

• A point \((x,y)\) reflects to \((x,-y)\).
• The path of light can be treated as a straight line by reflecting one point.
• The point of reflection lies on the x-axis \((y=0)\).

Solution Roadmap

• Reflect point \(Q(5,3)\) to \(Q'(5,-3)\).
• Find equation of line joining \(P(1,2)\) and \(Q'(5,-3)\).
• Find its intersection with x-axis.

Solution

Reflect \(Q(5,3)\) across x-axis: \[ Q'(5,-3) \]

Slope of line \(PQ'\): \[ \begin{aligned} m &= \frac{-3 - 2}{5 - 1} \\&= -\frac{5}{4} \end{aligned} \]

Equation through \(P(1,2)\): \[ y - 2 = -\frac{5}{4}(x - 1) \]

At point \(A\), \(y=0\): \[ \begin{aligned} -2 &= -\frac{5}{4}(x - 1)\\ \Rightarrow -8 &= -5(x - 1)\\ \Rightarrow 8 &= 5(x - 1)\\ \Rightarrow x &= \frac{13}{5} \end{aligned} \]

Hence, \[ A\left(\frac{13}{5}, 0\right) \]

Illustration

Final Answer

\[ A\left(\frac{13}{5},\,0\right) \]

Significance

This problem demonstrates the reflection principle, widely used in coordinate geometry and optics-based problems in JEE. It converts a reflection path into a straight-line problem, simplifying calculations significantly.

Theory

The perpendicular distance from a point \((x_1,y_1)\) to a line \(Ax+By+C=0\) is: \[ d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \]

• Convert the given equation into standard form.
• Compute distances from both symmetric points.
• Multiply and simplify using trigonometric identities.

Solution Roadmap

• Convert line to standard form.
• Apply distance formula for both points.
• Multiply expressions and simplify.

Solution

Given line: \[ \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1 \] Multiply by \(ab\): \[ bx\cos\theta + ay\sin\theta - ab = 0 \]

Distance from \(P\left(\sqrt{a^2-b^2},0\right)\): \[ d_1 = \frac{\left| b\sqrt{a^2-b^2}\cos\theta - ab \right|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} \]

Distance from \(Q\left(-\sqrt{a^2-b^2},0\right)\): \[ d_2 = \frac{\left| -b\sqrt{a^2-b^2}\cos\theta - ab \right|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} \]

Product: \[ d_1 d_2 = \frac{\left| (b\sqrt{a^2-b^2}\cos\theta - ab)(-b\sqrt{a^2-b^2}\cos\theta - ab)\right|} {b^2\cos^2\theta + a^2\sin^2\theta} \]

Using \((x-y)(-x-y)=y^2-x^2\): \[ = \frac{\left| a^2b^2 - b^2(a^2-b^2)\cos^2\theta \right|} {b^2\cos^2\theta + a^2\sin^2\theta} \]

Factor \(b^2\): \[ = \frac{b^2\left| a^2 - (a^2-b^2)\cos^2\theta \right|} {b^2\cos^2\theta + a^2\sin^2\theta} \]

Simplify: \[ a^2 - (a^2-b^2)\cos^2\theta = a^2\sin^2\theta + b^2\cos^2\theta \]

Hence: \[ d_1 d_2 = \frac{b^2(a^2\sin^2\theta + b^2\cos^2\theta)} {a^2\sin^2\theta + b^2\cos^2\theta} = b^2 \]

Illustration

Final Result

\[ b^2 \]

Significance

This is a standard proof combining coordinate geometry and trigonometric identities. It is highly important for board exams and JEE, especially for problems involving symmetry, distances, and algebraic simplification in geometry.

Theory

The shortest distance from a point to a line is along the perpendicular.

• First find the intersection point of the given two lines.
• Then find the slope of the target line.
• The required path is perpendicular to the target line.
• Use point–slope form to get the equation.

Solution Roadmap

• Find intersection point \(P\).
• Find slope of line \(6x-7y+8=0\).
• Find perpendicular slope.
• Form equation through \(P\).

Solution

Solve: \[ \begin{aligned} 2x - 3y &= -4,\\ 3x + 4y &= 5 \end{aligned} \]

Multiply: \[ \begin{aligned} 8x - 12y &= -16,\\ 9x + 12y &= 15 \end{aligned} \] Add: \[ \begin{aligned} 17x &= -1 \\\Rightarrow x &= -\frac{1}{17} \end{aligned} \]

Substitute: \[ \begin{aligned} 2\left(-\frac{1}{17}\right) - 3y &= -4\\ \Rightarrow y &= \frac{22}{17} \end{aligned} \]

Intersection point: \[ P\left(-\frac{1}{17}, \frac{22}{17}\right) \]

Given line: \[ \begin{aligned} 6x - 7y + 8 &= 0 \\\Rightarrow y &= \frac{6}{7}x + \frac{8}{7} \end{aligned} \] Slope = \(\frac{6}{7}\)

Perpendicular slope: \[ -\frac{7}{6} \]

Equation through \(P\): \[ y - \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right) \]

Multiply by 102: \[ \begin{aligned} 102y - 132 &= -119x - 7\\ \Rightarrow 119x + 102y - 125 &= 0 \end{aligned} \]

Illustration

Final Answer

\[ 119x + 102y - 125 = 0 \]

Significance

This problem highlights the shortest distance concept (perpendicular distance), which is extremely important in coordinate geometry. It is frequently tested in board exams and JEE, especially in optimization and geometric interpretation problems.