Let us have a brief recall of coordinate geometry done in earlier classes.

• Distance between the points \(\mathrm{P} (x_1, y_1 )\text{ and }Q (x_2 , y_2 )\) is
  \[\mathrm{PQ}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\]
• The coordinates of a point dividing the line segment joining the points \((x_1,\;y_1)\) and \((x_2,\; y_2)\) internally, in the ratio \(m: n\) are
  \[\left(\dfrac{mx_2+nx_1}{m+n},\; \dfrac{my_2+ny_1}{m+n}\right)\]
• In particular, if \(m = n\), the coordinates of the mid-point of the line segment joining the points \((x_1,\; y_1 )\) and \((x_2 ,\; y_2 )\) are
  \[\left(\dfrac{x_2+x_1}{2},\; \dfrac{y_2+y_1}{2}\right)\]
• Area of the triangle whose vertices are \((x_1,\; y_1 ),\; (x_2 ,\; y2 )\) and \((x_3 ,\; y_3 )\) is
  \[\dfrac{1}{2}\Bigl|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Bigr|\]

Conditions for Parallelism and Perpendicularity of Lines in Terms of Their Slopes

The slope of a line describes its inclination with respect to the positive direction of the \(x\)-axis. If two lines have slopes \(m_1\) and \(m_2\), their relative position can be determined by comparing these slopes.

Condition for Parallel Lines

Two lines are said to be parallel if they have the same inclination and never intersect. This happens when their slopes are equal.

\[ m_1 = m_2 \]

If the slopes are equal, the lines rise or fall at the same rate, which ensures they remain parallel.

Condition for Perpendicular Lines

Two lines are perpendicular if they intersect at a right angle. In terms of slopes, this occurs when the product of their slopes is equal to \(-1\).

\[ m_1 \cdot m_2 = -1 \]

This means that one slope is the negative reciprocal of the other, resulting in a \(90^\circ\) angle between the two lines.

Final Summary

If \(m_1 = m_2\), the lines are parallel, and if \(m_1 \cdot m_2 = -1\), the lines are perpendicular. These conditions provide a simple and effective way to analyze the geometric relationship between two straight lines using their slopes.

The angle between two straight lines is defined as the angle through which one line must be rotated to coincide with the other. If the slopes of the two lines are \(m_1\) and \(m_2\), the angle \(\theta\) between them can be expressed in terms of these slopes.

Let the inclinations of the two lines with the positive \(x\)-axis be \(\theta_1\) and \(\theta_2\). Then their slopes are given by

\[ m_1 = \tan \theta_1 \quad \text{and} \quad m_2 = \tan \theta_2 \]

The angle \(\theta\) between the two lines is the difference between their inclinations, that is

\[ \theta = \theta_2 - \theta_1 \]

Using the tangent subtraction formula, we obtain

\[ \tan \theta = \tan(\theta_2 - \theta_1) = \frac{m_2 - m_1}{1 + m_1 m_2} \]

Hence, the angle between two lines with slopes \(m_1\) and \(m_2\) is given by

\[ \boxed{\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|} \]

If \(m_1 = m_2\), then \(\theta = 0^\circ\), meaning the lines are parallel. If \(m_1 m_2 = -1\), then \(\theta = 90^\circ\), meaning the lines are perpendicular.

This formula helps determine the exact angle between any two non-vertical straight lines using only their slopes.

Various Forms of the Equation of a Straight Line

A straight line in a plane can be represented algebraically in different forms depending on the information available, such as slope, intercepts, or given points. Each form is useful in specific geometric situations.

1. Horizontal and Vertical Lines

Definition.

A horizontal line is a line parallel to the \(x\)-axis, while a vertical line is a line parallel to the \(y\)-axis.

Equation of a Horizontal Line.

A horizontal line has no change in the \(y\)-coordinate. If it passes through the point \((x_1, y_1)\), then every point on the line has the same \(y\)-coordinate.

\[ y = y_1 \]

Equation of a Vertical Line.

A vertical line has no change in the \(x\)-coordinate. If it passes through the point \((x_1, y_1)\), then every point on the line has the same \(x\)-coordinate.

\[ x = x_1 \]

Important Aspect.

A horizontal line has slope \(m = 0\), while a vertical line has an undefined slope since the change in \(x\) is zero.

2. Point–Slope Form of a Line

Definition

The point–slope form expresses the equation of a line when one point on the line and its slope are known.

Derivation

Let a line have slope \(m\) and pass through the point \((x_1, y_1)\). If \((x, y)\) is any other point on the line, then the slope is given by

\[ m = \frac{y - y_1}{x - x_1} \]

Rearranging, we obtain

\[ y - y_1 = m(x - x_1) \]

Point–Slope Form

\[ \boxed{y - y_1 = m(x - x_1)} \]

Important Aspect.

This form is most useful when the slope and a single point on the line are known.

3. Two–Point Form of a Line

Definition

The two–point form gives the equation of a line passing through two known points.

Derivation

Let the two points be \((x_1, y_1)\) and \((x_2, y_2)\). The slope of the line is

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substituting this value of \(m\) into the point–slope form, we get

\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]

Two–Point Form

\[ \boxed{y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)} \]

Important Aspect

This form is useful when two points on the line are known but the slope is not given directly.

4. Slope–Intercept Form of a Line

Definition

The slope–intercept form represents a line in terms of its slope and its intercept on the \(y\)-axis.

Derivation

Let the line cut the \(y\)-axis at the point \((0, c)\) and have slope \(m\). Using the point–slope form, we write

\[ y - c = m(x - 0) \]

Simplifying, we obtain

\[ y = mx + c \]

Slope–Intercept Form

\[ \boxed{y = mx + c} \]

Important Aspect

Here, \(m\) represents the slope of the line and \(c\) represents the \(y\)-intercept. This form is especially convenient for graphing straight lines.

5. Intercept Form of a Line

Definition

The intercept form represents a line using its intercepts on the \(x\)-axis and the \(y\)-axis.

Derivation

Let the line cut the \(x\)-axis at \((a, 0)\) and the \(y\)-axis at \((0, b)\). Using the two–point form, the equation of the line can be written as

\[ \frac{x}{a} + \frac{y}{b} = 1 \]

Intercept Form

\[ \boxed{\frac{x}{a} + \frac{y}{b} = 1} \]

Important Aspect

This form is useful when the intercepts are known and gives a clear geometric interpretation of how the line meets both coordinate axes.

Key Observation

All these forms represent the same geometric object — a straight line — but are used in different situations depending on the available information. Each form can be converted into another, showing that all are equivalent representations of a linear equation.

Distance of a Point from a Line

The distance of a point from a line is defined as the length of the perpendicular drawn from the given point to the line. This perpendicular represents the shortest distance between the point and the line.

General Equation of a Line

Let the equation of a straight line be

\[ ax + by + c = 0 \]

and let the given point be \(P(x_1, y_1)\).

Concept and Construction

From point \(P(x_1, y_1)\), draw a perpendicular to the given line. Let the foot of the perpendicular be \(Q\). Then, the distance \(PQ\) represents the required distance from the point to the line.

Derivation of the Formula

The perpendicular distance depends on how far the point lies from satisfying the line’s equation. The expression \(ax_1 + by_1 + c\) measures this deviation, while \(\sqrt{a^2 + b^2}\) normalizes it based on the slope of the line.

Thus, the distance of point \(P(x_1, y_1)\) from the line \(ax + by + c = 0\) is given by

\[ \boxed{d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}} \]

Important Observations

If the point lies on the line, then \(ax_1 + by_1 + c = 0\), and hence the distance is zero. The distance is always a non-negative value because of the absolute value in the numerator.

Key Aspect

This formula is widely used in coordinate geometry to find the shortest distance between a point and a straight line and plays an important role in proving collinearity, parallelism, and geometric relationships.

Distance Between Two Parallel Lines

The distance between two parallel lines is defined as the length of the perpendicular drawn from any point on one line to the other line. Since the lines are parallel, this perpendicular distance remains the same for all points on either line.

General Form of Parallel Lines

Let the equations of two parallel lines be

\[ ax + by + c_1 = 0 \]

and

\[ ax + by + c_2 = 0 \]

These lines are parallel because the coefficients of \(x\) and \(y\) are the same, which means they have the same slope.

Concept and Construction

Take any point \(P(x_1, y_1)\) on the first line \(ax + by + c_1 = 0\). The perpendicular distance from this point to the second line \(ax + by + c_2 = 0\) represents the distance between the two parallel lines.

Derivation of the Formula

Since point \(P(x_1, y_1)\) lies on the first line, it satisfies

\[ ax_1 + by_1 + c_1 = 0 \]

The perpendicular distance from point \(P\) to the second line is given by the distance of a point from a line formula:

\[ d = \frac{|ax_1 + by_1 + c_2|}{\sqrt{a^2 + b^2}} \]

Substituting \(ax_1 + by_1 = -c_1\), we get

\[ d = \frac{| -c_1 + c_2 |}{\sqrt{a^2 + b^2}} \]

which simplifies to

\[ \boxed{d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}} \]

Important Observations

The distance depends only on the constants \(c_1\) and \(c_2\), not on any specific point chosen on the lines. If \(c_1 = c_2\), the distance becomes zero, meaning the two lines coincide.

Key Aspect

This formula provides a direct method to find the shortest distance between two parallel lines and is widely used in coordinate geometry for solving geometric and analytical problems.

Find the slope of the lines:
(a) Passing through the points (3, – 2) and (–1, 4),
(b) Passing through the points (3, – 2) and (7, – 2),
(c) Passing through the points (3, – 2) and (3, 4),
(d) Making inclination of 60° with the positive direction of x-axis.

Solution

(a) The slope of a line passing through two distinct points is obtained using the relation \( m=\dfrac{y_2-y_1}{x_2-x_1} \). Substituting the coordinates of the given points, we evaluate the slope as follows

\[ \begin{aligned} (3,-2),(-1,4)\\ m=\dfrac{y_2-y_1}{x_2-x_1}\\ =\dfrac{4-(-2)}{-1-3}\\ =\dfrac{6}{-4}\\ =-\dfrac{3}{2} \end{aligned} \]

Thus, the required line is decreasing and its slope is \( -\dfrac{3}{2} \)

(b) Using the same slope formula for the given points, we observe that both points have equal \( y \)-coordinates, which indicates a horizontal line

\[ \begin{aligned} (3,-2),(7,-2)\\ m=\dfrac{y_2-y_1}{x_2-x_1}\\ =\dfrac{-2-(-2)}{7-3}\\ =\dfrac{0}{4}\\ =0 \end{aligned} \]

Hence, the slope of the line is zero, confirming that the line is parallel to the \( x \)-axis

(c) For these points, the \( x \)-coordinates are identical, which suggests a vertical line. Substituting the values in the slope formula gives

\[ \begin{aligned} (3,-2),(3,4)\\ m=\dfrac{y_2-y_1}{x_2-x_1}\\ =\dfrac{4-(-2)}{3-3}\\ =\dfrac{6}{0} \end{aligned} \]

Since division by zero is not defined, the slope does not exist. Therefore, the given line is vertical and its slope is undefined

(d) When a line makes an inclination \( \theta \) with the positive direction of the \( x \)-axis, its slope is given by \( m=\tan\theta \). For the given inclination, we have

\[ \begin{aligned} \theta=60^\circ\\ m=\tan 60^\circ\\ m=\sqrt{3} \end{aligned} \]

Hence, the slope of the line is \( \sqrt{3} \)

If the angle between two lines is \(\dfrac{\pi}{4}\) and slope of one of the lines is \(\dfrac{1}{2}\), find the slope of the other line.

Solution

The relation between the slopes of two lines and the angle between them is given by \( \tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right| \). Here, the slope of one line is known and the angle between the two lines is specified

\[ \begin{aligned} m_1=\dfrac{1}{2}\\ m_2=m\\ \theta=\dfrac{\pi}{4}\\ \tan\theta=\left|\dfrac{m_2-m_1}{1+m_2m_1}\right| \end{aligned} \]

Since \( \theta=\dfrac{\pi}{4} \), we have \( \tan 45^\circ=1 \). Substituting the known values, the required condition becomes

\[ \begin{aligned} 1=\left|\dfrac{m-\dfrac{1}{2}}{1+\dfrac{1}{2}m}\right| \end{aligned} \]

This absolute value equation gives two possible cases. First, when the expression inside the modulus is equal to \(1\), we obtain

\[ \begin{aligned} \dfrac{m-\dfrac{1}{2}}{1+\dfrac{1}{2}m}=1\\ m-\dfrac{1}{2}=1+\dfrac{1}{2}m\\ m-\dfrac{1}{2}m=1+\dfrac{1}{2}\\ \dfrac{1}{2}m=\dfrac{3}{2}\\ m=3 \end{aligned} \]

Secondly, when the expression inside the modulus is equal to \(-1\), we get

\[ \begin{aligned} \dfrac{m-\dfrac{1}{2}}{1+\dfrac{1}{2}m}=-1\\ m-\dfrac{1}{2}=-1-\dfrac{1}{2}m\\ m+\dfrac{1}{2}m=-1+\dfrac{1}{2}\\ \dfrac{3}{2}m=-\dfrac{1}{2}\\ m=-\dfrac{1}{3} \end{aligned} \]

Hence, the slope of the other line can be either \( 3 \) or \( -\dfrac{1}{3} \)

Line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of \(x\).

Solution

The slope of a line joining two points is given by \( m=\dfrac{y_2-y_1}{x_2-x_1} \). We first determine the slopes of the two given lines using their respective points

\[ \begin{aligned} (-2,6),(4,8)\\ m_1=\dfrac{y_2-y_1}{x_2-x_1}\\ =\dfrac{8-6}{4-(-2)}\\ =\dfrac{2}{6}\\ =\dfrac{1}{3} \end{aligned} \]

Next, we find the slope of the second line passing through the points \( (8,12) \) and \( (x,24) \)

\[ \begin{aligned} (8,12),(x,24)\\ m_2=\dfrac{y_2-y_1}{x_2-x_1}\\ =\dfrac{24-12}{x-8}\\ =\dfrac{12}{x-8} \end{aligned} \]

Since the two lines are perpendicular, the product of their slopes is equal to \(-1\). Using this condition, we obtain

\[ \begin{aligned} m_1m_2=-1\\ \dfrac{1}{3}\cdot\dfrac{12}{x-8}=-1\\ \dfrac{4}{x-8}=-1\\ x-8=-4\\ x=4 \end{aligned} \]

Hence, the required value of \( x \) is \( 4 \)

Find the equations of the lines parallel to axes and passing through (– 2, 3).

Solution

Line passing through \((-2, 3)\) and parallel to the coordinate axes

A line parallel to the \(y\)-axis has a constant \(x\)-coordinate. Since the given point is \((-2, 3)\), the equation of the line parallel to the \(y\)-axis is

\[ x = -2 \]

A line parallel to the \(x\)-axis has a constant \(y\)-coordinate. Since the given point is \((-2, 3)\), the equation of the line parallel to the \(x\)-axis is

\[ y = 3 \]

Thus, the required equations of the lines parallel to the axes and passing through the point \((-2, 3)\) are \(x = -2\) and \(y = 3\)

Find the equation of the line through (– 2, 3) with slope – 4.

Solution

The slope of the required line is given as \(m = -4\), and the line passes through the point \((-2, 3)\)

Using the point–slope form of a straight line, \(y - y_1 = m(x - x_1)\), we substitute \(m = -4\), \(x_1 = -2\), and \(y_1 = 3\)

\[ \begin{aligned} y - y_{1} &= m\left(x - x_{1}\right) \\ y - 3 &= -4\left(x + 2\right) \\ y - 3 &= -4x - 8 \\ y &= -4x - 8 + 3 \\ y &= -4x - 5 \end{aligned} \]

Hence, the equation of the line passing through \((-2, 3)\) with slope \(-4\) is \(y = -4x - 5\)

Write the equation of the line through the points (1, –1) and (3, 5).

Solution

The line passes through the points \((1, -1)\) and \((3, 5)\)

First, we find the slope of the line using the slope formula \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)

\[ \begin{aligned} m &= \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \dfrac{5 - \left(-1\right)}{3 - 1} \\ &= \dfrac{6}{2} \\ &= 3 \end{aligned} \]

Using the point–slope form of a straight line \(y - y_1 = m(x - x_1)\) and substituting \(m = 3\) and the point \((1, -1)\)

\[ \begin{aligned} y - y_{1} &= m\left(x - x_{1}\right) \\ y + 1 &= 3\left(x - 1\right) \\ y + 1 &= 3x - 3 \\ y &= 3x - 3 - 1 \\ y &= 3x - 4 \end{aligned} \]

Hence, the equation of the line passing through \((1, -1)\) and \((3, 5)\) is \(y = 3x - 4\)

Write the equation of the lines for which \(\tan \theta =\frac{1}{2}\), where \(\theta\) is the inclination of the line and (i) y-intercept is \(-\frac{3}{2}\) (ii) x-intercept is 4.

Solution

Given that \(\tan \theta = \frac{1}{2}\), the slope of the line is \(m = \frac{1}{2}\)

For part (i), the \(y\)-intercept is \(-\frac{3}{2}\). Using the slope–intercept form of a straight line \(y = mx + c\), we substitute \(m = \frac{1}{2}\) and \(c = -\frac{3}{2}\)

\[ \begin{aligned} \tan \theta &= \dfrac{1}{2} \\ m &= \dfrac{1}{2} \\ y &= mx + c \\ y &= \dfrac{1}{2}x - \dfrac{3}{2} \\ 2y - x + 3 &= 0 \end{aligned} \]

For part (ii), the \(x\)-intercept is \(4\), so the line passes through the point \((4, 0)\). Using the point–slope form \(y - y_1 = m(x - x_1)\), we substitute \(m = \frac{1}{2}\), \(x_1 = 4\), and \(y_1 = 0\)

\[ \begin{aligned} y &= m\left(x - a\right) \\ &= \dfrac{1}{2}\left(x - 4\right) \\ 2y &= x - 4 \\ 2y - x + 4 &= 0 \end{aligned} \]

Thus, the required equations of the lines are \(2y - x + 3 = 0\) and \(2y - x + 4 = 0\)

Find the equation of the line, which makes intercepts –3 and 2 on the x- and y-axes respectively.

Solution

The given line makes an \(x\)-intercept of \(-3\) and a \(y\)-intercept of \(2\)

Using the intercept form of a straight line, \(\dfrac{x}{a} + \dfrac{y}{b} = 1\), where \(a\) and \(b\) are the \(x\)- and \(y\)-intercepts respectively, we substitute \(a = -3\) and \(b = 2\)

\[ \begin{aligned} \dfrac{x}{-3} + \dfrac{y}{2} &= 1 \\ 2x - 3y &= -6 \\ 3y - 2x - 6 &= 0 \end{aligned} \]

Hence, the equation of the line with intercepts \(-3\) on the \(x\)-axis and \(2\) on the \(y\)-axis is \(3y - 2x - 6 = 0\)

Find the distance of the point (3, – 5) from the line \(3x – 4y –26 = 0\).

Solution

The given point is \((3, -5)\), and the line is \(3x - 4y - 26 = 0\)

To find the perpendicular distance of a point from a line, we use the formula \(d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\), where \(A\), \(B\), and \(C\) are coefficients of the line and \((x_1, y_1)\) is the given point

\[ \begin{aligned} d &= \dfrac{Ax_{1} + By_{1} + C}{\sqrt{A^{2} + B^{2}}} \\ d &= \dfrac{\left| 3x - 4y - 26 \right|}{\sqrt{3^{2} + \left(-4\right)^{2}}} \\ d &= \dfrac{\left| 3 \cdot 3 - 4 \cdot \left(-5\right) - 26 \right|}{\sqrt{9 + 16}} \\ d &= \dfrac{\left| 9 + 20 - 26 \right|}{\sqrt{25}} \\ d &= \dfrac{3}{5} \end{aligned} \]

Hence, the distance of the point \((3, -5)\) from the line \(3x - 4y - 26 = 0\) is \(\frac{3}{5}\)

Find the distance between the parallel lines \(3x – 4y +7 = 0\) and \(3x – 4y + 5 = 0\)

Solution

The given equations of the parallel lines are \(3x - 4y + 7 = 0\) and \(3x - 4y + 5 = 0\)

Since both lines have the same coefficients of \(x\) and \(y\), they are parallel. The distance between two parallel lines \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\) is given by \(d = \dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}\)

\[ \begin{aligned} 3x - 4y + 7 &= 0 \\ 3x - 4y + 5 &= 0 \end{aligned} \]

\[ \begin{aligned} d &= \dfrac{c_{1} - c_{2}}{\sqrt{a^{2} + b^{2}}} \\ d &= \dfrac{7 - 5}{\sqrt{3^{2} + \left(-4\right)^{2}}} \\ d &= \dfrac{2}{\sqrt{9 + 16}} \\ d &= \dfrac{2}{5} \end{aligned} \]

Hence, the distance between the two parallel lines is \(\dfrac{2}{5}\)

If the lines \(2x+y-3=0,\; 5x+ky-3=0\) and \(3x-y-2=0\) are concurrent, find the value of k.

Solution

Since the given lines are concurrent, they must pass through one common point \((x_1, y_1)\)

Let the common point of concurrency be \((x_1, y_1)\). Then it satisfies all three equations

\[ \begin{aligned} 2x + y - 3 &= 0 \\ 5x + ky - 3 &= 0 \\ 3x - y - 2 &= 0 \end{aligned} \]

Substituting \((x_1, y_1)\) into the first two equations and eliminating \(x_1\)

\[ \begin{aligned} (2x_{1} + y_{1} - 3 = 0)\times 5 \\ (5x_{1} + ky_{1} - 3 = 0)\times 2 \\ 10x_{1} + 5y_{1} - 15 &= 0 \\ 10x_{1} + 2ky_{1} - 6 &= 0 \\ (5 - 2k)y_{1} - 9 &= 0 \\ y_{1} &= \dfrac{9}{5 - 2k} \end{aligned} \]

Now using the first and third equations to eliminate \(x_1\)

\[ \begin{aligned} (3x_{1} - y_{1} - 2 = 0)\times 2 \\ (2x_{1} + y_{1} - 3 = 0)\times 3 \\ 6x_{1} - 2y_{1} - 4 &= 0 \\ 6x_{1} + 3y_{1} - 9 &= 0 \\ -5y_{1} + 5 &= 0 \\ y_{1} &= 1 \end{aligned} \]

Equating the two expressions for \(y_1\)

\[ \begin{aligned} y_{1} &= \dfrac{9}{5 - 2k} \\ 1 &= \dfrac{9}{5 - 2k} \\ 5 - 2k &= 9 \\ -2k &= 4 \\ k &= -2 \end{aligned} \]

Hence, the value of \(k\) is \(-2\)

Find the distance of the line \(4x – y = 0\) from the point P (4, 1) measured along the line making an angle of 135° with the positive x-axis.

Solution

The given line is

\[ 4x - y = 0 \]

The point \(P(4, 1)\) lies on the line making an angle of \(135^\circ\) with the positive \(x\)-axis. The slope of this line is found using \(m = \tan \theta\)

\[ \begin{aligned} m &= \tan 135^\circ \\ &= \tan \left(90^\circ + 45^\circ\right) \\ &= -\tan 45^\circ \\ &= -1 \end{aligned} \]

Using the point–slope form of a line through \(P(4, 1)\)

\[ \begin{aligned} y - 1 &= m(x - 4) \\ y - 1 &= -1(x - 4) \\ y - 1 &= -x + 4 \\ y + x - 5 &= 0 \end{aligned} \]

To find the required distance, we determine the point of intersection of the two lines

\[ \begin{aligned} 4x - y &= 0 \\ x + y - 5 &= 0 \end{aligned} \]

\[ \begin{aligned} y &= 4x \\ x + 4x - 5 &= 0 \\ 5x &= 5 \\ x &= 1 \\ y &= 5 - x \\ y &= 4 \\ (x, y) &= (1, 4) \end{aligned} \]

The required distance is the distance between the points \((1, 4)\) and \((4, 1)\)

\[ \begin{aligned} \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \\ = \sqrt{(4 - 1)^{2} + (1 - 4)^{2}} \\ = \sqrt{3^{2} + (-3)^{2}} \\ = \sqrt{9 + 9} \\ = 3\sqrt{2} \end{aligned} \]

Hence, the distance of the line \(4x - y = 0\) from the point \(P(4, 1)\), measured along the line inclined at \(135^\circ\), is \(3\sqrt{2}\)

Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line \(x – 3y + 4 = 0\).

Solution

We are given the point \(P(1, 2)\) and the line \(x - 3y + 4 = 0\), which acts as a plane mirror. To find the image of the point, we reflect it across the given line.

For a line \(Ax + By + C = 0\), the image \((x', y')\) of a point \((x_1, y_1)\) is given by

\[ \begin{aligned} x' &= x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2} \\ y' &= y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \end{aligned} \]

Here, \(A = 1\), \(B = -3\), \(C = 4\), and \((x_1, y_1) = (1, 2)\)

\[ \begin{aligned} Ax_1 + By_1 + C &= 1(1) - 3(2) + 4 \\ &= 1 - 6 + 4 \\ &= -1 \end{aligned} \]

\[ \begin{aligned} A^2 + B^2 &= 1^2 + (-3)^2 \\ &= 1 + 9 \\ &= 10 \end{aligned} \]

\[ \begin{aligned} x' &= 1 - \frac{2(1)(-1)}{10} \\ &= 1 + \frac{2}{10} \\ &= 1 + \frac{1}{5} \\ &= \frac{6}{5} \end{aligned} \]

\[ \begin{aligned} y' &= 2 - \frac{2(-3)(-1)}{10} \\ &= 2 - \frac{6}{10} \\ &= 2 - \frac{3}{5} \\ &= \frac{7}{5} \end{aligned} \]

Hence, the image of the point \((1, 2)\) in the line \(x - 3y + 4 = 0\) is

\[ \boxed{\left(\frac{6}{5}, \frac{7}{5}\right)} \]

Show that the area of the triangle formed by the lines \(y = m_1x + c_1,\; y = m_2x + c_2\) and \(x = 0\) is \(\dfrac{(c_1-c_2)^2}{2\mid m_1-m_2\mid}\)

Solution

We are given the three lines \(y = m_1x + c_1\), \(y = m_2x + c_2\), and \(x = 0\). These lines form a triangle whose area we need to determine.

First, we find the points of intersection with the line \(x = 0\)

\[ \begin{aligned} \text{For } y = m_1x + c_1 \text{ at } x = 0 &\Rightarrow y = c_1 \\ \text{Point } A &= (0, c_1) \end{aligned} \]

\[ \begin{aligned} \text{For } y = m_2x + c_2 \text{ at } x = 0 &\Rightarrow y = c_2 \\ \text{Point } B &= (0, c_2) \end{aligned} \]

Next, we find the point of intersection of the two lines \(y = m_1x + c_1\) and \(y = m_2x + c_2\)

\[ \begin{aligned} m_1x + c_1 &= m_2x + c_2 \\ (m_1 - m_2)x &= c_2 - c_1 \\ x &= \frac{c_2 - c_1}{m_1 - m_2} \end{aligned} \]

Substituting this value of \(x\) into \(y = m_1x + c_1\)

\[ \begin{aligned} y &= m_1\left(\frac{c_2 - c_1}{m_1 - m_2}\right) + c_1 \end{aligned} \]

Thus, the third vertex of the triangle is \(C\left(\frac{c_2 - c_1}{m_1 - m_2}, \, m_1\frac{c_2 - c_1}{m_1 - m_2} + c_1\right)\)

The base of the triangle lies along the line \(x = 0\), so its length is the distance between \(A(0, c_1)\) and \(B(0, c_2)\)

\[ \begin{aligned} AB &= |c_1 - c_2| \end{aligned} \]

The perpendicular distance from point \(C\) to the line \(x = 0\) is the absolute value of its \(x\)-coordinate, which gives the height of the triangle

\[ \begin{aligned} \text{Height} &= \left| \frac{c_2 - c_1}{m_1 - m_2} \right| \end{aligned} \]

Using the formula for the area of a triangle, \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)

\[ \begin{aligned} \text{Area} &= \frac{1}{2} \cdot |c_1 - c_2| \cdot \left| \frac{c_2 - c_1}{m_1 - m_2} \right| \\ &= \frac{(c_1 - c_2)^2}{2\,|m_1 - m_2|} \end{aligned} \]

Hence, the area of the triangle formed by the given lines is \[ \boxed{\frac{(c_1 - c_2)^2}{2\,|m_1 - m_2|}} \]

A line is such that its segment between the lines \(5x – y + 4 = 0\) and \(3x + 4y – 4 = 0\) is bisected at the point (1, 5). Obtain its equation.

Solution

Let the required line pass through the given midpoint \(M(1, 5)\) and have slope \(m\). Its equation can be written in point–slope form as

\[ \begin{aligned} y - 5 = m(x - 1) \end{aligned} \]

This line intersects the given lines \(5x - y + 4 = 0\) and \(3x + 4y - 4 = 0\) at points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) respectively.

Substituting \(y = m(x - 1) + 5\) into \(5x - y + 4 = 0\)

\[ \begin{aligned} 5x - \left[m(x - 1) + 5\right] + 4 &= 0 \\ 5x - mx + m - 5 + 4 &= 0 \\ (5 - m)x + m - 1 &= 0 \\ x_1 &= \frac{1 - m}{5 - m} \end{aligned} \]

Then

\[ \begin{aligned} y_1 &= m\left(\frac{1 - m}{5 - m} - 1\right) + 5 \end{aligned} \]

Now substituting \(y = m(x - 1) + 5\) into \(3x + 4y - 4 = 0\)

\[ \begin{aligned} 3x + 4\left[m(x - 1) + 5\right] - 4 &= 0 \\ 3x + 4mx - 4m + 20 - 4 &= 0 \\ (3 + 4m)x - 4m + 16 &= 0 \\ x_2 &= \frac{4m - 16}{3 + 4m} \end{aligned} \]

Then

\[ \begin{aligned} y_2 &= m\left(\frac{4m - 16}{3 + 4m} - 1\right) + 5 \end{aligned} \]

Since \(M(1, 5)\) is the midpoint of segment \(AB\), we apply the midpoint formula

\[ \begin{aligned} \frac{x_1 + x_2}{2} &= 1 \end{aligned} \]

Substituting the expressions for \(x_1\) and \(x_2\)

\[ \begin{aligned} \frac{\frac{1 - m}{5 - m} + \frac{4m - 16}{3 + 4m}}{2} &= 1 \end{aligned} \]

Solving this equation gives

\[ \begin{aligned} m &= 1 \end{aligned} \]

Substituting \(m = 1\) into the line equation \(y - 5 = m(x - 1)\)

\[ \begin{aligned} y - 5 &= x - 1 \\ y &= x + 4 \end{aligned} \]

Hence, the equation of the required line is \[ \boxed{y = x + 4} \]

Show that the path of a moving point such that its distances from two lines \(3x – 2y = 5\) and \(3x + 2y = 5\) are equal is a straight line.

Solution

Let the moving point be \(P(x, y)\). Its distances from the two given lines \(3x - 2y = 5\) and \(3x + 2y = 5\) are equal.

Writing the lines in general form, we have

\[ \begin{aligned} 3x - 2y - 5 &= 0 \\ 3x + 2y - 5 &= 0 \end{aligned} \]

The distance of a point \((x, y)\) from a line \(Ax + By + C = 0\) is given by \(\dfrac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\)

Since the distances from both lines are equal, we write

\[ \begin{aligned} \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} &= \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} \end{aligned} \]

Since both denominators are equal, they cancel out

\[ \begin{aligned} |3x - 2y - 5| &= |3x + 2y - 5| \end{aligned} \]

This gives two possible cases

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

\[ \begin{aligned} -2y &= 2y \\ 4y &= 0 \\ y &= 0 \end{aligned} \]

The second case is

\[ \begin{aligned} 3x - 2y - 5 &= -(3x + 2y - 5) \end{aligned} \]

\[ \begin{aligned} 3x - 2y - 5 &= -3x - 2y + 5 \\ 6x &= 10 \\ x &= \frac{5}{3} \end{aligned} \]

Thus, the locus of the moving point is represented by the equations \(y = 0\) and \(x = \frac{5}{3}\), which are straight lines.

Hence, the path of the moving point is a straight line.