GSEB Class 11 Maths Solutions Chapter 10 Straight Lines Exercise 10.3

Question 1. Reduce the following equations into slope-intercept form and find their slopes and the intercept:
(i) \( x + 7y = 0 \)
(ii) \( 6x + 3y - 5 = 0 \)
(iii) \( y = 0 \)
Answer:
(i) For the equation \( x + 7y = 0 \):
\( 7y = -x \)
\( y = -\frac{1}{7}x + 0 \)
Comparing this with \( y = mx + c \), we get:
Slope \( m = -\frac{1}{7} \)
Y-intercept \( c = 0 \)
(ii) For the equation \( 6x + 3y - 5 = 0 \):
\( 3y = -6x + 5 \)
\( y = -2x + \frac{5}{3} \)
Comparing this with \( y = mx + c \), we get:
Slope \( m = -2 \)
Y-intercept \( c = \frac{5}{3} \)
(iii) For the equation \( y = 0 \):
\( y = 0x + 0 \)
Comparing this with \( y = mx + c \), we get:
Slope \( m = 0 \)
Y-intercept \( c = 0 \)
In simple words: We change each equation to the form \( y = mx + c \), where \( m \) is the slope (how steep the line is) and \( c \) is the y-intercept (where the line crosses the y-axis). Then, we simply read off the values of \( m \) and \( c \).

Exam Tip: Remember that slope-intercept form is \( y = mx + c \). Make sure to isolate \( y \) on one side and simplify the coefficients to correctly identify \( m \) and \( c \).

Question 2. Reduce the following equations into intercepts form and find their intercepts on the axes.
(i) \( 3x + 2y - 12 = 0 \)
(ii) \( 4x - 3y = 6 \)
(iii) \( 3y + 2 = 0 \)
Answer:
(i) For the equation \( 3x + 2y - 12 = 0 \):
First, move the constant term to the right side: \( 3x + 2y = 12 \).
Now, divide the entire equation by the constant term (12) to make the right side 1:
\( \frac{3x}{12} + \frac{2y}{12} = \frac{12}{12} \)
\( \frac{x}{4} + \frac{y}{6} = 1 \)
Comparing this with the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \), we find:
X-intercept \( a = 4 \)
Y-intercept \( b = 6 \)
(ii) For the equation \( 4x - 3y = 6 \):
The constant term is already on the right side. Divide the entire equation by 6:
\( \frac{4x}{6} - \frac{3y}{6} = \frac{6}{6} \)
\( \frac{2x}{3} - \frac{y}{2} = 1 \)
To match the standard form \( \frac{x}{a} + \frac{y}{b} = 1 \), rewrite \( \frac{2x}{3} \) as \( \frac{x}{3/2} \) and \( -\frac{y}{2} \) as \( \frac{y}{-2} \):
\( \frac{x}{3/2} + \frac{y}{-2} = 1 \)
Comparing, we get:
X-intercept \( a = \frac{3}{2} \)
Y-intercept \( b = -2 \)
(iii) For the equation \( 3y + 2 = 0 \):
First, move the constant term to the right side: \( 3y = -2 \).
Divide the equation by -2:
\( \frac{3y}{-2} = \frac{-2}{-2} \)
\( \frac{y}{-2/3} = 1 \)
This can also be written as \( \frac{x}{0} + \frac{y}{-2/3} = 1 \) to show the lack of an x-intercept in a format similar to intercept form.
Comparing, we find:
X-intercept \( a = 0 \) (indicating the line is parallel to the x-axis and does not intersect it, or it passes through the origin which is not the case here for `y=-2/3`)
Y-intercept \( b = -\frac{2}{3} \)
In simple words: We rewrite each line's equation into the "intercept form" \( \frac{x}{a} + \frac{y}{b} = 1 \). Here, \( a \) tells us where the line crosses the x-axis, and \( b \) tells us where it crosses the y-axis. We get this form by making the right side of the equation equal to 1.

Exam Tip: For lines parallel to an axis (like \( 3y + 2 = 0 \)), one of the intercepts will be undefined or expressed as 0 (as shown for x-intercept here) when forcing it into the \( \frac{x}{a} + \frac{y}{b} = 1 \) format. Remember to manage negative signs carefully for the intercepts.

Question 3. Reduce the following equations into the normal form. Find their perpendicular distance from the origin and angle between perpendicular and positive direction of x-axis.
(i) \( x - \sqrt{3}y + 8 = 0 \)
(ii) \( y - 2 = 0 \)
(iii) \( x - y = 4 \)
Answer:
The normal form of a line is \( x \cos \omega + y \sin \omega = p \), where \( p \) is the perpendicular distance from the origin to the line, and \( \omega \) is the angle that the normal (perpendicular from origin) makes with the positive x-axis.

(i) For the equation \( x - \sqrt{3}y + 8 = 0 \):
First, ensure the constant term \( C \) is positive when the equation is written as \( Ax + By + C = 0 \). Here, \( C=8 \), which is positive. So, we rewrite the equation as \( -x + \sqrt{3}y = 8 \).
Compare this with \( Ax + By = C \), so \( A = -1 \), \( B = \sqrt{3} \), \( C = 8 \).
Calculate \( r = \sqrt{A^2 + B^2} = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2 \).
Divide the equation \( -x + \sqrt{3}y = 8 \) by \( r = 2 \):
\( -\frac{1}{2}x + \frac{\sqrt{3}}{2}y = \frac{8}{2} \)
\( -\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4 \)
Comparing with \( x \cos \omega + y \sin \omega = p \):
\( \cos \omega = -\frac{1}{2} \) and \( \sin \omega = \frac{\sqrt{3}}{2} \)
These values indicate that \( \omega \) is in the second quadrant.
Therefore, \( \omega = 120^\circ \) or \( \frac{2\pi}{3} \) radians.
The perpendicular distance from the origin is \( p = 4 \).

(ii) For the equation \( y - 2 = 0 \):
Rewrite as \( y = 2 \). This is a horizontal line.
This can be expressed in the normal form as \( x \cos \frac{\pi}{2} + y \sin \frac{\pi}{2} = 2 \).
Comparing with \( x \cos \omega + y \sin \omega = p \):
\( \cos \omega = \cos \frac{\pi}{2} = 0 \) and \( \sin \omega = \sin \frac{\pi}{2} = 1 \).
Therefore, \( \omega = \frac{\pi}{2} \) radians (or \( 90^\circ \)).
The perpendicular distance from the origin is \( p = 2 \).

(iii) For the equation \( x - y = 4 \):
Here, \( A = 1 \), \( B = -1 \), \( C = 4 \).
Calculate \( r = \sqrt{A^2 + B^2} = \sqrt{(1)^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2} \).
Divide the equation \( x - y = 4 \) by \( r = \sqrt{2} \):
\( \frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = \frac{4}{\sqrt{2}} \)
\( \frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = 2\sqrt{2} \)
Comparing with \( x \cos \omega + y \sin \omega = p \):
\( \cos \omega = \frac{1}{\sqrt{2}} \) and \( \sin \omega = -\frac{1}{\sqrt{2}} \)
These values mean \( \omega \) is in the fourth quadrant.
Therefore, \( \omega = 360^\circ - 45^\circ = 315^\circ \) or \( \frac{7\pi}{4} \) radians.
The perpendicular distance from the origin is \( p = 2\sqrt{2} \).
In simple words: We change the line's equation into a special "normal form" \( x \cos \omega + y \sin \omega = p \). In this form, \( p \) is the shortest distance from the origin (the point (0,0)) to the line, and \( \omega \) is the angle that this shortest distance line makes with the positive x-axis. We find \( p \) and \( \omega \) for each equation.

Exam Tip: When converting to normal form \( x \cos \omega + y \sin \omega = p \), always make sure \( p \) is positive. If the constant term `C` in \( Ax+By=C \) is negative, multiply the entire equation by -1 before dividing by \( \sqrt{A^2+B^2} \). Be careful with the quadrant of \( \omega \) based on the signs of \( \cos \omega \) and \( \sin \omega \).

Question 4. Find the distance of the point (- 1, 1) from the line 12(x + 6) = 5(y − 2).
Answer:
First, we need to rewrite the given line equation in the general form \( Ax + By + C = 0 \).
The given equation is \( 12(x + 6) = 5(y - 2) \).
Distribute the numbers:
\( 12x + 72 = 5y - 10 \)
Move all terms to one side:
\( 12x - 5y + 72 + 10 = 0 \)
\( 12x - 5y + 82 = 0 \)
Now, we have the line in general form with \( A=12 \), \( B=-5 \), and \( C=82 \).
The given point is \( (x_1, y_1) = (-1, 1) \).
The formula for the perpendicular distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:
\( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)
Substitute the values into the formula:
\( d = \frac{|12(-1) + (-5)(1) + 82|}{\sqrt{12^2 + (-5)^2}} \)
\( d = \frac{|-12 - 5 + 82|}{\sqrt{144 + 25}} \)
\( d = \frac{|65|}{\sqrt{169}} \)
\( d = \frac{65}{13} \)
\( d = 5 \)
Thus, the distance of the point \( (-1, 1) \) from the line is 5 units.
In simple words: We first change the line's equation to a standard form. Then, we use a special formula that takes the point's coordinates and the line's coefficients to calculate the shortest distance between the point and the line.

Exam Tip: Always convert the line equation to the standard form \( Ax + By + C = 0 \) before applying the distance formula. Remember that distance is always a positive value, hence the absolute value in the numerator.

Question 5. Find the points on the x-axis, whose distances from \( \frac{x}{3} + \frac{y}{4} = 1 \) are 4 units.
Answer:
First, convert the given line equation into the general form \( Ax + By + C = 0 \).
The equation is \( \frac{x}{3} + \frac{y}{4} = 1 \).
Multiply by the least common multiple of 3 and 4, which is 12:
\( 12 \left( \frac{x}{3} \right) + 12 \left( \frac{y}{4} \right) = 12(1) \)
\( 4x + 3y = 12 \)
Rewrite in general form:
\( 4x + 3y - 12 = 0 \)
Let the point on the x-axis be \( (x_1, 0) \), since any point on the x-axis has a y-coordinate of 0.
We are given that the distance from \( (x_1, 0) \) to the line \( 4x + 3y - 12 = 0 \) is 4 units.
Using the distance formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \):
\( 4 = \frac{|4(x_1) + 3(0) - 12|}{\sqrt{4^2 + 3^2}} \)
\( 4 = \frac{|4x_1 - 12|}{\sqrt{16 + 9}} \)
\( 4 = \frac{|4x_1 - 12|}{\sqrt{25}} \)
\( 4 = \frac{|4x_1 - 12|}{5} \)
Multiply both sides by 5:
\( 20 = |4x_1 - 12| \)
This absolute value equation gives two possibilities:
Case 1: \( 4x_1 - 12 = 20 \)
\( 4x_1 = 20 + 12 \)
\( 4x_1 = 32 \)
\( x_1 = \frac{32}{4} \)
\( x_1 = 8 \)
Case 2: \( 4x_1 - 12 = -20 \)
\( 4x_1 = -20 + 12 \)
\( 4x_1 = -8 \)
\( x_1 = \frac{-8}{4} \)
\( x_1 = -2 \)
Therefore, the required points on the x-axis are \( (8, 0) \) and \( (-2, 0) \).
In simple words: We want to find points on the x-axis that are exactly 4 units away from the given line. First, we convert the line's equation into its general form. Since the points are on the x-axis, their y-coordinate is 0. We then use the distance formula, setting the distance to 4, and solve for the x-coordinate. This gives us two possible x-coordinates, and thus two points.

Exam Tip: When solving equations involving absolute values, remember to consider both the positive and negative cases. Points on the x-axis always have a y-coordinate of zero, which simplifies the application of the distance formula.

Question 6. Find the distance between the parallel lines:
(i) \( 15x + 8y - 34 = 0 \) and \( 15x + 8y + 31 = 0 \).
(ii) \( l(x + y) + p = 0 \) and \( lx + ly - r = 0 \).
Answer:
(i) For the lines \( 15x + 8y - 34 = 0 \) and \( 15x + 8y + 31 = 0 \):
The lines are in the form \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \).
Here, \( A = 15 \), \( B = 8 \), \( C_1 = -34 \), and \( C_2 = 31 \).
The distance between two parallel lines is given by the formula:
\( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \)
Substitute the values:
\( d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}} \)
\( d = \frac{|-65|}{\sqrt{225 + 64}} \)
\( d = \frac{65}{\sqrt{289}} \)
\( d = \frac{65}{17} \)
Alternatively, we can find a point on one line and calculate its distance to the other line.
Let's take \( 15x + 8y - 34 = 0 \). If \( y = 0 \), then \( 15x = 34 \implies x = \frac{34}{15} \).
So, a point on the first line is \( (\frac{34}{15}, 0) \).
Distance from this point to \( 15x + 8y + 31 = 0 \):
\( d = \frac{|15(\frac{34}{15}) + 8(0) + 31|}{\sqrt{15^2 + 8^2}} \)
\( d = \frac{|34 + 0 + 31|}{\sqrt{225 + 64}} \)
\( d = \frac{|65|}{\sqrt{289}} = \frac{65}{17} \)

(ii) For the lines \( l(x + y) + p = 0 \) and \( lx + ly - r = 0 \):
Rewrite the first line as \( lx + ly + p = 0 \).
The second line is \( lx + ly - r = 0 \).
These lines are in the form \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \).
Here, \( A = l \), \( B = l \), \( C_1 = p \), and \( C_2 = -r \).
Using the distance formula:
\( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \)
\( d = \frac{|p - (-r)|}{\sqrt{l^2 + l^2}} \)
\( d = \frac{|p + r|}{\sqrt{2l^2}} \)
Assuming \( l > 0 \), \( \sqrt{2l^2} = l\sqrt{2} \).
\( d = \frac{|p + r|}{l\sqrt{2}} \)
In simple words: For parallel lines, the distance between them is constant. We use a formula that takes the constant terms of the two line equations and the coefficients of x and y (which are the same for parallel lines) to calculate this distance.

Exam Tip: Ensure that the parallel lines are written in the same general form, \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \), before applying the distance formula \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \). This means the coefficients A and B must be identical.

Question 7. Find the equation of a line parallel to the line \( 3x - 4y + 2 = 0 \) and passing through the point (- 2, 3).
Answer:
We are given the line \( 3x - 4y + 2 = 0 \).
To find the slope of this line, rearrange it into slope-intercept form \( y = mx + c \):
\( -4y = -3x - 2 \)
\( y = \frac{-3}{-4}x - \frac{2}{-4} \)
\( y = \frac{3}{4}x + \frac{1}{2} \)
The slope of the given line is \( m_1 = \frac{3}{4} \).
Since the required line is parallel to this line, it will have the same slope.
So, the slope of the parallel line is \( m_2 = \frac{3}{4} \).
The parallel line passes through the point \( (x_1, y_1) = (-2, 3) \).
Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \):
\( y - 3 = \frac{3}{4}(x - (-2)) \)
\( y - 3 = \frac{3}{4}(x + 2) \)
Multiply both sides by 4 to remove the fraction:
\( 4(y - 3) = 3(x + 2) \)
\( 4y - 12 = 3x + 6 \)
Rearrange into the general form \( Ax + By + C = 0 \):
\( 0 = 3x - 4y + 6 + 12 \)
\( 3x - 4y + 18 = 0 \)
Alternatively, any line parallel to \( 3x - 4y + 2 = 0 \) can be written as \( 3x - 4y + k = 0 \).
Since this line passes through \( (-2, 3) \), substitute these values into the equation:
\( 3(-2) - 4(3) + k = 0 \)
\( -6 - 12 + k = 0 \)
\( -18 + k = 0 \)
\( k = 18 \)
Substitute \( k = 18 \) back into the parallel line equation:
\( 3x - 4y + 18 = 0 \)
In simple words: To find a parallel line, we know it must have the same "steepness" (slope) as the original line. We find the slope of the first line. Then, using that slope and the given point it passes through, we can write the equation for the new parallel line.

Exam Tip: Remember that parallel lines have identical slopes. An easy way to find the equation of a parallel line is to use the form \( Ax + By + k = 0 \) if the original line is \( Ax + By + C = 0 \), and then solve for \( k \) using the given point.

Question 8. Find the equation of the line perpendicular to the line \( x - 7y + 5 = 0 \) and having x-intercept 3.
Answer:
The required line has an x-intercept of 3, meaning it passes through the point \( (3, 0) \).
First, find the slope of the given line \( x - 7y + 5 = 0 \).
Rearrange into slope-intercept form \( y = mx + c \):
\( -7y = -x - 5 \)
\( y = \frac{-x}{-7} - \frac{5}{-7} \)
\( y = \frac{1}{7}x + \frac{5}{7} \)
The slope of the given line is \( m_1 = \frac{1}{7} \).
Since the required line is perpendicular to this line, its slope \( m_2 \) will be the negative reciprocal of \( m_1 \).
\( m_2 = -\frac{1}{m_1} = -\frac{1}{1/7} = -7 \).
Now we have the slope \( m_2 = -7 \) and a point \( (x_1, y_1) = (3, 0) \) through which the required line passes.
Using the point-slope form \( y - y_1 = m_2(x - x_1) \):
\( y - 0 = -7(x - 3) \)
\( y = -7x + 21 \)
Rearrange into the general form \( Ax + By + C = 0 \):
\( 7x + y - 21 = 0 \)
Alternatively, a line perpendicular to \( Ax + By + C = 0 \) is of the form \( Bx - Ay + k = 0 \).
For the given line \( x - 7y + 5 = 0 \) (where \( A=1, B=-7 \)), a perpendicular line would be \( (-7)x - (1)y + k = 0 \), which simplifies to \( -7x - y + k = 0 \) or \( 7x + y - k = 0 \). We can write it as \( 7x + y = k \).
Since this line passes through \( (3, 0) \), substitute these values to find \( k \):
\( 7(3) + 0 = k \)
\( 21 = k \)
So, the equation of the required line is \( 7x + y = 21 \).
In simple words: We need to find a line that cuts another line at a 90-degree angle and crosses the x-axis at 3. First, we find the "steepness" (slope) of the given line. Then, we calculate the negative reciprocal to get the slope of our perpendicular line. Finally, using this new slope and the point (3,0) where it crosses the x-axis, we write its equation.

Exam Tip: Perpendicular lines have slopes that are negative reciprocals of each other (i.e., \( m_1 m_2 = -1 \)). Always remember that an x-intercept of \( a \) means the line passes through the point \( (a, 0) \).

Question 9. Find the angles between the lines \( \sqrt{3}x + y = 1 \) and \( x + \sqrt{3}y = 1 \).
Answer:
First, find the slopes of both lines.
For the first line: \( \sqrt{3}x + y = 1 \)
Rearrange to slope-intercept form \( y = mx + c \):
\( y = -\sqrt{3}x + 1 \)
The slope of the first line is \( m_1 = -\sqrt{3} \).
For the second line: \( x + \sqrt{3}y = 1 \)
Rearrange to slope-intercept form:
\( \sqrt{3}y = -x + 1 \)
\( y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \)
The slope of the second line is \( m_2 = -\frac{1}{\sqrt{3}} \).
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by the formula:
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \)
Substitute the values of \( m_1 \) and \( m_2 \):
\( \tan \theta = \left| \frac{-\sqrt{3} - \left(-\frac{1}{\sqrt{3}}\right)}{1 + (-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)} \right| \)
\( \tan \theta = \left| \frac{-\sqrt{3} + \frac{1}{\sqrt{3}}}{1 + 1} \right| \)
Simplify the numerator:
\( -\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{-\sqrt{3} \times \sqrt{3} + 1}{\sqrt{3}} = \frac{-3 + 1}{\sqrt{3}} = \frac{-2}{\sqrt{3}} \)
Substitute back into the \( \tan \theta \) formula:
\( \tan \theta = \left| \frac{\frac{-2}{\sqrt{3}}}{2} \right| \)
\( \tan \theta = \left| -\frac{2}{\sqrt{3}} \times \frac{1}{2} \right| \)
\( \tan \theta = \left| -\frac{1}{\sqrt{3}} \right| \)
\( \tan \theta = \frac{1}{\sqrt{3}} \)
This implies that the acute angle \( \theta = 30^\circ \) or \( \frac{\pi}{6} \) radians.
The obtuse angle between the lines would be \( 180^\circ - 30^\circ = 150^\circ \) or \( \pi - \frac{\pi}{6} = \frac{5\pi}{6} \) radians.
Thus, the angles between the lines are \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \).
In simple words: We find how steep each line is (its slope). Then, we use a specific formula involving these slopes to figure out the angle where the two lines cross. Since there are always two angles where lines cross (an acute and an obtuse one), we find both.

Exam Tip: Remember the formula for the angle between two lines, \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \). The absolute value ensures you find the acute angle first. If the lines are perpendicular, \( 1 + m_1 m_2 = 0 \), making \( \tan \theta \) undefined, meaning \( \theta = 90^\circ \).

Question 10. The line through the points (h, 3) and (4, 1) intersects the line \( 7x - 9y - 19 = 0 \) at right angles. Find the value of h?
Answer:
Let the first line pass through points \( P(h, 3) \) and \( Q(4, 1) \).
The slope of the line PQ, \( m_{PQ} \), is given by:
\( m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{4 - h} = \frac{-2}{4 - h} \)
The second line is \( 7x - 9y - 19 = 0 \).
To find its slope, \( m_{AB} \), rearrange to slope-intercept form \( y = mx + c \):
\( -9y = -7x + 19 \)
\( y = \frac{7}{9}x - \frac{19}{9} \)
The slope of the second line is \( m_{AB} = \frac{7}{9} \).
Since the two lines intersect at right angles, they are perpendicular.
The product of their slopes must be -1: \( m_{PQ} \times m_{AB} = -1 \).
\( \left( \frac{-2}{4 - h} \right) \times \left( \frac{7}{9} \right) = -1 \)
\( \frac{-14}{9(4 - h)} = -1 \)
Multiply both sides by -1 to cancel the negative signs:
\( \frac{14}{9(4 - h)} = 1 \)
\( 14 = 9(4 - h) \)
\( 14 = 36 - 9h \)
Rearrange to solve for \( h \):
\( 9h = 36 - 14 \)
\( 9h = 22 \)
\( h = \frac{22}{9} \)
Therefore, the value of \( h \) is \( \frac{22}{9} \).
In simple words: We have two lines that cross each other at a perfect right angle. One line goes through two points, one of which has a missing coordinate 'h'. We find the "steepness" (slope) of both lines. Since they are perpendicular, we know their slopes multiplied together equal -1. We use this fact to solve for the missing 'h' value.

Exam Tip: When given two points, use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). For perpendicular lines, the product of their slopes is -1. Ensure you handle algebraic manipulations, especially with negative signs, carefully.

Question 11. Prove that the line through the point \( (x_1, y_1) \) and parallel to the line \( Ax + By + C = 0 \) is \( A(x - x_1) + B(y - y_1) = 0 \).
Answer:
**Proof:**
We are given a line with the equation \( Ax + By + C = 0 \).
To find the slope of this line, rearrange it into slope-intercept form \( y = mx + c \):
\( By = -Ax - C \)
\( y = -\frac{A}{B}x - \frac{C}{B} \)
The slope of this line is \( m = -\frac{A}{B} \).
Since the required line is parallel to \( Ax + By + C = 0 \), it will have the same slope, \( m' = -\frac{A}{B} \).
The required line passes through the point \( (x_1, y_1) \).
Using the point-slope form of a line, \( y - y_1 = m'(x - x_1) \):
\( y - y_1 = -\frac{A}{B}(x - x_1) \)
Multiply both sides by \( B \) to eliminate the fraction:
\( B(y - y_1) = -A(x - x_1) \)
Move the term from the right side to the left side:
\( A(x - x_1) + B(y - y_1) = 0 \)
Thus, the equation of the line through \( (x_1, y_1) \) and parallel to \( Ax + By + C = 0 \) is \( A(x - x_1) + B(y - y_1) = 0 \).
**Alternative Method:**
Any line parallel to \( Ax + By + C = 0 \) can be written in the form \( Ax + By + K = 0 \) for some constant \( K \).
Let this be equation (1): \( Ax + By + K = 0 \).
Since this parallel line passes through the point \( (x_1, y_1) \), we can substitute these coordinates into equation (1):
\( Ax_1 + By_1 + K = 0 \)
From this, we can find the value of \( K \):
\( K = -(Ax_1 + By_1) \)
Now, substitute this value of \( K \) back into equation (1):
\( Ax + By - (Ax_1 + By_1) = 0 \)
\( Ax + By - Ax_1 - By_1 = 0 \)
Rearrange by grouping terms with \( A \) and \( B \):
\( A(x - x_1) + B(y - y_1) = 0 \)
Both methods prove the given statement.
In simple words: We want to prove a formula for a line that is parallel to another line and passes through a specific point. We start by finding the "steepness" (slope) of the original line. Since parallel lines have the same steepness, we use this slope along with the given point to write the new line's equation using the point-slope form. After rearranging, it matches the formula we needed to prove.

Exam Tip: For proofs, clearly state your assumptions and use standard formulas (like slope formula and point-slope form) step-by-step. The alternative method (using \( Ax + By + K = 0 \)) is often quicker for parallel and perpendicular lines.

Question 12. Two lines passing through the point (2, 3) intersect each other at an angle of 60°. If the slope of one line is 2, find the equation of the other line.
Answer:
Let the slope of the first line be \( m_1 = 2 \).
Let the slope of the other line be \( m \).
The angle of intersection between the two lines is \( \theta = 60^\circ \).
The formula for the angle between two lines is:
\( \tan \theta = \left| \frac{m_1 - m}{1 + m_1 m} \right| \)
Substitute the known values:
\( \tan 60^\circ = \left| \frac{2 - m}{1 + 2m} \right| \)
\( \sqrt{3} = \left| \frac{2 - m}{1 + 2m} \right| \)
This gives us two cases:

**Case 1: Positive value**
\( \sqrt{3} = \frac{2 - m}{1 + 2m} \)
\( \sqrt{3}(1 + 2m) = 2 - m \)
\( \sqrt{3} + 2\sqrt{3}m = 2 - m \)
Move terms with \( m \) to one side and constants to the other:
\( 2\sqrt{3}m + m = 2 - \sqrt{3} \)
\( m(2\sqrt{3} + 1) = 2 - \sqrt{3} \)
\( m = \frac{2 - \sqrt{3}}{2\sqrt{3} + 1} \)
To rationalize the denominator, multiply by the conjugate \( (2\sqrt{3} - 1) \):
\( m = \frac{(2 - \sqrt{3})(2\sqrt{3} - 1)}{(2\sqrt{3} + 1)(2\sqrt{3} - 1)} \)
\( m = \frac{4\sqrt{3} - 2 - 6 + \sqrt{3}}{12 - 1} \)
\( m = \frac{5\sqrt{3} - 8}{11} \)
The equation of the line passing through \( (2, 3) \) with slope \( m = \frac{5\sqrt{3} - 8}{11} \) is \( y - 3 = \frac{5\sqrt{3} - 8}{11}(x - 2) \).
\( 11(y - 3) = (5\sqrt{3} - 8)(x - 2) \)
\( 11y - 33 = (5\sqrt{3} - 8)x - 2(5\sqrt{3} - 8) \)
\( 11y - 33 = (5\sqrt{3} - 8)x - 10\sqrt{3} + 16 \)
Rearranging to general form \( Ax + By + C = 0 \):
\( (5\sqrt{3} - 8)x - 11y + 16 + 33 - 10\sqrt{3} = 0 \)
\( (5\sqrt{3} - 8)x - 11y + 49 - 10\sqrt{3} = 0 \)

**Case 2: Negative value**
\( -\sqrt{3} = \frac{2 - m}{1 + 2m} \)
\( -\sqrt{3}(1 + 2m) = 2 - m \)
\( -\sqrt{3} - 2\sqrt{3}m = 2 - m \)
Move terms with \( m \) to one side and constants to the other:
\( m - 2\sqrt{3}m = 2 + \sqrt{3} \)
\( m(1 - 2\sqrt{3}) = 2 + \sqrt{3} \)
\( m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \)
To rationalize the denominator, multiply by the conjugate \( (1 + 2\sqrt{3}) \):
\( m = \frac{(2 + \sqrt{3})(1 + 2\sqrt{3})}{(1 - 2\sqrt{3})(1 + 2\sqrt{3})} \)
\( m = \frac{2 + 4\sqrt{3} + \sqrt{3} + 6}{1 - 12} \)
\( m = \frac{8 + 5\sqrt{3}}{-11} \)
\( m = -\frac{8 + 5\sqrt{3}}{11} \)
The equation of the line passing through \( (2, 3) \) with slope \( m = -\frac{8 + 5\sqrt{3}}{11} \) is \( y - 3 = -\frac{8 + 5\sqrt{3}}{11}(x - 2) \).
\( 11(y - 3) = -(8 + 5\sqrt{3})(x - 2) \)
\( 11y - 33 = -(8 + 5\sqrt{3})x + 2(8 + 5\sqrt{3}) \)
\( 11y - 33 = -(8 + 5\sqrt{3})x + 16 + 10\sqrt{3} \)
Rearranging to general form \( Ax + By + C = 0 \):
\( (8 + 5\sqrt{3})x + 11y - 33 - 16 - 10\sqrt{3} = 0 \)
\( (8 + 5\sqrt{3})x + 11y - 49 - 10\sqrt{3} = 0 \)
Therefore, the two possible equations for the other line are:
1. \( (5\sqrt{3} - 8)x - 11y + 49 - 10\sqrt{3} = 0 \)
2. \( (8 + 5\sqrt{3})x + 11y - 49 - 10\sqrt{3} = 0 \)
*(Note: The solution in the source has simplified/rearranged constants slightly differently but the core form matches.)*
In simple words: We have one line's steepness (slope) and know it crosses another line at a 60-degree angle, both passing through the same point. We use a formula that relates the slopes and the angle between them. Since a 60-degree angle can result from two different slopes, we solve for both possibilities. For each possible slope, we then write the equation of the line using the given point.

Exam Tip: Remember that the absolute value in the angle formula \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \) leads to two possible values for \( m_2 \). You must consider both the positive and negative cases to find all possible equations for the other line. Be thorough with radical arithmetic.

Question 13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Answer:
Let the two given points be \( A(-1, 2) \) and \( B(3, 4) \).
A right bisector (or perpendicular bisector) is a line that is perpendicular to the line segment AB and passes through its midpoint.

**Step 1: Find the midpoint of the line segment AB.**
The midpoint \( M \) of a segment with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
\( M = \left( \frac{-1 + 3}{2}, \frac{2 + 4}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3) \)

**Step 2: Find the slope of the line segment AB.**
The slope \( m_{AB} \) is given by \( \frac{y_2 - y_1}{x_2 - x_1} \).
\( m_{AB} = \frac{4 - 2}{3 - (-1)} = \frac{2}{3 + 1} = \frac{2}{4} = \frac{1}{2} \)

**Step 3: Find the slope of the right bisector.**
Since the right bisector is perpendicular to AB, its slope \( m_{\text{bisector}} \) will be the negative reciprocal of \( m_{AB} \).
\( m_{\text{bisector}} = -\frac{1}{m_{AB}} = -\frac{1}{1/2} = -2 \)

**Step 4: Find the equation of the right bisector.**
The right bisector passes through the midpoint \( M(1, 3) \) and has a slope of \( -2 \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 3 = -2(x - 1) \)
\( y - 3 = -2x + 2 \)
Rearrange into the general form \( Ax + By + C = 0 \):
\( 2x + y - 3 - 2 = 0 \)
\( 2x + y - 5 = 0 \)
Thus, the equation of the right bisector is \( 2x + y - 5 = 0 \).
In simple words: A "right bisector" is a line that cuts another line segment exactly in half and at a 90-degree angle. First, we find the middle point of the segment. Then, we find how steep the segment is (its slope). The right bisector's steepness will be the opposite and upside-down of the segment's steepness. Finally, we use this new steepness and the middle point to write the equation of the right bisector.

Exam Tip: Always remember the two key properties of a right bisector: it passes through the midpoint of the line segment, and it is perpendicular to the line segment. Make sure to calculate both the midpoint and the perpendicular slope accurately.

Question 14. Find the co-ordinates of the foot of the perpendicular from a point (- 1, 3) to the line \( 3x - 4y - 16 = 0 \).
Answer:
Let the given point be \( P(-1, 3) \) and the given line be \( L_1: 3x - 4y - 16 = 0 \).
The foot of the perpendicular is the point where a line passing through \( P \) and perpendicular to \( L_1 \) intersects \( L_1 \).

**Step 1: Find the slope of the given line \( L_1 \).**
Rearrange \( 3x - 4y - 16 = 0 \) into slope-intercept form \( y = mx + c \):
\( -4y = -3x + 16 \)
\( y = \frac{3}{4}x - 4 \)
The slope of \( L_1 \) is \( m_1 = \frac{3}{4} \).

**Step 2: Find the slope of the perpendicular line \( L_2 \).**
Since \( L_2 \) is perpendicular to \( L_1 \), its slope \( m_2 \) is the negative reciprocal of \( m_1 \).
\( m_2 = -\frac{1}{m_1} = -\frac{1}{3/4} = -\frac{4}{3} \).

**Step 3: Find the equation of the perpendicular line \( L_2 \).**
Line \( L_2 \) passes through \( P(-1, 3) \) and has a slope of \( -\frac{4}{3} \).
Using the point-slope form \( y - y_1 = m_2(x - x_1) \):
\( y - 3 = -\frac{4}{3}(x - (-1)) \)
\( y - 3 = -\frac{4}{3}(x + 1) \)
Multiply by 3:
\( 3(y - 3) = -4(x + 1) \)
\( 3y - 9 = -4x - 4 \)
Rearrange to general form:
\( 4x + 3y - 9 + 4 = 0 \)
\( L_2: 4x + 3y - 5 = 0 \).

**Step 4: Find the point of intersection of \( L_1 \) and \( L_2 \).**
We need to solve the system of equations:
1. \( 3x - 4y = 16 \)
2. \( 4x + 3y = 5 \)
We can use the cross-multiplication method or substitution/elimination.
Using cross-multiplication for \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \)
For \( 3x - 4y - 16 = 0 \): \( a_1=3, b_1=-4, c_1=-16 \)
For \( 4x + 3y - 5 = 0 \): \( a_2=4, b_2=3, c_2=-5 \)
\( \frac{x}{(-4)(-5) - (3)(-16)} = \frac{y}{(-16)(4) - (-5)(3)} = \frac{1}{(3)(3) - (4)(-4)} \)
\( \frac{x}{20 - (-48)} = \frac{y}{-64 - (-15)} = \frac{1}{9 - (-16)} \)
\( \frac{x}{20 + 48} = \frac{y}{-64 + 15} = \frac{1}{9 + 16} \)
\( \frac{x}{68} = \frac{y}{-49} = \frac{1}{25} \)
From this:
\( x = \frac{68}{25} \)
\( y = -\frac{49}{25} \)
The coordinates of the foot of the perpendicular are \( \left( \frac{68}{25}, -\frac{49}{25} \right) \).
In simple words: We want to find the exact spot on a line where a perpendicular line, starting from a given point, touches it. First, we find the steepness (slope) of the given line. Then, we find the slope of a line that would be perpendicular to it. Using this perpendicular slope and the given point, we write the equation of this new perpendicular line. Finally, we find where these two lines cross, which gives us the required coordinates.

Exam Tip: The "foot of the perpendicular" is simply the point of intersection between the original line and the line drawn perpendicular to it from the given point. Use standard formulas for slope of perpendicular lines and solving simultaneous equations (e.g., cross-multiplication) to find the intersection point.

Question 15. The perpendicular from the origin to the line \( y = mx + c \) meets it at the point (- 1, 2). Find the value of m and c.
Answer:
Let the given line be \( L: y = mx + c \).
The perpendicular from the origin \( O(0, 0) \) to line \( L \) meets it at point \( M(-1, 2) \).
This means the line segment OM is perpendicular to line L, and point M lies on line L.

**Step 1: Find the slope of the line segment OM.**
The slope of OM, \( m_{OM} \), is given by \( \frac{y_2 - y_1}{x_2 - x_1} \).
\( m_{OM} = \frac{2 - 0}{-1 - 0} = \frac{2}{-1} = -2 \).

**Step 2: Use the perpendicularity condition to find \( m \).**
Since OM is perpendicular to line L, the product of their slopes is -1.
\( m \times m_{OM} = -1 \)
\( m \times (-2) = -1 \)
\( m = \frac{-1}{-2} \)
\( m = \frac{1}{2} \).

**Step 3: Use the fact that M lies on line L to find \( c \).**
The point \( M(-1, 2) \) lies on the line \( y = mx + c \). Substitute its coordinates and the value of \( m \) we found:
\( 2 = \left(\frac{1}{2}\right)(-1) + c \)
\( 2 = -\frac{1}{2} + c \)
To solve for \( c \), add \( \frac{1}{2} \) to both sides:
\( c = 2 + \frac{1}{2} \)
\( c = \frac{4}{2} + \frac{1}{2} \)
\( c = \frac{5}{2} \)
Therefore, the value of \( m \) is \( \frac{1}{2} \) and the value of \( c \) is \( \frac{5}{2} \).
In simple words: We have a line (whose equation has unknown \( m \) and \( c \)) and a point where a perpendicular line from the origin touches it. First, we find the steepness (slope) of the line from the origin to that point. Since this line is perpendicular to our main line, we can find the slope \( m \) of our main line. Then, because the point also lies on our main line, we can use its coordinates and the newly found \( m \) to calculate the missing \( c \) value.

Exam Tip: Understand that the "perpendicular from the origin to the line" means you have a segment connecting \( (0,0) \) to the given point, and this segment is normal to the line. This allows you to use both the slope and point-on-line conditions.

Question 16. If p and q are the lengths of perpendiculars from the origin to the lines \( x\cos \theta - y\sin \theta = k\cos 2\theta \) and \( x \sec \theta + y \csc \theta = k \) respectively, prove that \( p^2 + 4q^2 = k^2 \).
Answer:
**Part 1: Find \( p \)**
The first line is \( x\cos \theta - y\sin \theta = k\cos 2\theta \).
This equation is already in the normal form \( x\cos \omega + y\sin \omega = p' \), where \( \cos \omega = \cos \theta \), \( \sin \omega = -\sin \theta \), and \( p' = k\cos 2\theta \).
The length of the perpendicular from the origin \( (0,0) \) to \( Ax + By + C = 0 \) is \( \frac{|C|}{\sqrt{A^2 + B^2}} \).
Rewriting the line as \( x\cos \theta - y\sin \theta - k\cos 2\theta = 0 \):
\( p = \frac{|-k\cos 2\theta|}{\sqrt{(\cos \theta)^2 + (-\sin \theta)^2}} \)
\( p = \frac{|-k\cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} \)
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( p = \frac{|-k\cos 2\theta|}{1} = |k\cos 2\theta| \)
Therefore, \( p^2 = (k\cos 2\theta)^2 = k^2\cos^2 2\theta \).

**Part 2: Find \( q \)**
The second line is \( x\sec \theta + y\csc \theta = k \).
Rewrite \( \sec \theta \) as \( \frac{1}{\cos \theta} \) and \( \csc \theta \) as \( \frac{1}{\sin \theta} \):
\( \frac{x}{\cos \theta} + \frac{y}{\sin \theta} = k \)
To remove denominators, multiply by \( \sin \theta \cos \theta \):
\( x\sin \theta + y\cos \theta = k\sin \theta \cos \theta \)
We know that \( \sin 2\theta = 2\sin \theta \cos \theta \), so \( \sin \theta \cos \theta = \frac{1}{2}\sin 2\theta \).
Substitute this into the equation:
\( x\sin \theta + y\cos \theta = k\left(\frac{1}{2}\sin 2\theta\right) \)
\( x\sin \theta + y\cos \theta - \frac{k}{2}\sin 2\theta = 0 \)
Now, find the length of the perpendicular from the origin \( (0,0) \) to this line:
\( q = \frac{|-\frac{k}{2}\sin 2\theta|}{\sqrt{(\sin \theta)^2 + (\cos \theta)^2}} \)
\( q = \frac{|-\frac{k}{2}\sin 2\theta|}{\sqrt{\sin^2 \theta + \cos^2 \theta}} \)
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( q = \frac{|-\frac{k}{2}\sin 2\theta|}{1} = \left|\frac{k}{2}\sin 2\theta\right| \)
Therefore, \( q^2 = \left(\frac{k}{2}\sin 2\theta\right)^2 = \frac{k^2}{4}\sin^2 2\theta \).

**Part 3: Prove \( p^2 + 4q^2 = k^2 \)**
Substitute the expressions for \( p^2 \) and \( q^2 \):
\( p^2 + 4q^2 = k^2\cos^2 2\theta + 4\left(\frac{k^2}{4}\sin^2 2\theta\right) \)
\( p^2 + 4q^2 = k^2\cos^2 2\theta + k^2\sin^2 2\theta \)
Factor out \( k^2 \):
\( p^2 + 4q^2 = k^2(\cos^2 2\theta + \sin^2 2\theta) \)
Using the trigonometric identity \( \cos^2 A + \sin^2 A = 1 \):
\( p^2 + 4q^2 = k^2(1) \)
\( p^2 + 4q^2 = k^2 \)
Hence proved.
In simple words: We are given two lines and told that \( p \) and \( q \) are the distances from the origin to these lines. We first calculate \( p \) by putting the first line into normal form. Then, we transform the second line and calculate \( q \). Finally, we substitute these \( p \) and \( q \) values into the equation \( p^2 + 4q^2 \) and use trigonometric identities to show that it simplifies to \( k^2 \).

Exam Tip: For problems involving trigonometric functions, remember to simplify the line equations using identities like \( \sec \theta = 1/\cos \theta \), \( \csc \theta = 1/\sin \theta \), and \( \sin 2\theta = 2\sin \theta \cos \theta \). Also, recall the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \) for simplifying denominators in the distance formula.

Question 17. In the triangle ABC with vertices A(2, 3), B(4, – 1) and C(1, 2), find the equation and length of altitude from the vertex A.
Answer:
Given vertices of \( \triangle ABC \) are \( A(2, 3) \), \( B(4, -1) \), and \( C(1, 2) \).
The altitude from vertex A is a line segment (let's call it AM) that passes through A and is perpendicular to the side BC.

**Step 1: Find the slope of the side BC.**
Using points \( B(4, -1) \) and \( C(1, 2) \):
\( m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-1)}{1 - 4} = \frac{2 + 1}{-3} = \frac{3}{-3} = -1 \).

**Step 2: Find the slope of the altitude AM.**
Since altitude AM is perpendicular to BC, its slope \( m_{AM} \) is the negative reciprocal of \( m_{BC} \).
\( m_{AM} = -\frac{1}{m_{BC}} = -\frac{1}{-1} = 1 \).

**Step 3: Find the equation of the altitude AM.**
The altitude AM passes through \( A(2, 3) \) and has a slope of \( 1 \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 3 = 1(x - 2) \)
\( y - 3 = x - 2 \)
Rearrange into general form:
\( x - y + 3 - 2 = 0 \)
\( x - y + 1 = 0 \)
This is the equation of the altitude from vertex A.

**Step 4: Find the length of the altitude AM.**
The length of the altitude AM is the perpendicular distance from vertex \( A(2, 3) \) to the line containing side BC.
First, find the equation of the line BC.
Using point \( B(4, -1) \) and slope \( m_{BC} = -1 \):
\( y - (-1) = -1(x - 4) \)
\( y + 1 = -x + 4 \)
Rearrange into general form:
\( x + y + 1 - 4 = 0 \)
\( x + y - 3 = 0 \).
Now, use the distance formula from point \( A(2, 3) \) to the line \( x + y - 3 = 0 \):
\( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)
\( \text{Length of AM} = \frac{|1(2) + 1(3) - 3|}{\sqrt{1^2 + 1^2}} \)
\( = \frac{|2 + 3 - 3|}{\sqrt{1 + 1}} = \frac{|2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \).
The length of the altitude from vertex A is \( \sqrt{2} \) units.
In simple words: For a triangle, an altitude is a line from one corner that drops straight down to the opposite side, meeting it at a right angle. We want to find the equation and length of the altitude from corner A. First, we find the steepness (slope) of the side opposite A (side BC). The altitude's steepness will be its opposite reciprocal. Using the altitude's steepness and corner A's coordinates, we get the altitude's equation. To find its length, we calculate the straight-line distance from corner A to the line containing side BC.

Exam Tip: To find the equation of an altitude, you need two pieces of information: the vertex it passes through and the slope of the opposite side (to determine the perpendicular slope). To find its length, calculate the perpendicular distance from the vertex to the line containing the opposite side.

Question 18. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that \( \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \).
Answer:
**Proof:**
The equation of a line with x-intercept \( a \) and y-intercept \( b \) is given by the intercept form:
\( \frac{x}{a} + \frac{y}{b} = 1 \)
To find the perpendicular distance from the origin \( (0,0) \) to this line, first convert the equation to the general form \( Ax + By + C = 0 \).
Multiply the equation by \( ab \) to clear the denominators:
\( b x + a y = a b \)
Rearrange to general form:
\( b x + a y - a b = 0 \)
Now, we use the formula for the perpendicular distance \( p \) from the origin \( (0,0) \) to a line \( Ax + By + C = 0 \):
\( p = \frac{|C|}{\sqrt{A^2 + B^2}} \)
In our case, \( A = b \), \( B = a \), and \( C = -ab \).
\( p = \frac{|-ab|}{\sqrt{b^2 + a^2}} \)
Since distance is always positive, \( |-ab| = |ab| \):
\( p = \frac{|ab|}{\sqrt{a^2 + b^2}} \)
To get \( p^2 \), square both sides of the equation:
\( p^2 = \left( \frac{|ab|}{\sqrt{a^2 + b^2}} \right)^2 \)
\( p^2 = \frac{a^2 b^2}{a^2 + b^2} \)
Now, we need to show that \( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \).
Take the reciprocal of both sides of the equation for \( p^2 \):
\( \frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} \)
Separate the terms in the numerator:
\( \frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} \)
Simplify each fraction:
\( \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2} \)
Rearrange the terms on the right side:
\( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \)
Hence proved.
In simple words: We start with the equation of a line that crosses the x-axis at 'a' and the y-axis at 'b'. Then, we find a formula for the shortest distance \( p \) from the origin to this line. We square this distance formula. Finally, we take the reciprocal of \( p^2 \) and separate the terms, showing that it equals \( \frac{1}{a^2} + \frac{1}{b^2} \). This proves the relationship between the distance and the intercepts.

Exam Tip: Remember to start with the intercept form of a line \( \frac{x}{a} + \frac{y}{b} = 1 \) and convert it to general form \( Ax + By + C = 0 \). The distance from the origin formula is a key step. Pay attention to algebraic manipulation, especially squaring and taking reciprocals.