Samacheer Kalvi Class 10 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5.3

Question 1. Find the equation of a straight line passing through the mid-point of a line segment joining the points (1, -5), (4, 2) and parallel to (i) X axis (ii) Y axis
Answer:
First, find the mid-point of the line segment joining (1, -5) and (4, 2). The mid-point formula is \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
Mid-point \( = \left(\frac{1+4}{2}, \frac{-5+2}{2}\right) \)
Mid-point \( = \left(\frac{5}{2}, \frac{-3}{2}\right) \)

(i) Equation of a line parallel to the X-axis:
A line parallel to the X-axis always has a slope of 0. Its equation is in the form \( y = k \), where \( k \) is the y-coordinate of any point on the line.
Using the mid-point \( \left(\frac{5}{2}, \frac{-3}{2}\right) \), the y-coordinate is \( \frac{-3}{2} \).
So, the equation is \( y = \frac{-3}{2} \)
Multiply both sides by 2 to clear the fraction:
\( 2y = -3 \)
\( 2y + 3 = 0 \). This is the required equation.

(ii) Equation of a line parallel to the Y-axis:
A line parallel to the Y-axis has an undefined slope. Its equation is in the form \( x = h \), where \( h \) is the x-coordinate of any point on the line.
Using the mid-point \( \left(\frac{5}{2}, \frac{-3}{2}\right) \), the x-coordinate is \( \frac{5}{2} \).
So, the equation is \( x = \frac{5}{2} \)
Multiply both sides by 2 to clear the fraction:
\( 2x = 5 \)
\( 2x - 5 = 0 \). This is the required equation.
In simple words: First, find the middle point between the two given points. Then, for a line parallel to the X-axis, use the y-value of the mid-point. For a line parallel to the Y-axis, use the x-value of the mid-point. These values will help you write the equations.

🎯 Exam Tip: Remember that a line parallel to the X-axis always has the form \( y = \text{constant} \), and a line parallel to the Y-axis always has the form \( x = \text{constant} \). The constant is the corresponding coordinate of any point the line passes through.

Question 2. The equation of a straight line is \( 2(x - y) + 5 = 0 \). Find its slope, inclination and intercept on the Y axis.
Answer:
The given equation is \( 2(x - y) + 5 = 0 \).
First, we need to rewrite this equation in the standard slope-intercept form, which is \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept.
Expand the equation:
\( 2x - 2y + 5 = 0 \)
Move the \( x \) term and the constant to the right side:
\( -2y = -2x - 5 \)
Divide the entire equation by -2 to solve for \( y \):
\( y = \frac{-2x}{-2} + \frac{-5}{-2} \)
\( y = x + \frac{5}{2} \)

Now we can find the required values:
Slope (\( m \)): By comparing \( y = x + \frac{5}{2} \) with \( y = mx + c \), we see that \( m = 1 \).

Y-intercept (\( c \)): The y-intercept is the constant term when the equation is in \( y = mx + c \) form. So, \( c = \frac{5}{2} \).

Inclination (\( \theta \)): The inclination is the angle the line makes with the positive X-axis. We know that \( m = \tan \theta \).
Since \( m = 1 \), we have \( \tan \theta = 1 \).
We know that \( \tan 45^\circ = 1 \).
So, \( \theta = 45^\circ \). The line makes a 45-degree angle with the x-axis.
In simple words: To find the slope, inclination, and y-intercept, change the given line equation into the form "y equals mx plus c". Here, 'm' is the slope, 'c' is the y-intercept, and the angle whose tangent is 'm' is the inclination.

🎯 Exam Tip: Always convert the linear equation into the slope-intercept form \( y = mx + c \) to easily identify the slope \( m \) and the y-intercept \( c \). Remember that the inclination \( \theta \) is found using \( \tan \theta = m \).

Question 3. Find the equation of a line whose inclination is 30° and making an intercept -3 on the Y axis.
Answer:
We are given the inclination \( \theta = 30^\circ \).
The slope (\( m \)) of a line is given by \( m = \tan \theta \).
So, \( m = \tan 30^\circ \).
We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Thus, the slope of the line is \( m = \frac{1}{\sqrt{3}} \).

We are also given that the y-intercept (\( c \)) is -3.
The equation of a line in slope-intercept form is \( y = mx + c \).
Substitute the values of \( m \) and \( c \) into the equation:
\( y = \frac{1}{\sqrt{3}}x - 3 \)
To remove the fraction and simplify, multiply the entire equation by \( \sqrt{3} \):
\( \sqrt{3}y = x - 3\sqrt{3} \)
Rearrange the terms to get the standard form \( Ax + By + C = 0 \):
\( 0 = x - \sqrt{3}y - 3\sqrt{3} \)
So, the equation of the line is \( x - \sqrt{3}y - 3\sqrt{3} = 0 \). This line perfectly matches the given conditions.
In simple words: First, use the given angle (inclination) to find the slope of the line. Then, with the slope and the y-intercept (where it crosses the 'y' line), write the equation of the line. Make sure to clear any fractions to get a clean equation.

🎯 Exam Tip: Always know the standard trigonometric values for common angles like 30°, 45°, and 60°. Remember the slope-intercept form \( y = mx + c \) is very useful when given slope and y-intercept.

Question 4. Find the slope and y intercept of \( \sqrt { 3 }x + (1 - \sqrt { 3 })y = 3 \).
Answer:
The given equation of the line is \( \sqrt { 3 }x + (1 - \sqrt { 3 })y = 3 \).
To find the slope and y-intercept, we convert this equation into the slope-intercept form \( y = mx + c \).
First, isolate the term with \( y \):
\( (1 - \sqrt { 3 })y = -\sqrt { 3 }x + 3 \)
Now, divide the entire equation by \( (1 - \sqrt { 3 }) \):
\( y = \frac{-\sqrt{3}}{1 - \sqrt{3}}x + \frac{3}{1 - \sqrt{3}} \)

Slope (\( m \)): The slope is the coefficient of \( x \).
\( m = \frac{-\sqrt{3}}{1 - \sqrt{3}} \)
To rationalize the denominator, multiply the numerator and denominator by the conjugate of \( (1 - \sqrt { 3 }) \), which is \( (1 + \sqrt { 3 }) \):
\( m = \frac{-\sqrt{3}}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \)
\( m = \frac{-\sqrt{3}(1 + \sqrt{3})}{(1)^2 - (\sqrt{3})^2} \)
\( m = \frac{-\sqrt{3} - 3}{1 - 3} \)
\( m = \frac{-\sqrt{3} - 3}{-2} \)
\( m = \frac{\sqrt{3} + 3}{2} \)

Y-intercept (\( c \)): The y-intercept is the constant term.
\( c = \frac{3}{1 - \sqrt{3}} \)
Rationalize the denominator by multiplying by the conjugate \( (1 + \sqrt{3}) \):
\( c = \frac{3}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \)
\( c = \frac{3(1 + \sqrt{3})}{1 - 3} \)
\( c = \frac{3 + 3\sqrt{3}}{-2} \)
\( c = -\frac{3 + 3\sqrt{3}}{2} \)
So, the slope of the line is \( \frac{3 + \sqrt{3}}{2} \) and the y-intercept is \( -\frac{3 + 3\sqrt{3}}{2} \).
In simple words: To find the slope and y-intercept, rearrange the equation into the form "y equals mx plus c". The number next to 'x' will be the slope, and the other number will be the y-intercept. Remember to make the bottom part of any fraction a whole number by multiplying.

🎯 Exam Tip: Always remember to rationalize the denominator when you have a square root in the bottom part of a fraction. This is done by multiplying both the numerator and denominator by the conjugate of the denominator.

Question 5. Find the value of 'a', if the line through (-2,3) and (8,5) is perpendicular to \( y = ax + 2 \)
Answer:
First, let's find the slope of the line passing through the points \( (-2, 3) \) and \( (8, 5) \).
The formula for the slope \( m_1 \) using two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( m_1 = \frac{y_2 - y_1}{x_2 - x_1} \).
\( m_1 = \frac{5 - 3}{8 - (-2)} \)
\( m_1 = \frac{2}{8 + 2} \)
\( m_1 = \frac{2}{10} \)
\( m_1 = \frac{1}{5} \)

Next, find the slope of the second line, \( y = ax + 2 \).
This equation is already in the slope-intercept form \( y = mx + c \).
By comparing, the slope \( m_2 \) of this line is \( a \).

We are told that the two lines are perpendicular to each other.
For two lines to be perpendicular, the product of their slopes must be -1.
So, \( m_1 \times m_2 = -1 \). This is a key property of perpendicular lines.
Substitute the slopes we found:
\( \frac{1}{5} \times a = -1 \)
\( \frac{a}{5} = -1 \)
Multiply both sides by 5 to solve for \( a \):
\( a = -1 \times 5 \)
\( a = -5 \)
Therefore, the value of \( a \) is -5.
In simple words: First, find how steep the first line is (its slope) using the two given points. The second line's steepness (its slope) is 'a'. Since the lines cross each other at a right angle, multiply their slopes together, and the answer must be -1. Use this to find 'a'.

🎯 Exam Tip: Remember the condition for perpendicular lines: the product of their slopes is -1. For parallel lines, their slopes are equal. Always clearly identify the slopes of both lines before applying the condition.

Question 6. The hill in the form of a right triangle has its foot at (19,3). The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Answer:
The foot of the hill is given as a point \( (x_1, y_1) = (19, 3) \).
The inclination of the hill to the ground is \( \theta = 45^\circ \).

The slope (\( m \)) of the line representing the hill is given by \( m = \tan \theta \).
\( m = \tan 45^\circ \).
We know that \( \tan 45^\circ = 1 \).
So, the slope of the hill (line) is \( m = 1 \). A 45-degree angle means the climb is steady, one unit up for one unit across.

Now, we have a point \( (19, 3) \) and the slope \( m = 1 \). We can use the point-slope form of a linear equation: \( y - y_1 = m(x - x_1) \).
Substitute the values:
\( y - 3 = 1(x - 19) \)
\( y - 3 = x - 19 \)
Rearrange the equation into the standard form \( Ax + By + C = 0 \):
\( 0 = x - y - 19 + 3 \)
\( 0 = x - y - 16 \)
So, the equation of the line joining the foot and top of the hill is \( x - y - 16 = 0 \).
In simple words: We know where the hill starts and how steep it is (its angle). First, find the slope from the angle. Then, use the starting point and the slope to write the equation for the line that shows the hill's path.

🎯 Exam Tip: When given a point and inclination, first calculate the slope using \( m = \tan \theta \), and then use the point-slope form \( y - y_1 = m(x - x_1) \) to find the equation of the line. Be careful with calculations involving negative signs.

Question 7. Find the equation of a line through the given pair of points.
(i) \( (2,\frac{2}{3}) \) and \( (\frac{-1}{2}, -2) \)
(ii) (2,3) and (-7,-1)
Answer:
The equation of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) can be found using the two-point formula:
\( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \)

(i) For the points \( (2, \frac{2}{3}) \) and \( (\frac{-1}{2}, -2) \):
Let \( (x_1, y_1) = (2, \frac{2}{3}) \) and \( (x_2, y_2) = (\frac{-1}{2}, -2) \).
Substitute these values into the formula:
\( \frac{y - \frac{2}{3}}{-2 - \frac{2}{3}} = \frac{x - 2}{\frac{-1}{2} - 2} \)
Simplify the denominators:
\( -2 - \frac{2}{3} = \frac{-6 - 2}{3} = \frac{-8}{3} \)
\( \frac{-1}{2} - 2 = \frac{-1 - 4}{2} = \frac{-5}{2} \)
So, the equation becomes:
\( \frac{\frac{3y - 2}{3}}{\frac{-8}{3}} = \frac{x - 2}{\frac{-5}{2}} \)
\( \frac{3y - 2}{-8} = \frac{2(x - 2)}{-5} \)
Cross-multiply:
\( -5(3y - 2) = -8 \times 2(x - 2) \)
\( -15y + 10 = -16(x - 2) \)
\( -15y + 10 = -16x + 32 \)
Move all terms to one side to get the standard form \( Ax + By + C = 0 \):
\( 16x - 15y + 10 - 32 = 0 \)
\( 16x - 15y - 22 = 0 \). This is the equation of the line.

(ii) For the points \( (2, 3) \) and \( (-7, -1) \):
Let \( (x_1, y_1) = (2, 3) \) and \( (x_2, y_2) = (-7, -1) \).
Substitute these values into the formula:
\( \frac{y - 3}{-1 - 3} = \frac{x - 2}{-7 - 2} \)
Simplify the denominators:
\( \frac{y - 3}{-4} = \frac{x - 2}{-9} \)
Cross-multiply:
\( -9(y - 3) = -4(x - 2) \)
\( -9y + 27 = -4x + 8 \)
Move all terms to one side:
\( 4x - 9y + 27 - 8 = 0 \)
\( 4x - 9y + 19 = 0 \). This is the equation of the line.
In simple words: When you have two points, you can use a special formula to find the line that connects them. Plug the x and y values of both points into the formula. Then, do the math steps carefully to simplify the equation.

🎯 Exam Tip: Be very careful with fractions and negative signs when using the two-point formula. It's often helpful to simplify the denominators first before cross-multiplication.

Question 8. A cat is located at the point(-6, -4) in xy plane. A bottle of milk is kept at (5,11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.
Answer:
The shortest path between two points is a straight line. We need to find the equation of the line connecting the cat's position \( (-6, -4) \) and the milk bottle's position \( (5, 11) \).
We can use the two-point formula for a line: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \).
Let \( (x_1, y_1) = (-6, -4) \) and \( (x_2, y_2) = (5, 11) \).
Substitute the coordinates into the formula:
\( \frac{y - (-4)}{11 - (-4)} = \frac{x - (-6)}{5 - (-6)} \)
\( \frac{y + 4}{11 + 4} = \frac{x + 6}{5 + 6} \)
\( \frac{y + 4}{15} = \frac{x + 6}{11} \)
Now, cross-multiply to solve for the equation:
\( 11(y + 4) = 15(x + 6) \)
\( 11y + 44 = 15x + 90 \)
Rearrange the terms to get the standard form \( Ax + By + C = 0 \):
\( 0 = 15x - 11y + 90 - 44 \)
\( 0 = 15x - 11y + 46 \)
So, the equation of the path the cat needs to take is \( 15x - 11y + 46 = 0 \). This equation describes the shortest, straight path.
In simple words: The shortest way from one point to another is a straight line. We have two points: the cat's spot and the milk's spot. Use the two-point formula to find the equation of the line connecting these two points. This line is the shortest path.

🎯 Exam Tip: Always remember that "shortest distance" implies a straight line. Use the two-point formula to find the equation of the line, and be careful with arithmetic, especially when dealing with negative coordinates.

Question 9. Find the equation of the median and altitude of AABC through A where the vertices are A(6,2), B(-5, -1) and C(1,9).
Answer:
Given vertices are A(6,2), B(-5, -1), and C(1,9).

(i) To find the equation of the median from A:
A median from a vertex connects that vertex to the midpoint of the opposite side. Here, we need the median from A to the midpoint of BC.
Let D be the midpoint of BC. Using the midpoint formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \):
Midpoint D \( = \left(\frac{-5+1}{2}, \frac{-1+9}{2}\right) \)
Midpoint D \( = \left(\frac{-4}{2}, \frac{8}{2}\right) \)
Midpoint D \( = (-2, 4) \)
Now we need the equation of the line passing through A(6,2) and D(-2,4). Use the two-point formula:
\( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \)
\( \frac{y - 2}{4 - 2} = \frac{x - 6}{-2 - 6} \)
\( \frac{y - 2}{2} = \frac{x - 6}{-8} \)
Cross-multiply:
\( -8(y - 2) = 2(x - 6) \)
\( -8y + 16 = 2x - 12 \)
Divide the entire equation by 2 to simplify:
\( -4y + 8 = x - 6 \)
Rearrange into standard form \( Ax + By + C = 0 \):
\( 0 = x + 4y - 6 - 8 \)
\( x + 4y - 14 = 0 \). This is the equation of the median AD.

(ii) To find the equation of the altitude from A:
An altitude from a vertex is a line segment perpendicular to the opposite side. Here, the altitude from A is perpendicular to BC.
First, find the slope of BC. Let B = \( (-5, -1) \) and C = \( (1, 9) \).
Slope of BC (\( m_{BC} \)) \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - (-1)}{1 - (-5)} \)
\( m_{BC} = \frac{9 + 1}{1 + 5} = \frac{10}{6} = \frac{5}{3} \)
Since the altitude from A is perpendicular to BC, its slope (\( m_{alt} \)) will be the negative reciprocal of \( m_{BC} \).
\( m_{alt} = -\frac{1}{m_{BC}} = -\frac{1}{\frac{5}{3}} = -\frac{3}{5} \)
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point A(6,2) and slope \( m_{alt} = -\frac{3}{5} \).
\( y - 2 = -\frac{3}{5}(x - 6) \)
Multiply both sides by 5:
\( 5(y - 2) = -3(x - 6) \)
\( 5y - 10 = -3x + 18 \)
Rearrange into standard form \( Ax + By + C = 0 \):
\( 3x + 5y - 10 - 18 = 0 \)
\( 3x + 5y - 28 = 0 \). This is the equation of the altitude from A.
In simple words: For the median from A, find the middle point of side BC, then find the line connecting A to that middle point. For the altitude from A, find the slope of side BC. The altitude will have a slope that is the negative opposite of BC's slope. Then, use point A and this new slope to find the altitude's equation.

🎯 Exam Tip: Clearly differentiate between median and altitude. A median connects to a midpoint, while an altitude is perpendicular to the opposite side. Always use the correct formulas for midpoint and perpendicular slopes.

Question 10. Find the equation of a straight line which has slope \( \frac{-5}{4} \) and passing through the point (-1,2).
Answer:
We are given the slope (\( m \)) of the line as \( m = \frac{-5}{4} \).
We are also given a point that the line passes through, \( (x_1, y_1) = (-1, 2) \).

We can use the point-slope form of a linear equation, which is \( y - y_1 = m(x - x_1) \). This form is very useful when you have a point and the slope.
Substitute the given values into the formula:
\( y - 2 = \frac{-5}{4}(x - (-1)) \)
\( y - 2 = \frac{-5}{4}(x + 1) \)
To eliminate the fraction, multiply both sides of the equation by 4:
\( 4(y - 2) = -5(x + 1) \)
\( 4y - 8 = -5x - 5 \)
Rearrange the terms to get the standard form \( Ax + By + C = 0 \):
\( 5x + 4y - 8 + 5 = 0 \)
\( 5x + 4y - 3 = 0 \)
This is the required equation of the straight line. It describes all points on the line that passes through (-1,2) with a slope of \( \frac{-5}{4} \).
In simple words: When you know how steep a line is (its slope) and one point it passes through, you can use a special formula called the "point-slope form" to find its equation. Just put the numbers in and simplify.

🎯 Exam Tip: The point-slope form \( y - y_1 = m(x - x_1) \) is highly efficient when you have a single point and the slope. Always simplify the equation to the standard form \( Ax + By + C = 0 \) unless specified otherwise.

Question 11. You are downloading a song. The percenty (in decimal form) of mega bytes remaining to get downloaded in x seconds is given by \( y = -0.1x + 1 \).
(i) graph the equation.
(ii) find the total MB of the song.
(iii) after how many seconds will 75% of the song gets downloaded?
(iv) after how many seconds the song will be downloaded completely?
Answer:
The equation given is \( y = -0.1x + 1 \), where \( y \) is the remaining percentage (in decimal) and \( x \) is time in seconds.

(i) To graph the equation \( y = -0.1x + 1 \):
This is a linear equation. We can find two points and draw a straight line through them.
- When \( x = 0 \) (at the start of download), \( y = -0.1(0) + 1 = 1 \). So, point (0, 1).
- To find where \( y = 0 \) (download complete), \( 0 = -0.1x + 1 \implies 0.1x = 1 \implies x = 10 \). So, point (10, 0).
Plot these two points (0,1) and (10,0) on a graph. The x-axis represents time in seconds, and the y-axis represents the remaining percentage (as a decimal). Draw a straight line connecting these two points. The line will go downwards from (0,1) to (10,0), showing the remaining percentage decreasing over time.

(ii) To find the total MB of the song:
The total MB is represented by the initial amount when the download starts, which is at \( x = 0 \) seconds. This is also the y-intercept.
Substitute \( x = 0 \) into the equation:
\( y = -0.1(0) + 1 \)
\( y = 1 \)
Since \( y \) is the percentage in decimal form, \( y = 1 \) means 100% (or 1 mega byte if we interpret \( y \) as actual MB remaining out of a total of 1 MB). The problem implies \( y \) is a fraction of the total song, and at \( x=0 \), \( y=1 \) means the full song (1 "unit" or 1 mega byte) is remaining. So the total size of the song is 1 MB.

(iii) After how many seconds will 75% of the song get downloaded?
If 75% of the song is downloaded, then 100% - 75% = 25% of the song remains.
As a decimal, 25% is 0.25. So, we set \( y = 0.25 \).
Substitute \( y = 0.25 \) into the equation:
\( 0.25 = -0.1x + 1 \)
Subtract 1 from both sides:
\( 0.25 - 1 = -0.1x \)
\( -0.75 = -0.1x \)
Divide both sides by -0.1:
\( x = \frac{-0.75}{-0.1} \)
\( x = 7.5 \) seconds.
It takes 7.5 seconds for 75% of the song to download.

(iv) After how many seconds will the song be downloaded completely?
When the song is downloaded completely, the remaining percentage \( y \) will be 0.
Substitute \( y = 0 \) into the equation:
\( 0 = -0.1x + 1 \)
Add \( 0.1x \) to both sides:
\( 0.1x = 1 \)
Divide both sides by 0.1:
\( x = \frac{1}{0.1} \)
\( x = 10 \) seconds.
The song will be downloaded completely in 10 seconds.
In simple words: This equation shows how much of a song is left to download as time passes. To draw the graph, find points for when the download starts and when it finishes. The total song size is how much is left at the very beginning. To find when a certain percentage is downloaded, calculate how much is left, then put that number into the equation and solve for time. When the song is fully done, nothing is left, so set 'y' to zero and find the time.

🎯 Exam Tip: For problems involving linear equations like this, understanding what each variable represents is crucial. Pay close attention to whether the question asks for the amount downloaded or the amount *remaining*.

Question 12. Find the equation of a line whose intercepts on the x and y axes are given below.
(i) 4, -6
(ii) -5, -4
Answer:
The intercept form of a linear equation is \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) is the x-intercept and \( b \) is the y-intercept.

(i) Given x-intercept \( a = 4 \) and y-intercept \( b = -6 \):
Substitute these values into the intercept form:
\( \frac{x}{4} + \frac{y}{-6} = 1 \)
This can be written as:
\( \frac{x}{4} - \frac{y}{6} = 1 \)
To eliminate the fractions, find the Least Common Multiple (LCM) of the denominators (4 and 6), which is 12. Multiply the entire equation by 12:
\( 12 \left(\frac{x}{4}\right) - 12 \left(\frac{y}{6}\right) = 12(1) \)
\( 3x - 2y = 12 \)
Rearrange into standard form \( Ax + By + C = 0 \):
\( 3x - 2y - 12 = 0 \). This is the equation of the line.

(ii) Given x-intercept \( a = -5 \) and y-intercept \( b = \frac{3}{4} \):
Substitute these values into the intercept form:
\( \frac{x}{-5} + \frac{y}{\frac{3}{4}} = 1 \)
This can be rewritten as:
\( \frac{x}{-5} + \frac{4y}{3} = 1 \)
To eliminate the fractions, find the LCM of the denominators (5 and 3), which is 15. Multiply the entire equation by 15:
\( 15 \left(\frac{x}{-5}\right) + 15 \left(\frac{4y}{3}\right) = 15(1) \)
\( -3x + 20y = 15 \)
Rearrange into standard form \( Ax + By + C = 0 \):
\( -3x + 20y - 15 = 0 \)
Multiplying by -1 (optional, to make the leading coefficient positive):
\( 3x - 20y + 15 = 0 \). This is the equation of the line.
In simple words: When you know where a line crosses the 'x' axis and where it crosses the 'y' axis, you can use a special formula called the "intercept form." Just put the crossing points into the formula and then get rid of any fractions to make the equation simple.

🎯 Exam Tip: The intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \) is the most direct method when given x and y intercepts. Always simplify the equation to integer coefficients by multiplying by the LCM of the denominators.

Question 13. Find the intercepts made by the following lines on the coordinate axes.
(i) \( 3x - 2y - 6 = 0 \)
(ii) \( 4x + 3y + 12 = 0 \)
Answer:
To find the x-intercept, set \( y = 0 \) in the equation and solve for \( x \).
To find the y-intercept, set \( x = 0 \) in the equation and solve for \( y \).

(i) For the line \( 3x - 2y - 6 = 0 \):
X-intercept (set \( y = 0 \)):
\( 3x - 2(0) - 6 = 0 \)
\( 3x - 6 = 0 \)
\( 3x = 6 \)
\( x = \frac{6}{3} \)
\( x = 2 \). The x-intercept is 2.

Y-intercept (set \( x = 0 \)):
\( 3(0) - 2y - 6 = 0 \)
\( -2y - 6 = 0 \)
\( -2y = 6 \)
\( y = \frac{6}{-2} \)
\( y = -3 \). The y-intercept is -3.

(ii) For the line \( 4x + 3y + 12 = 0 \):
X-intercept (set \( y = 0 \)):
\( 4x + 3(0) + 12 = 0 \)
\( 4x + 12 = 0 \)
\( 4x = -12 \)
\( x = \frac{-12}{4} \)
\( x = -3 \). The x-intercept is -3.

Y-intercept (set \( x = 0 \)):
\( 4(0) + 3y + 12 = 0 \)
\( 3y + 12 = 0 \)
\( 3y = -12 \)
\( y = \frac{-12}{3} \)
\( y = -4 \). The y-intercept is -4.
In simple words: To find where a line crosses the 'x' axis (x-intercept), imagine the 'y' value is zero and solve for 'x'. To find where it crosses the 'y' axis (y-intercept), imagine the 'x' value is zero and solve for 'y'.

🎯 Exam Tip: This is a fundamental concept. Always remember that for the x-intercept, \( y=0 \), and for the y-intercept, \( x=0 \). It's a quick way to find two points on any linear graph.

Question 14. Find the equation of a straight line.
(i) passing through (1,-4) and has intercepts which are in the ratio 2 : 5
(ii) passing through (-8, 4) and making equal intercepts on the coordinate axes.
Answer:
(i) Passing through (1,-4) and intercepts in the ratio 2:5.
Let the x-intercept be \( a \) and the y-intercept be \( b \). Given that \( a:b = 2:5 \).
So, we can write \( a = 2k \) and \( b = 5k \) for some constant \( k \).
The intercept form of the equation of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Substitute \( a = 2k \) and \( b = 5k \):
\( \frac{x}{2k} + \frac{y}{5k} = 1 \)
Since the line passes through the point (1, -4), substitute \( x = 1 \) and \( y = -4 \) into the equation:
\( \frac{1}{2k} + \frac{-4}{5k} = 1 \)
\( \frac{1}{2k} - \frac{4}{5k} = 1 \)
To combine the fractions, find a common denominator, which is \( 10k \):
\( \frac{5}{10k} - \frac{8}{10k} = 1 \)
\( \frac{5 - 8}{10k} = 1 \)
\( \frac{-3}{10k} = 1 \)
\( -3 = 10k \)
\( k = -\frac{3}{10} \)
Now, find the actual intercepts:
\( a = 2k = 2 \left(-\frac{3}{10}\right) = -\frac{6}{10} = -\frac{3}{5} \)
\( b = 5k = 5 \left(-\frac{3}{10}\right) = -\frac{15}{10} = -\frac{3}{2} \)
Substitute these intercept values back into the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \):
\( \frac{x}{-\frac{3}{5}} + \frac{y}{-\frac{3}{2}} = 1 \)
\( -\frac{5x}{3} - \frac{2y}{3} = 1 \)
Multiply the entire equation by 3:
\( -5x - 2y = 3 \)
Rearrange into standard form:
\( 5x + 2y + 3 = 0 \). This is the equation of the line.

(ii) Passing through (-8, 4) and making equal intercepts on the coordinate axes.
If the intercepts are equal, then \( a = b \).
The intercept form of the equation of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Substitute \( b = a \):
\( \frac{x}{a} + \frac{y}{a} = 1 \)
Multiply the entire equation by \( a \):
\( x + y = a \)
Since the line passes through the point (-8, 4), substitute \( x = -8 \) and \( y = 4 \) into this equation:
\( -8 + 4 = a \)
\( -4 = a \)
So, the intercepts are \( a = -4 \) and \( b = -4 \).
Substitute \( a = -4 \) back into \( x + y = a \):
\( x + y = -4 \)
Rearrange into standard form:
\( x + y + 4 = 0 \). This is the equation of the line.
In simple words: (i) If the x and y intercepts have a certain ratio, write them as '2k' and '5k'. Use the intercept formula, and then put the given point's x and y values into the equation to find 'k'. Once you have 'k', find the actual intercepts and write the final line equation. (ii) If the x and y intercepts are the same, call them both 'a'. Write the intercept formula, then use the given point to find the value of 'a'. After that, write the complete equation of the line.

🎯 Exam Tip: For intercept problems, the key is knowing the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \). When a line passes through a point, substitute its coordinates into the equation to find any unknown parameters (like \( k \) or \( a \)).