Welcome, future engineers, to a deep dive into one of the fundamental concepts of Coordinate Geometry:
Intercepts of a Line on Coordinate Axes. This concept might seem simple on the surface, but understanding its nuances is crucial for tackling more complex problems in lines, conic sections, and even 3D geometry.
Let's begin our exploration!
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1. What are Intercepts? - The Basics
Imagine a straight line extending infinitely in both directions. Now, picture our familiar Cartesian coordinate system with its X-axis and Y-axis. When this line crosses or "cuts" these axes, the points where it does so are called its
intercept points. The directed distances from the origin to these intercept points are simply called the
intercepts.
These intercepts provide vital information about the line's position relative to the coordinate axes and are frequently used in various mathematical problems.
1.1. The X-intercept
The
X-intercept of a line is the point where the line crosses the X-axis.
*
Key Property: Any point on the X-axis has its Y-coordinate equal to zero. Therefore, the X-intercept point will always be of the form
(x, 0).
*
How to find it: To find the X-intercept of a line given its equation, simply substitute
y = 0 into the equation and solve for
x.
*
Geometric Interpretation: It tells us how far the line is from the Y-axis when it crosses the X-axis.
1.2. The Y-intercept
The
Y-intercept of a line is the point where the line crosses the Y-axis.
*
Key Property: Any point on the Y-axis has its X-coordinate equal to zero. Therefore, the Y-intercept point will always be of the form
(0, y).
*
How to find it: To find the Y-intercept of a line given its equation, simply substitute
x = 0 into the equation and solve for
y.
*
Geometric Interpretation: It tells us how far the line is from the X-axis when it crosses the Y-axis.
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2. Finding Intercepts from Different Forms of a Line's Equation
Let's explore how to determine the intercepts depending on the form in which the line's equation is given.
2.1. General Form of a Straight Line: Ax + By + C = 0
This is the most common and versatile form. Let the equation of a line be
Ax + By + C = 0, where A, B, and C are real numbers, and A and B are not both zero.
*
To find the X-intercept:
Set
y = 0:
Ax + B(0) + C = 0
Ax + C = 0
Ax = -C
x = -C/A (provided A β 0)
So, the X-intercept point is
(-C/A, 0).
*
To find the Y-intercept:
Set
x = 0:
A(0) + By + C = 0
By + C = 0
By = -C
y = -C/B (provided B β 0)
So, the Y-intercept point is
(0, -C/B).
Example 1: Find the intercepts of the line
3x - 4y + 12 = 0.
Step-by-step solution:
- For X-intercept: Set y = 0.
3x - 4(0) + 12 = 0
3x + 12 = 0
3x = -12
x = -4
The X-intercept point is (-4, 0).
- For Y-intercept: Set x = 0.
3(0) - 4y + 12 = 0
-4y + 12 = 0
-4y = -12
y = 3
The Y-intercept point is (0, 3).
2.2. Slope-Intercept Form: y = mx + c
This form explicitly gives us the Y-intercept!
* Here,
m is the slope of the line, and
c is the
Y-intercept.
So, the Y-intercept point is directly
(0, c).
* To find the X-intercept:
Set
y = 0:
0 = mx + c
mx = -c
x = -c/m (provided m β 0)
So, the X-intercept point is
(-c/m, 0).
Example 2: Find the intercepts of the line
y = 2x + 5.
Step-by-step solution:
- Y-intercept: From the form y = mx + c, we can directly see that c = 5.
So, the Y-intercept point is (0, 5).
- X-intercept: Set y = 0.
0 = 2x + 5
2x = -5
x = -5/2
The X-intercept point is (-5/2, 0).
2.3. Intercept Form of a Straight Line: x/a + y/b = 1
This form is explicitly designed to show the intercepts, and it's incredibly useful, especially for
JEE problems.
Let the equation of a line be
x/a + y/b = 1.
*
To find the X-intercept:
Set
y = 0:
x/a + 0/b = 1
x/a = 1
x = a
So, the X-intercept point is
(a, 0). Here,
'a' is the length of the X-intercept (from origin).
*
To find the Y-intercept:
Set
x = 0:
0/a + y/b = 1
y/b = 1
y = b
So, the Y-intercept point is
(0, b). Here,
'b' is the length of the Y-intercept (from origin).
This is why it's called the intercept form! It directly provides the intercepts.
Derivation: General Form to Intercept Form
Let the general equation of a line be Ax + By + C = 0.
Assuming C β 0 (otherwise, the line passes through the origin, and the intercepts are (0,0)), we can rearrange it:
Ax + By = -C
Divide the entire equation by -C (assuming -C β 0):
Ax/(-C) + By/(-C) = (-C)/(-C)
x/(-C/A) + y/(-C/B) = 1
Comparing this with x/a + y/b = 1, we get:
a = -C/A (X-intercept)
b = -C/B (Y-intercept)
This confirms our formulas derived from the general form.
Example 3: Convert the line
3x - 4y = 12 into intercept form and find its intercepts.
Step-by-step solution:
- The RHS must be 1. So, divide the entire equation by 12:
(3x)/12 - (4y)/12 = 12/12
x/4 - y/3 = 1
To match the standard form x/a + y/b = 1, we write it as:
x/4 + y/(-3) = 1
- Comparing with x/a + y/b = 1:
a = 4
b = -3
- So, the X-intercept point is (4, 0) and the Y-intercept point is (0, -3).
2.4. Other Forms (Point-Slope, Two-Point, Normal Form)
For other forms, the easiest way to find intercepts is usually to convert them to the general form (Ax + By + C = 0) or the slope-intercept form (y = mx + c) first, and then apply the methods discussed above.
*
Point-Slope Form: y - y1 = m(x - x1)
Rearrange to
y = mx - mx1 + y1 (Slope-intercept form where
c = y1 - mx1).
Alternatively, expand to
mx - y + (y1 - mx1) = 0 (General form).
*
Two-Point Form: (y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)
Cross-multiply and rearrange to the general form.
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3. Special Cases and JEE-level Considerations
Understanding these special scenarios is crucial for JEE.
3.1. Lines Passing Through the Origin
If a line passes through the origin (0,0), its equation will be of the form
y = mx or
Ax + By = 0 (i.e.,
C = 0 in general form).
In this case:
* Setting
y = 0 gives
mx = 0 => x = 0. So, X-intercept is
(0, 0).
* Setting
x = 0 gives
y = 0. So, Y-intercept is
(0, 0).
Both intercepts are at the origin. The "intercept form"
x/a + y/b = 1 cannot represent lines passing through the origin because it would imply
a=0 and
b=0, leading to division by zero.
3.2. Lines Parallel to the Coordinate Axes
*
Line Parallel to the X-axis: Equation is
y = k (where
k is a constant).
*
X-intercept: Setting
y = 0 gives
k = 0.
* If
k = 0, the line is the X-axis itself, and it has infinitely many X-intercepts.
* If
k β 0, there is no X-intercept (the line never crosses the X-axis).
*
Y-intercept: Setting
x = 0 gives
y = k. So, the Y-intercept point is
(0, k).
*
Line Parallel to the Y-axis: Equation is
x = k (where
k is a constant).
*
X-intercept: Setting
y = 0 gives
x = k. So, the X-intercept point is
(k, 0).
*
Y-intercept: Setting
x = 0 gives
k = 0.
* If
k = 0, the line is the Y-axis itself, and it has infinitely many Y-intercepts.
* If
k β 0, there is no Y-intercept (the line never crosses the Y-axis).
3.3. Length of Intercept vs. Intercept Coordinate
In problems, when "X-intercept" or "Y-intercept" is mentioned, it typically refers to the coordinate value (e.g., 'a' for X-intercept or 'b' for Y-intercept in
x/a + y/b = 1). However, sometimes questions might ask for the "length of the intercept cut off by the axes". In such cases, they refer to the positive magnitude of 'a' and 'b', i.e.,
|a| and
|b| respectively. This is important for clarity.
3.4. Area of Triangle Formed by a Line and the Coordinate Axes
This is a frequently tested concept in JEE.
Consider a line in intercept form:
x/a + y/b = 1.
It intersects the X-axis at
A(a, 0) and the Y-axis at
B(0, b). The origin is
O(0, 0).
These three points form a right-angled triangle OAB, with the right angle at the origin.
The base of this triangle can be considered as the length along the X-axis, which is
|a|.
The height of this triangle can be considered as the length along the Y-axis, which is
|b|.
Therefore, the
Area of ΞOAB = (1/2) Γ base Γ height = (1/2) Γ |a| Γ |b|.
Example 4 (JEE Focus): A line passes through the point
(2, 3) and forms a triangle of area 12 square units with the coordinate axes. Find the equation of the line.
Step-by-step solution:
- Let the equation of the line in intercept form be x/a + y/b = 1.
Here, 'a' is the X-intercept and 'b' is the Y-intercept.
- The line passes through (2, 3). So, substitute x=2 and y=3 into the equation:
2/a + 3/b = 1 --- (1)
- The area of the triangle formed with the coordinate axes is 12 sq units.
So, (1/2) |a| |b| = 12
This implies |ab| = 24.
Thus, ab = 24 or ab = -24.
- From (1), multiply by ab to clear denominators:
2b + 3a = ab
- Case 1: ab = 24 (This means 'a' and 'b' have the same sign. Since (2,3) is in the first quadrant, both 'a' and 'b' must be positive for the triangle to enclose the point in the first quadrant, though not strictly required.)
Substitute ab = 24 into 2b + 3a = ab:
2b + 3a = 24
From ab = 24, we have b = 24/a. Substitute this into 2b + 3a = 24:
2(24/a) + 3a = 24
48/a + 3a = 24
Multiply by a:
48 + 3aΒ² = 24a
3aΒ² - 24a + 48 = 0
Divide by 3:
aΒ² - 8a + 16 = 0
(a - 4)Β² = 0
So, a = 4.
If a = 4, then b = 24/4 = 6.
The equation of the line is x/4 + y/6 = 1.
Multiply by 12: 3x + 2y = 12.
- Case 2: ab = -24 (This means 'a' and 'b' have opposite signs)
Substitute ab = -24 into 2b + 3a = ab:
2b + 3a = -24
From ab = -24, we have b = -24/a. Substitute this into 2b + 3a = -24:
2(-24/a) + 3a = -24
-48/a + 3a = -24
Multiply by a:
-48 + 3aΒ² = -24a
3aΒ² + 24a - 48 = 0
Divide by 3:
aΒ² + 8a - 16 = 0
Use the quadratic formula a = [-b Β± sqrt(bΒ² - 4ac)] / 2a:
a = [-8 Β± sqrt(8Β² - 4(1)(-16))] / 2(1)
a = [-8 Β± sqrt(64 + 64)] / 2
a = [-8 Β± sqrt(128)] / 2
a = [-8 Β± 8β2] / 2
a = -4 Β± 4β2
If a = -4 + 4β2, then b = -24 / (-4 + 4β2) = -6 / (-1 + β2) = 6 / (1 - β2) = 6(1 + β2) / (1 - 2) = -6(1 + β2).
Equation: x/(-4 + 4β2) + y/(-6(1 + β2)) = 1
If a = -4 - 4β2, then b = -24 / (-4 - 4β2) = 6 / (1 + β2) = 6(β2 - 1) / (2 - 1) = 6(β2 - 1).
Equation: x/(-4 - 4β2) + y/(6(β2 - 1)) = 1
- Final Answer: There are three possible lines. The simplest one is 3x + 2y = 12. The other two are x/(-4 + 4β2) + y/(-6(1 + β2)) = 1 and x/(-4 - 4β2) + y/(6(β2 - 1)) = 1.
This example demonstrates how different forms of equations and conceptual understanding of intercepts are combined in JEE-level problems. Pay special attention to the possibility of multiple solutions and handling the signs of intercepts for area calculations.
By mastering the concept of intercepts and their applications, you'll build a strong foundation for many advanced topics in coordinate geometry!