Consider the following lines in the $XY$-plane:
$L_1 : 5x - 2y = 1$,
$L_2 :$ the line passing through $(0, 1)$ and $(100, 101)$,
$L_3 :$ the line passing through $(1, 11)$ and parallel to the vector $-\hat{i} + 2\hat{j}$.
Let $A = (L_1 \cap L_2) \cup (L_2 \cap L_3) \cup (L_3 \cap L_1)$. What is the total number of elements of $A$?
Show Hint
If the slopes of three lines are all distinct, they are not parallel.
They will form a triangle with exactly 3 intersection points unless they are concurrent.
Simply finding the intersection of two lines and verifying if it lies on the third is a highly efficient way to confirm non-concurrency.
• Step 1 : Understanding the Question:
In this problem, we are given three distinct straight lines, \(L_1\)., \(L_2\)., and \(L_3\). in the Cartesian plane.
The set \(A\) is defined as the union of the pairwise intersection points of these three lines.
Specifically, \(A\) contains all points where at least two of these lines intersect.
We need to determine the cardinality of \(A\)., which is the total number of unique intersection points.
• Step 2 : Key Formula or Approach:
First, we determine the explicit equations of all three lines.
The slope of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
The slope-intercept form or point-slope form is used to write down the equation of the line.
If the slopes of the three lines are distinct, then no two lines are parallel, and they will intersect pairwise in exactly three points unless they are concurrent.
If they are concurrent, they intersect at a single common point, and the cardinality of \(A\) would be 1.
• Step 3 : Detailed Explanation:
Let us find the equation and slope of each line:
The equation of the first line is:
\[ L_1: 5x - 2y = 1 \]
The slope of \(L_1\) is \(m_1 = \frac{5}{2} = 2.5\).
The second line \(L_2\) passes through the points \((0, 1)\) and \((100, 101)\).
The slope of \(L_2\) is:
\[ m_2 = \frac{101 - 1}{100 - 0} = \frac{100}{100} = 1 \]
Using the point-slope form with point \((0, 1)\)., the equation of \(L_2\) is:
\[ y - 1 = 1(x - 0) \implies x - y = -1 \]
The third line \(L_3\) passes through \((1, 11)\) and is parallel to the vector \(-\hat{i} + 2\hat{j}\).
The slope of \(L_3\) is:
\[ m_3 = \frac{2}{-1} = -2 \]
Using the point-slope form with point \((1, 11)\)., the equation of \(L_3\) is:
\[ y - 11 = -2(x - 1) \implies y - 11 = -2x + 2 \implies 2x + y = 13 \]
Let us analyze the slopes of the three lines:
\[ m_1 = \frac{5}{2}, \quad m_2 = 1, \quad m_3 = -2 \]
Since all three slopes are distinct, no two lines are parallel.
This guarantees that each pair of lines has exactly one unique intersection point.
Now, we check if the three lines are concurrent by finding the intersection of \(L_2\) and \(L_3\) and checking if it lies on \(L_1\).
From the equation of \(L_2\)., we have:
\[ y = x + 1 \]
Substitute this into the equation of \(L_3\):
\[ 2x + (x + 1) = 13 \implies 3x + 1 = 13 \implies 3x = 12 \implies x = 4 \]
Then, the corresponding \(y\)-coordinate is:
\[ y = 4 + 1 = 5 \]
Thus, \(L_2 \cap L_3 = \{(4, 5)\}\).
We substitute this point into the equation of \(L_1\):
\[ 5(4) - 2(5) = 20 - 10 = 10 \neq 1 \]
Since the point \((4, 5)\) does not satisfy the equation of \(L_1\)., the three lines are not concurrent.
Therefore, the three pairwise intersection points:
\[ P_{12} = L_1 \cap L_2, \quad P_{23} = L_2 \cap L_3, \quad P_{31} = L_3 \cap L_1 \]
are distinct.
Consequently, the set \(A = \{P_{12}, P_{23}, P_{31}\}\) has exactly 3 elements.
• Step 4 : Final Answer:
The total number of elements of \(A\) is 3.
This corresponds to option (A).
