Triangle Perpendiculars Problem Solution
Problem Statement
The vertices of a triangle are \( (2, 1) \), \( (5, 2) \), and \( (4, 4) \). Find the lengths of the perpendiculars from these vertices to their opposite sides. [IIT 1962]
Options:
Detailed Solution
We need to find the lengths of the perpendiculars (altitudes) from each vertex to the line containing the opposite side in a triangle with vertices \( A(2, 1) \), \( B(5, 2) \), and \( C(4, 4) \). This is a coordinate geometry problem, and we’ll use the perpendicular distance formula. Let’s dive in!
Step 1: Understand the Problem
For a triangle \( ABC \): - Perpendicular from \( A(2, 1) \) to side \( BC \). - Perpendicular from \( B(5, 2) \) to side \( AC \). - Perpendicular from \( C(4, 4) \) to side \( AB \).
We’ll find the equations of lines \( BC \), \( AC \), and \( AB \), then calculate the perpendicular distances from the opposite vertices.
Step 2: Formula for Perpendicular Distance
The distance from a point \( (x_0, y_0) \) to a line \( ax + by + c = 0 \) is:
\[ \text{Distance} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]We’ll use this for each calculation.
Step 3: Perpendicular from \( A(2, 1) \) to \( BC \)
Find the equation of line \( BC \):
Points: \( B(5, 2) \), \( C(4, 4) \).
Slope of \( BC \):
\[ m_{BC} = \frac{4 - 2}{4 - 5} = \frac{2}{-1} = -2 \]Using point-slope form with \( B(5, 2) \):
\[ y - 2 = -2 (x - 5) \] \[ y - 2 = -2x + 10 \] \[ y = -2x + 12 \]Convert to standard form:
\[ 2x + y - 12 = 0 \]Verify: For \( C(4, 4) \): \( 2(4) + 4 - 12 = 8 + 4 - 12 = 0 \). It holds.
Distance from \( A(2, 1) \) to \( 2x + y - 12 = 0 \):
Here, \( a = 2 \), \( b = 1 \), \( c = -12 \), \( (x_0, y_0) = (2, 1) \).
\[ \text{Distance} = \frac{|2 \cdot 2 + 1 \cdot 1 - 12|}{\sqrt{2^2 + 1^2}} = \frac{|4 + 1 - 12|}{\sqrt{4 + 1}} = \frac{|-7|}{\sqrt{5}} = \frac{7}{\sqrt{5}} \]So, the perpendicular from \( A \) to \( BC \) is \( \frac{7}{\sqrt{5}} \).
Step 4: Perpendicular from \( B(5, 2) \) to \( AC \)
Find the equation of line \( AC \):
Points: \( A(2, 1) \), \( C(4, 4) \).
Slope of \( AC \):
\[ m_{AC} = \frac{4 - 1}{4 - 2} = \frac{3}{2} \]Using point-slope form with \( A(2, 1) \):
\[ y - 1 = \frac{3}{2} (x - 2) \] \[ y - 1 = \frac{3}{2}x - 3 \] \[ y = \frac{3}{2}x - 2 \]Standard form (multiply by 2 to clear fraction):
\[ 3x - 2y - 4 = 0 \]Verify: For \( C(4, 4) \): \( 3(4) - 2(4) - 4 = 12 - 8 - 4 = 0 \). Correct.
Distance from \( B(5, 2) \) to \( 3x - 2y - 4 = 0 \):
Here, \( a = 3 \), \( b = -2 \), \( c = -4 \), \( (x_0, y_0) = (5, 2) \).
\[ \text{Distance} = \frac{|3 \cdot 5 + (-2) \cdot 2 - 4|}{\sqrt{3^2 + (-2)^2}} = \frac{|15 - 4 - 4|}{\sqrt{9 + 4}} = \frac{|7|}{\sqrt{13}} = \frac{7}{\sqrt{13}} \]So, the perpendicular from \( B \) to \( AC \) is \( \frac{7}{\sqrt{13}} \).
Step 5: Perpendicular from \( C(4, 4) \) to \( AB \)
Find the equation of line \( AB \):
Points: \( A(2, 1) \), \( B(5, 2) \).
Slope of \( AB \):
\[ m_{AB} = \frac{2 - 1}{5 - 2} = \frac{1}{3} \]Using point-slope form with \( A(2, 1) \):
\[ y - 1 = \frac{1}{3} (x - 2) \] \[ y - 1 = \frac{1}{3}x - \frac{2}{3} \] \[ y = \frac{1}{3}x + \frac{1}{3} \]Standard form (multiply by 3):
\[ x - 3y + 1 = 0 \]Verify: For \( B(5, 2) \): \( 5 - 3(2) + 1 = 5 - 6 + 1 = 0 \). Correct.
Distance from \( C(4, 4) \) to \( x - 3y + 1 = 0 \):
Here, \( a = 1 \), \( b = -3 \), \( c = 1 \), \( (x_0, y_0) = (4, 4) \).
\[ \text{Distance} = \frac{|1 \cdot 4 + (-3) \cdot 4 + 1|}{\sqrt{1^2 + (-3)^2}} = \frac{|4 - 12 + 1|}{\sqrt{1 + 9}} = \frac{|-7|}{\sqrt{10}} = \frac{7}{\sqrt{10}} \]So, the perpendicular from \( C \) to \( AB \) is \( \frac{7}{\sqrt{10}} \).
Step 6: Verification with Area (Optional Check)
Let’s confirm using the triangle’s area. Area formula with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} | x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) | \]For \( A(2, 1) \), \( B(5, 2) \), \( C(4, 4) \):
\[ \text{Area} = \frac{1}{2} | 2 (2 - 4) + 5 (4 - 1) + 4 (1 - 2) | = \frac{1}{2} | 2(-2) + 5(3) + 4(-1) | = \frac{1}{2} | -4 + 15 - 4 | = \frac{1}{2} \cdot 7 = \frac{7}{2} \]Area = \( \frac{7}{2} \). Now, \( \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} \):
- Base \( BC = \sqrt{(5-4)^2 + (2-4)^2} = \sqrt{5} \), height = \( \frac{7}{\sqrt{5}} \): \[ \frac{1}{2} \cdot \sqrt{5} \cdot \frac{7}{\sqrt{5}} = \frac{7}{2} \]
- Base \( AC = \sqrt{(2-4)^2 + (1-4)^2} = \sqrt{13} \), height = \( \frac{7}{\sqrt{13}} \): \[ \frac{1}{2} \cdot \sqrt{13} \cdot \frac{7}{\sqrt{13}} = \frac{7}{2} \]
- Base \( AB = \sqrt{(2-5)^2 + (1-2)^2} = \sqrt{10} \), height = \( \frac{7}{\sqrt{10}} \): \[ \frac{1}{2} \cdot \sqrt{10} \cdot \frac{7}{\sqrt{10}} = \frac{7}{2} \]
All match the area, confirming our distances!
Step 7: Match with Options
Our perpendiculars are:
- From \( A \) to \( BC \): \( \frac{7}{\sqrt{5}} \)
- From \( B \) to \( AC \): \( \frac{7}{\sqrt{13}} \)
- From \( C \) to \( AB \): \( \frac{7}{\sqrt{10}} \)
Comparing with options:
- a) \( \frac{7}{\sqrt{5}}, \frac{7}{\sqrt{13}}, \frac{7}{\sqrt{6}} \) – Third term incorrect.
- b) \( \frac{7}{\sqrt{6}}, \frac{7}{\sqrt{8}}, \frac{7}{\sqrt{10}} \) – First two incorrect.
- c) \( \frac{7}{\sqrt{5}}, \frac{7}{\sqrt{8}}, \frac{7}{\sqrt{15}} \) – Last two incorrect.
- d) \( \frac{7}{\sqrt{5}}, \frac{7}{\sqrt{13}}, \frac{7}{\sqrt{10}} \) – Matches exactly!
Correct Answer: Option d
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