Exercise 4.2 Practice
Area of a Triangle using Determinants
Q1
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Find the area of the triangle with vertices at the points given in each of the following:
(i) \( (2, 3), (-1, 0), (2, -4) \)
(ii) \( (-2, -3), (3, 2), (-1, -8) \)
(i) \( (2, 3), (-1, 0), (2, -4) \)
(ii) \( (-2, -3), (3, 2), (-1, -8) \)
Formula
Area \( \Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \).
Using determinant: \( \Delta = \frac{1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \).
Using determinant: \( \Delta = \frac{1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \).
Part (i) Calculation
\( \Delta = \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)| \)
\( = \frac{1}{2} |2(4) + (-1)(-7) + 2(3)| = \frac{1}{2} |8 + 7 + 6| = \frac{21}{2} \) sq units.
\( = \frac{1}{2} |2(4) + (-1)(-7) + 2(3)| = \frac{1}{2} |8 + 7 + 6| = \frac{21}{2} \) sq units.
Part (ii) Calculation
\( \Delta = \frac{1}{2} |-2(2 - (-8)) + 3(-8 - (-3)) + (-1)(-3 - 2)| \)
\( = \frac{1}{2} |-2(10) + 3(-5) + (-1)(-5)| = \frac{1}{2} |-20 - 15 + 5| = \frac{1}{2} |-30| = 15 \) sq units.
\( = \frac{1}{2} |-2(10) + 3(-5) + (-1)(-5)| = \frac{1}{2} |-20 - 15 + 5| = \frac{1}{2} |-30| = 15 \) sq units.
Q2
00:00
Show that the points \( A(1, 4) \), \( B(3, -2) \), and \( C(-3, 16) \) are collinear.
Concept
Points are collinear if the area of the triangle formed by them is zero.
Calculation
\( \Delta = \frac{1}{2} |1(-2 - 16) + 3(16 - 4) + (-3)(4 - (-2))| \)
\( = \frac{1}{2} |1(-18) + 3(12) - 3(6)| \)
\( = \frac{1}{2} |-18 + 36 - 18| = \frac{1}{2} |0| = 0 \).
\( = \frac{1}{2} |1(-18) + 3(12) - 3(6)| \)
\( = \frac{1}{2} |-18 + 36 - 18| = \frac{1}{2} |0| = 0 \).
Since Area = 0, the points are collinear.
Q3
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Find the value of \( k \) if the area of the triangle is 4 sq. units and vertices are:
(i) \( (k, 0), (4, 0), (0, 2) \)
(ii) \( (-2, 0), (0, 4), (0, k) \)
(i) \( (k, 0), (4, 0), (0, 2) \)
(ii) \( (-2, 0), (0, 4), (0, k) \)
Note
Since area is absolute, we use \( \pm 4 \). So, Determinant Value = \( \pm 8 \) (because of the 1/2 factor).
Part (i)
\( |k(0 - 2) + 4(2 - 0) + 0(0 - 0)| = \pm 8 \)
\( |-2k + 8| = \pm 8 \).
Case 1: \( -2k + 8 = 8 \Rightarrow -2k = 0 \Rightarrow k = 0 \).
Case 2: \( -2k + 8 = -8 \Rightarrow -2k = -16 \Rightarrow k = 8 \).
\( |-2k + 8| = \pm 8 \).
Case 1: \( -2k + 8 = 8 \Rightarrow -2k = 0 \Rightarrow k = 0 \).
Case 2: \( -2k + 8 = -8 \Rightarrow -2k = -16 \Rightarrow k = 8 \).
Part (ii)
\( |-2(4 - k) + 0 + 0| = \pm 8 \)
\( |-8 + 2k| = \pm 8 \).
Case 1: \( -8 + 2k = 8 \Rightarrow 2k = 16 \Rightarrow k = 8 \).
Case 2: \( -8 + 2k = -8 \Rightarrow 2k = 0 \Rightarrow k = 0 \).
\( |-8 + 2k| = \pm 8 \).
Case 1: \( -8 + 2k = 8 \Rightarrow 2k = 16 \Rightarrow k = 8 \).
Case 2: \( -8 + 2k = -8 \Rightarrow 2k = 0 \Rightarrow k = 0 \).
k = 0, 8 for both cases (coincidentally).
Q4
00:00
(i) Find the equation of the line joining \( (3, 1) \) and \( (9, 3) \) using determinants.
(ii) Find the equation of the line joining \( (1, 2) \) and \( (3, 6) \) using determinants.
(ii) Find the equation of the line joining \( (1, 2) \) and \( (3, 6) \) using determinants.
Method
Let generic point be \( (x, y) \). The area of triangle with \( (x, y) \) and given points must be 0.
Part (i) Solution
\( \begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0 \)
Expanding: \( x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0 \)
\( -2x + 6y = 0 \Rightarrow x = 3y \).
Expanding: \( x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0 \)
\( -2x + 6y = 0 \Rightarrow x = 3y \).
Part (ii) Solution
\( \begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0 \)
Expanding: \( x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0 \)
\( -4x + 2y = 0 \Rightarrow 2y = 4x \Rightarrow y = 2x \).
Expanding: \( x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0 \)
\( -4x + 2y = 0 \Rightarrow 2y = 4x \Rightarrow y = 2x \).
Q5
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If area of triangle is 35 sq units with vertices \( (2, -6), (5, 4), \) and \( (k, 4) \), then \( k \) is:
(A) 12 (B) -2 (C) -12, -2 (D) 12, -2
(A) 12 (B) -2 (C) -12, -2 (D) 12, -2
Setup
\( \frac{1}{2} |2(4 - 4) + 5(4 - (-6)) + k(-6 - 4)| = \pm 35 \).
\( |2(0) + 5(10) - 10k| = 70 \).
\( |50 - 10k| = 70 \).
\( |2(0) + 5(10) - 10k| = 70 \).
\( |50 - 10k| = 70 \).
Solve Cases
Case 1: \( 50 - 10k = 70 \Rightarrow -10k = 20 \Rightarrow k = -2 \).
Case 2: \( 50 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12 \).
Case 2: \( 50 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12 \).
Correct Option: (D) 12, -2