NCERT Class X Chapter 7: Coordinate Geometry Example 1
NCERT Class X Chapter 7: Coordinate Geometry Example 1
Question:
Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.
Given:
The points are \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \).
To Find:
Whether the points form a triangle and, if so, the type of triangle formed.
Formula:
Distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \):
$$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$Solution:
Step 1: Let the points be \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \).
Step 2: Find the length of \( AB \) using the distance formula.
$$ AB = \sqrt{( -2 - 3 )^2 + ( -3 - 2 )^2} = \sqrt{ ( -5 )^2 + ( -5 )^2 } = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} $$Step 3: Find the length of \( BC \).
$$ BC = \sqrt{ (2 - ( -2 ))^2 + (3 - ( -3 ))^2 } = \sqrt{ (4)^2 + (6)^2 } = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} $$Step 4: Find the length of \( CA \).
$$ CA = \sqrt{ (3 - 2)^2 + (2 - 3)^2 } = \sqrt{ (1)^2 + ( -1 )^2 } = \sqrt{1 + 1} = \sqrt{2} $$Step 5: Check if the points are non-collinear (i.e., if they form a triangle).
The three distances are \( 5\sqrt{2} \), \( 2\sqrt{13} \), and \( \sqrt{2} \), all different and none add up to the third. So, the points form a triangle.
Step 6: Determine the type of triangle by comparing the sides.
Let us check if the triangle is right-angled or isosceles.
$$ (AB)^2 + (CA)^2 = (5\sqrt{2})^2 + (\sqrt{2})^2 = 50 + 2 = 52 \\ (BC)^2 = (2\sqrt{13})^2 = 4 \times 13 = 52 $$Since \( (AB)^2 + (CA)^2 = (BC)^2 \), the triangle is right-angled at \( A \). Also, the two sides \( AB \) and \( BC \) are not equal, but the triangle has a right angle.
Step 7: Conclusion.
The points form a right-angled triangle at \( A \).
Result:
Yes, the points (3, 2), (–2, –3) and (2, 3) form a right-angled triangle at \( A \).
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