# Example 1 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

Last updated at Aug. 16, 2021 by Teachoo

Examples

Example 1
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Example 2 Important

Example 3 Important

Example 4

Example 5 Important

Example 6

Example 7 Important

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Example 9 Important

Example 10 Important

Example 11 Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Important Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Deleted for CBSE Board 2023 Exams

Chapter 7 Class 10 Coordinate Geometry

Serial order wise

Last updated at Aug. 16, 2021 by Teachoo

Example 1 Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed. Let the three points be P(3, 2), Q(−2, −3) & R(2, 3) We find the distances PQ, QR, and PR Calculating PQ x1 = 3 , y1 = 2 x2 = −2 , y2 = −3 PQ = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −2 −3)2+(−3 −2)2) = √((−5)2+(−5)2) = √((5)2+(5)2) = √(2(5)2) = √2 × 5 = 5√𝟐 = 5 × 1.414 = 7.07 Calculating QR x1 = −2 , y1 = −3 x2 = 2 , y2 = 3 QR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 2 −(−2))2+(3 −(−3))2) = √(( 2+2)2+(3+3)2) = √(( 4)2+(6)2) = √(16+36) = √52 = √(4 × 13) = 𝟐√𝟏𝟑 = 7.21 Calculating PR x1 = 3 , y1 = 2 x2 = 2 , y2 = 3 PR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 3 −2)2+(2 −3)2) = √((1)2+(−1)2) = √(( 1)2+(1)2) = √(1+1) = √𝟐 = 1.141 Hence, PQ = 7.07 , QR = 7. 21, PR = 1.41 Since the sum of any two of these distances is greater than the third distance Therefore, P, Q, R form a triangle Finding which type of Triangle Since, PQ = √50 , QR = √52, PR = √2 So, PQ2 + PR2 = (√50)2 + (√2)2 = 50 + 2 = 52 = (QR)2 Therefore, PQ2 + PR2 = QR2 So, PQR is a right angled triangle