Math, asked by tnaina514, 9 months ago

using distance formula:A (-3,2) B(1,-2) C(9,-10) but in explation​

Answers

Answered by shashu200417
0

Step-by-step explanation:

given,

given, A(-3,2) ,B(1,-2) ,C(9,-10)

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64 =√128

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64 =√128 distance between A and C=√(9+3)^2+(-10-2)^2

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64 =√128 distance between A and C=√(9+3)^2+(-10-2)^2 =√144+144

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64 =√128 distance between A and C=√(9+3)^2+(-10-2)^2 =√144+144 =√288

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64 =√128 distance between A and C=√(9+3)^2+(-10-2)^2 =√144+144 =√288therefore,AB+BC=AC

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64 =√128 distance between A and C=√(9+3)^2+(-10-2)^2 =√144+144 =√288therefore,AB+BC=AC √32+√128=√288

given, A(-3,2) ,B(1,-2) ,C(9,-10) distance between A and B=√(1+3)^2+(-2-2)^2 =√16+16 =√32 distance between B and C=√(9-1)^2+(-10+2)^2 =√64+64 =√128 distance between A and C=√(9+3)^2+(-10-2)^2 =√144+144 =√288therefore,AB+BC=AC √32+√128=√288 SINCE AB+BC is not equal to AC

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