Find the distance between the points P(-6,7) and Q(-1,-5)
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Answered by
9
distance between PQ
= √[(x2 - x1)^2 + (y2 - y1 )^2 ]
= √[-1-(-6)] ^2+ [-5-7]^2
= √[-1+6]^2+ [-12]^2
= √[(5)^2 + 144]
= √(25+144)
= √169
= 13 units..
= √[(x2 - x1)^2 + (y2 - y1 )^2 ]
= √[-1-(-6)] ^2+ [-5-7]^2
= √[-1+6]^2+ [-12]^2
= √[(5)^2 + 144]
= √(25+144)
= √169
= 13 units..
Answered by
23
Heya,
pq = √ (x2-X1)^2 + (y2-y1)^2
√ (-1+6)^2+(-5-7)^2
√(5)^2+(-12)^2
√25 +144
√169
= 13 units Answer..
HOPE IT HELPS ^_^
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