if the distance between the points (3,a) and (6,1) is 5 ,find the value of a?
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√(x2-x1)^2+(y2-y1)^2=5
(6-3)^2+(1-a)^2=25
3(3)+1+a*a+2a=25
a^2+2a-15=0
(a*a)+5a-3a-15=0
a(a+5)-3(a+5)=0
(a-3)(a+5)=0
a=3
a=-5
(6-3)^2+(1-a)^2=25
3(3)+1+a*a+2a=25
a^2+2a-15=0
(a*a)+5a-3a-15=0
a(a+5)-3(a+5)=0
(a-3)(a+5)=0
a=3
a=-5
sivaprasad2:
per answer to mujhe kisine bola ki 5 ,-3
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