find the value of k in the slope on the line joining the point k,4 and -3-4 is 1/2
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slope of line=(y₂-y₁)÷(x₂-x₁)
(-4-4)÷(-3-k)=1÷2
(-8)÷(-3-k)=1÷2
(8)÷(3+k)=1÷2
3+k=16
k=16-3
k=13
(-4-4)÷(-3-k)=1÷2
(-8)÷(-3-k)=1÷2
(8)÷(3+k)=1÷2
3+k=16
k=16-3
k=13
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SOLUTION:-
given point ,
(k,4) and (-3,-4) slope = 1/2
by slope line joing equetion = (y2-y1)/(x2-x1)
here x1=k, x2= -3, y1=4, y2= -4
acourding to qeution:-
(-4-4)/(-3+k) = 1/2
-3+k = -16
k = -16+3
[ k = 13]ans
i hope it is helpfull of u
given point ,
(k,4) and (-3,-4) slope = 1/2
by slope line joing equetion = (y2-y1)/(x2-x1)
here x1=k, x2= -3, y1=4, y2= -4
acourding to qeution:-
(-4-4)/(-3+k) = 1/2
-3+k = -16
k = -16+3
[ k = 13]ans
i hope it is helpfull of u
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