IF POINTS (2K-3,K+2) lies on the line of equatiob 2x+3y+15=0.Find the value of K
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Answered by
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Answer:
Since the point lies on the graph, it must satisfy the equation.
So, 2(2k-3) + 3(k+2) + 15 = 0
4k - 6 + 3k + 6 + 15 = 0
7k = -15
k = -15/7
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Answered by
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According to the question,
The points (2K-3,K+2) lies on the graph of the equation 2x+3y+15=0.
So it should satisfy the equation
Here,
X=2K-3
Y=K+2
Therefore,
2 (2K - 3) + 3 (K + 2) + 15 = 0
=) 4K-6+3K+6+15=0
=) 7K+15=0
=) 7K= -15
Therefore, K= - 15/7
The answer is K= - 15/7
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