Math, asked by sharmadhruv14nov, 1 year ago

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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Answers

Answered by thesmartlerner

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 PravinRatta

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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