Question

for what value of m will the point (2,3) lies on the curve described by x^2 + y^2+mx+2y-30​

Answer 1

Answer:

Step-by-step explanation:

Given that (2, 3) lie on x² + y² + mx +2y - 30=0

So,

$\left(2\right)^{2}+\left(3\right)^{2}+m\left(2\right)+2\left(3\right)-30=0$

$\implies\,4+9+2m+6-30=0$

$\implies\,19+2m-30=0$

$\implies\,2m-11=0$

$\implies\,m=\dfrac{11}{2}$

Answer 2

Step-by-step explanation:

Given that (2, 3) lie on x² + y² + mx +2y - 30=0

So,

\left(2\right)^{2}+\left(3\right)^{2}+m\left(2\right)+2\left(3\right)-30=0(2)

2

+(3)

2

+m(2)+2(3)−30=0

\implies\,4+9+2m+6-30=0⟹4+9+2m+6−30=0

\implies\,19+2m-30=0⟹19+2m−30=0

\implies\,2m-11=0⟹2m−11=0

\implies\,m=\dfrac{11}{2}⟹m=

2

11