Question

if the midpoint of the line joining 3, 4 and k, 7 is X, Y and 2 X + 2 Y + 1 is equal to zerofind the value of k

Answer 1

Answer:

$Value \:of \: k = -15$

Step-by-step explanation:

$The \: midpoint \: of \: line \: joining \\two \: points \: (x_{1},y_{1}) \:and \: (x_{2},y_{2})\:is \\\left(\frac{x_{1}+x_{2}}{2},\frac{x_{1}+x_{2}}{2}\right)$

$The \: midpoint \: of \: line \: joining \\two \: points \: (3,4) \:and \: (k,7)=(X,Y)\:(given)$

$\implies \left(\frac{3+k}{2},\frac{4+7}{2}\right)=(X,Y)$

$\implies \left(\frac{3+k}{2},\frac{11}{2}\right)=(X,Y)$

$\implies \frac{3+k}{2}=X\:---(1) \\ \: \frac{11}{2}=Y\:---(2)$

$But , \: 2X + 2Y + 1 = 0 \: (given)$

$\implies 2\times \frac{3+k}{2}+2\times \frac{11}{2}+1=0\\[From \:(1)\:and\: (2)]$

$\implies 3+k+11+1=0$

$\implies k + 15 = 0$

$\implies k = -15$

Therefore,

$Value \:of \: k = -15$

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Answer 2

Answer:

Step-by-step explanation:

Answer:

Value \:of \: k = -15

Step-by-step explanation:

The \: midpoint \: of \: line \: joining \\two \: points \: (x_{1},y_{1}) \:and \: (x_{2},y_{2})\:is \\\left(\frac{x_{1}+x_{2}}{2},\frac{x_{1}+x_{2}}{2}\right)

The \: midpoint \: of \: line \: joining \\two \: points \: (3,4) \:and \: (k,7)=(X,Y)\:(given)

\implies \left(\frac{3+k}{2},\frac{4+7}{2}\right)=(X,Y)

\implies \left(\frac{3+k}{2},\frac{11}{2}\right)=(X,Y)

\implies \frac{3+k}{2}=X\:---(1) \\ \: \frac{11}{2}=Y\:---(2)

But , \: 2X + 2Y + 1 = 0 \: (given)

\implies 2\times \frac{3+k}{2}+2\times \frac{11}{2}+1=0\\[From \:(1)\:and\: (2)]

\implies 3+k+11+1=0

\implies k + 15 = 0

\implies k = -15

Therefore,

Value \:of \: k = -15