Question

find the equation of the line that is parallel to 2x+5y-7=0 and passes through the mid point of the segment joining (2,7) and(-4,1)

Answer 1

Answer:

The equation of the line  parallel to 2x+5y-7=0 is 2x+5y-18=0

Step-by-step explanation:

$\bf\;Concept:$

$\text{Equation of any line parallel to ax+by+c=0 and passing through the point (x_1,y_1) is }\\\\\boxed{\bf\;ax+by=ax_1+by_1}$

Given line is 2x+5y-7=0

The midpoint of the line joining (2,7) and (-4,1) is

$\bf(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(\frac{2-4)}{2},\frac{7+1}{2})$

$(-1,4)$

The equation of the line parallel to 2x+5y-7=0 and passes through (-1,4) is

$2x+5y=2x_1+5y_1$

$\text{Here,}(x_1,y_1)=(-1,4)$

$2x+5y=2(-1)+5(4)$

$2x+5y=-2+20$

$\implies\;\boxed{\bf\;2x+5y-18=0}$

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Answer 2

Answer:

Answer:

The equation of the line parallel to 2x+5y-7=0 is 2x+5y-18=0

Step-by-step explanation:

\bf\;Concept:Concept:

\begin{gathered}\text{Equation of any line parallel to ax+by+c=0 and passing through the point $(x_1,y_1)$ is }\\\\\boxed{\bf\;ax+by=ax_1+by_1}\end{gathered}

Equation of any line parallel to ax+by+c=0 and passing through the point (x

1

,y

1

) is

ax+by=ax

1

+by

1

Given line is 2x+5y-7=0

The midpoint of the line joining (2,7) and (-4,1) is

\bf(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})(

2

x

1

+x

2

,

2

y

1

+y

2

)

(\frac{2-4)}{2},\frac{7+1}{2})(

2

2−4)

,

2

7+1

)

(-1,4)(−1,4)

The equation of the line parallel to 2x+5y-7=0 and passes through (-1,4) is

2x+5y=2x_1+5y_12x+5y=2x

1

+5y

1

\text{Here,}(x_1,y_1)=(-1,4)Here,(x

1

,y

1

)=(−1,4)

2x+5y=2(-1)+5(4)2x+5y=2(−1)+5(4)

2x+5y=-2+202x+5y=−2+20

\implies\;\boxed{\bf\;2x+5y-18=0}⟹

2x+5y−18=0