Question

find the equation of the line parallel to 2x+5y -9=0 and passing through the mid point of the line segment joining A(2,7) and B(-4,1)​

Answer 1

EXPLANATION.

Equation of the line parallel to 2x + 5y - 9 = 0

mid point of line segment = A ( 2,7) and B ( -4,1)

$\sf : \implies \: equation \: of \: parallel \: lines \: = 2x + 5y - 9 = 0 \\ \\ \sf : \implies \: slope \: of \: parallel \: lines \: = \frac{ - a}{b} \\ \\ \sf : \implies \: slope \: of \: the \: line \: = 2x + 5y - 9 = 0 \: \: is \: = \frac{ - 2}{5}$

$\sf : \implies \: mid \: points \: are \: A(2 , 7) \: \: and \: \: B ( - 4 , 1) \\ \\ \sf : \implies \: coordinates \: \:are \: = ( \frac{2 - 4}{2} , \frac{7 + 1}{2} ) \\ \\ \sf : \implies \: coordinates \: are \: = ( - 1 , 4)$

$\sf : \implies \: equation \: of \: line \: \implies \: (y - y_{1}) = m(x - x_{1}) \\ \\ \sf : \implies \: (y - 4) = \frac{ - 2}{5}(x + 1) \\ \\ \sf : \implies \: 5(y - 4) = - 2(x + 1) \\ \\ \sf : \implies \: 5y - 20 = - 2x - 2 \\ \\ \sf : \implies \: 5y + 2x = 18$

$\sf : \implies \: \green{{ \underline{equation \: of \: line \: = 5y + 2x = 18}}}$

Answer 2

2x + 5y = 0


5y = 2x + 9


$\: \: \: \: \: \: \: \: \: \: \: \: \: \: y = \frac{2}{5} x + \frac{9}{5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = mx + c \: \: \: m - \: slofe$

$\red{ \boxed{ \green{m = \frac{2}{5} }}}$

$\red{ \boxed{ \green{ m_{1}m_{2} = - 1} }}$

$\: m_{2} \times - \frac{ 2}{5} = - 1 = m_{2} = - 1x + \frac{5}{2} = \red{ \boxed{ \frac{5}{2}}}$

$\big(2,7 \big), \big( - 4,1 \big)$

$\sf \: mid \: point \bigg( \frac{ x_{1} +x_{2} }{2} , \frac{y_ + y_{2} }{ 2} \bigg) = \bigg( \frac{2 + ( - 4)}{2} , \frac{7 + 1}{2} \bigg)$

$\bigg( \frac{ - 2}{2} , \frac{8}{2} \bigg) = \bigg( - 1,4 \bigg)$

$\sf \: poin t \: \: and \: \: slope$

$\bigg(y - y_{1} \bigg) = m \bigg(x + x_{1} \bigg)$

$\color{red} \bigg(y - 4 \bigg) = \frac{5}{2} \bigg(x + 1 \bigg)$


$\color{red}2y - 8 = 5x + 5$


$\blue{ \fbox{ \color{blue}5x - 2y + 13 = 0}}$