Question

Find the equation of the line that passes through (–3, 2) and the intersection of the lines x–2y=0 and 3x+y+5=0.

Answer 1

Answer:

$\frac{19}{7}$x+$\frac{11}{7}$y+5=0

Step-by-step explanation:

the intersection of x-2y=0 and 3x+y+5 is ($\frac{-10}{7}$;$\frac{-5}{7}$)

=> the line : $\frac{19}{7}$x+$\frac{11}{7}$y+5=0

Answer 2

Answer:

$Point \ of \ intersection = (\frac{-10}{7} , \frac{-5}{7})\\\\Equation \ of \ line : y = -\frac{19}{11}x - \frac{35}{11}$

Step-by-step explanation:

Find intersection of the given lines :

x - 2y = 0 => x = 2y ----------- ( 1 )

3x + y = - 5 -------------------- ( 2 )

Substitute ( 1 ) in ( 2 ) :

                               3x + y = - 5

                               3 ( 2y ) + y = - 5

                                6y + y = - 5

                                  7y = - 5

                                    $y = -\frac{5}{7}$

Substitute y in ( 1 ) :

                            x = 2y

                           $x = 2 \times \frac{-5}{7} = - \frac{10}{7}$

$Therefore , \ point \ of \ intersection\ is ( -\frac{10}{7}, -\frac{5}{7} )$

To find the equation of the line passing through ( - 3, 2) and point of intersection :

Standard equation of a line : y = mx + b , where m is the slope, b is the y intercept.

So step 1 : Find slope , m:

 $slope, m = \frac{y_2 - y_1}{x_2 - x_1}$       $[ \ where \ (x_1, y_ 1 ) = ( -3, 2 ) \ and \ (x_2, y_ 2 ) = ( \frac{-10}{7} , \frac{-5}{7}) \ ]$

              $= \frac{\frac{-5}{7}-(2)}{\frac{-10}{7} - (-3)}\\\\= \frac{-5- 14}{-10 + 21}\\\\=\frac{-19}{11}\\\\=-\frac{19}{11}$

Step 2 : Equation of the line :

$(y - y _1) = m (x - x_1)$

$(y - 2 ) = -\frac{19}{11}(x -( -3))\\\\(y - 2 ) = -\frac{19}{11} (x+ 3)\\\\y = -\frac{19}{11} (x+ 3) + 2\\\\ y = -\frac{19}{11}x +(-\frac{19}{11} \times 3) + 2\\\\y= - \frac{19}[11}x +(\frac{-57}{11} + 2)\\\\y= - \frac{19}{11}x +(\frac{-57+ 22}{11})\\\\y= - \frac{19}{11}x +(\frac{-35}{11})$