Question

find the equation of the line through the the intersection of the line 2x+3y-4=0 and x-5y=7 tjat has its x-intersept equal to -4​

Answer 1

Answer:

10x + 93y + 40 = 0

Step-by-step explanation:

Two steps:

1. get point P that is the intersection of the two given lines
2. get the equation of the line through P and Q=(-4,0).

Step 1.

Say P=(x,y).  As P is on the two lines, its coordinates satisfy both equations, so we just have to solve the system of simultaneous equations:

2x + 3y = 4    ... (1)

 x - 5y = 7    ... (2)

From (2), x = 5y + 7.  Substituting this into (1) gives

2(5y+7) + 3y = 4

=> 10y + 14 + 3y = 4

=> 13y = -10

=> y = -10/13

and x = 5y + 7 = -50/13 + 7  = -50/13 + 91/13 = 41/13.

So P = ( 41 / 13,  -10 / 13 ).

Step 2.

The line through the two points P = ( 41 / 13, -10 / 13 ) and Q = ( -4, 0 ) is given by...

( y - 0 ) / ( x - (-4) )  =  ( -10/13 - 0 ) / ( 41/13 - (-4) )

=> y / ( x + 4 )  =  ( -10/13 ) / ( 41/13 + 4 )

=> y / ( x + 4 )  =  -10 / ( 41 + 52 )           [ multiplied numerator and denominator by 13 ]

=> y / ( x + 4 ) = -10 / 93

=> 93y = -10 ( x + 4 )

=> 93y = -10x - 40

=> 10x + 93y + 40 = 0