Question

The equation of the line passing through the point of intersection of the line 4x – 3y – 1 = 0 and 5x – 2y – 3 = 0 and parallel to the line 2x – 3y + 2 = 0 is

Answer 1

Answer:

explanation

Step-by-step explanation:

Thanks for the A2A!

First, it asks to find the intersection of 2x−3y−5=0 and 7x−5y−2=0

To do this, we equate them.

2x−3y−5=7x−5y−2 get them equal

14x−21y−35=14x−10y−4 get x’s ready to cancel

−11y−31=0

y=−31/11

Then we substitute that in to the original equations.

2x−3(−31/11)−5=0

We solve that to get x=−19/11

We now have the point of intersection, (−19/11,−31/11)

Now we need it parallel to 2x−3y+14=0

The gradient of that line is 2/3.

Therefore, the line is of the form y=mx+c , where m = 2/3 and passes through the point above.

Therefore we get y=2x/3+c which passes through (−19/11,−31/11).

Substitute to get −31/11=−38/33+c.

Therefore c=−5/3 .

Therefore, the equation of the line passing through the intersection of those two lines, and parrallel to the third, is y=2x/3−5/3 , or 2x−3y−5=0.