Question #16314

Find the equation of the line that passes through the intersection of the line 5x - 3y + 2 = 0 and 3x - 4y + 1 = 0 and having slope of 3.

Expert's answer

Find the intersection of the lines 5x - 3y + 2 = 0 and 3x - 4y + 1 = 0.

3*first equation – 5*second equation:

(3*5x – 3*3y + 3*2) – (5*3x – 5*4y + 5*1) = 3*0 – 5*0

11y + 1 = 0 => y = -1/11

X = 4y/3 – 1/3 = -4/33 – 11/33 = -5/11.

The slope of the line is 3, so, the equation of the line looks: y = 3x + b.

The point (x, y) = (-5/11, -1/11) belongs to the line. So, -1/11 = -3*5/11 + b => b = 14/11

So, the equation is y = 3x + 14/11

3*first equation – 5*second equation:

(3*5x – 3*3y + 3*2) – (5*3x – 5*4y + 5*1) = 3*0 – 5*0

11y + 1 = 0 => y = -1/11

X = 4y/3 – 1/3 = -4/33 – 11/33 = -5/11.

The slope of the line is 3, so, the equation of the line looks: y = 3x + b.

The point (x, y) = (-5/11, -1/11) belongs to the line. So, -1/11 = -3*5/11 + b => b = 14/11

So, the equation is y = 3x + 14/11

## Comments

## Leave a comment