the equation of the line perpendicular to line 3x+2y=6, which passes the intersection point between lines 2x+3y-1=0 and 4x+7y+1=0 will cross Y-axis at point?

Question

anonymous · Answer

Find the intersection point.
Line perpendicular to 3x+2y = 6 is 2x-3y = k.
Use intersection point to find k.
For Y axis, x = 0. Put x  and fnd the required point.

jack1 · Answer

yep, that will work @ShailKumar , but try to give a few more details to make sure they understand all of your steps:

1.Find the intersection point {using simultaneous equations}
so equation 1 is 2x+3y-1=0
and equation 2 is 4x+7y+1=0
so use {the substitution method} / whichever method u like to solve for the intersection point.

2.Line perpendicular to 3x+2y = 6 is 2x-3y = k.  {because the gradient is always the negative reciprocal}

Use intersection point to find k. {by subbing in the values you found in place of "x"and "y".

For Y intercept point axis, substitute x = 0 {into the this equation, once you've solved for k: 
2x-3y = k  } 

This will allow you to find the y value where the line crosses the Y axis..

anonymous · Answer

yaa i got it!! thank u :)