Thread: the equation of a straight line

the equation of stralight line is 3x+2y=11 find the equation of the parrallel line passing through the point 2,5

i got y=2/3x +7 1/3

or is it y=2/3x -4.5

Originally Posted by coyoteflare

the equation of stralight line is 3x+2y=11 find the equation of the parrallel line passing through the point 2,5

i got y=2/3x +7 1/3

or is it y=2/3x -4.5

$\displaystyle 3x+2y=11$

$\displaystyle 2y=-3x+11$

$\displaystyle y=-\frac{3}{2}x+\frac{11}{2}$

Slope(m) = $\displaystyle -\frac{3}{2}$

Use slope-intercept form of the linear equation: y=mx+b and (2, 5) and solve for b.

$\displaystyle y=mx+b$

$\displaystyle 5=-\frac{3}{2}(2)+b$

$\displaystyle 5=-3+b$

$\displaystyle 8=b$

Substituting back into y=mx+b, we have:

$\displaystyle y=-\frac{3}{2}x+8$

Originally Posted by coyoteflare

the equation of stralight line is 3x+2y=11 find the equation of the parrallel line passing through the point 2,5

i got y=2/3x +7 1/3

or is it y=2/3x -4.5

The LHS of the equation of the parallel must be 3x + 2y too. To calculate the RHS you only have to plug in the coordinates of the given point. Thus the equation of the parallel is:

$\displaystyle \boxed{3x+2y=16}$

Originally Posted by masters

$\displaystyle 3x+2y=11$

$\displaystyle 2y=-3x+11$

$\displaystyle y=-\frac{3}{2}x+\frac{11}{2}$

Slope(m) = $\displaystyle -\frac{3}{2}$

Use slope-intercept form of the linear equation: y=mx+b and (2, 5) and solve for b.

$\displaystyle y=mx+b$

$\displaystyle 5=-\frac{3}{2}(2)+b$

$\displaystyle 5=-3+b$

$\displaystyle 8=b$

Substituting back into y=mx+b, we have:

$\displaystyle y=-\frac{3}{2}x+8$

..perpendicular.. sorry not parrallel...

getting confused cus im trying to do binomials too

Originally Posted by coyoteflare

..perpendicular.. sorry not parrallel...

getting confused cus im trying to do binomials too

In that case, just change the slope to $\displaystyle \frac{2}{3}$ which is the negative reciprocal of $\displaystyle -\frac{3}{2}$ and then proceed the same way.

$\displaystyle 5=\frac{2}{3}(2)+b$

$\displaystyle 5=\frac{4}{3}+b$

$\displaystyle \frac{11}{3}=b$

The equation of the perpendicular is $\displaystyle y=\frac{2}{3}x+\frac{11}{3}$