the equation of a straight line

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October 13th 2008, 11:05 AM
coyoteflare

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
October 13th 2008, 11:29 AM
masters

Quote:

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

$3x+2y=11$

$2y=-3x+11$

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

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

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

$y=mx+b$

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

$5=-3+b$

$8=b$

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

$y=-\frac{3}{2}x+8$
October 13th 2008, 11:29 AM
earboth

Quote:

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:

$\boxed{3x+2y=16}$
October 13th 2008, 11:35 AM
coyoteflare

i feel so stupid sorry i meant..

Quote:

Originally Posted by masters

$3x+2y=11$

$2y=-3x+11$

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

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

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

$y=mx+b$

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

$5=-3+b$

$8=b$

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

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

..perpendicular.. sorry not parrallel...

getting confused cus im trying to do binomials too
October 13th 2008, 11:56 AM
masters

Quote:

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 $\frac{2}{3}$ which is the negative reciprocal of $-\frac{3}{2}$ and then proceed the same way.

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

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

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

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

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