Find the equation of the line (in y = mx + c form) that:

has gradient 2, passing through the intersection of the lines with equations y= 3x - 5 and y= -2x + 5.

Show any working out...thanks!

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- Feb 28th 2009, 04:05 PMJoker37Finding the equation of a straight line
Find the equation of the line (in y = mx + c form) that:

has gradient 2, passing through the intersection of the lines with equations y= 3x - 5 and y= -2x + 5.

Show any working out...thanks! - Feb 28th 2009, 04:18 PMstapel
To find where the lines intersect, solve the

**system of equations**that they represent.

Since they're already both solved for "y=", I'd use substitution. So set it up like so:

. . . . .$\displaystyle 3x\, -\, 5\, =\, -2x\, +\, 5$

. . . . .$\displaystyle 3x\, +\, 2x\, =\, 5\, +\, 5$

...and so forth. This will give you the point $\displaystyle (x_1,\, y_1)$ of intersection.

You are already given the slope (or gradient): $\displaystyle m\, =\, 2$. Plug this information into whichever**straight-line equation**you prefer. Since you have a point and a slope, you might want to start with the point-slope form:

. . . . .$\displaystyle y\, -\, y_1\, =\, m(x\, -\, x_1)$

Multiply the slope through the parentheses, and then solve for "y=". The result is the equation you need.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you! :D - Feb 28th 2009, 05:57 PMJoker37
3x - 5 = -2x + 5 y = -2(2) + 5

5x = 10 = -4 + 5

x = 2 = 1

y - 1 = 2(x - 2)

y - 1 = 2x - 4

y = 2x - 3

Thanks! Earlier I tried this method but I made a rearragement error.

i.e at the start I went:

3x - 5 = -2x + 5

5x - 25 = 0

instead what's in the above. Happens when I'm tired!(Sleepy)