# Find the equation of the line which: passes through (3,4) and is perpendicular to y=x+2

Hello i need some help on this: Find the equation of the line which: passes through (3,4) and is perpendicular to y=x+2 the answer is y=-x+7 using y=mx+c i didn't have trouble solving parallels and straight lines but i couldn't get it here

• Do you know the relationship between slopes of perpendicular lines? – J. W. Tanner Apr 26 at 16:55

If the line has to be perpendicular to $$y=x+2$$ it has to have a 90 degree angle to it. Because it is a line with slope 1, your equation has to have slope -1. So we have $$y=-1x+n$$.

Now it has to pass through point (3,4). If you put those x and y values in the equation, you get: $$4=-1*3+n$$ $$4=n-3$$ $$7=n$$ So we have $$n=7$$ and $$m=-1$$: $$y=-x+7$$

• How did u get -1 in the first slope? – Yoda Apr 26 at 17:17
• There is a theorem in plane analytic geometry that says that two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1\cdot m_2 = -1$ (with some waffling about lines parallel or perpendicular to the axes because then the slope is either 0 or infinite). If you don't know that theorem, you should review it and perhaps prove it for yourself. – NickD Apr 26 at 18:35
• so the m2 is reciprocal to -1 for example if we have (y=-2x+c) we do -1/2? – Yoda Apr 27 at 7:12

Two lines with slopes $$m_1,m_2$$ resp. are perpendicular if $$m_1\cdot m_2=-1$$.

The slope of the first line is $$m_1=1$$.

Therefore $$m_2=\frac{-1}{1}\\ m_2=-1$$

So the slope of the second line must be $$-1$$. Also the line passes through $$(3,4)$$. Now the equation can be easily found.