Question 208331
write an equation of the line perpendicular to the graph of x-2y-6=0 and passing through A(-3,2)<br>

Well from this question we have the slope, just not in the current form, and a point that the line will travel through so we will use the point slope formula.  We can also use another method which I will use after I find the answer using the point slope formula.  <br>

SO we have the equation of a line here in standard form.  To find its slope we need to put it into slope-intercept form(i.e. solve for y).<br>

x-2y-6=0
x-2y-6+6=0+6
x-2y=6
x-x-2y=6-x
-2y=6-x
-2y/-2 = (6-x)/-2
y = -3+1/2x
y = 1/2x-3

so now we know that y = 1/2x-3 and slope intercept form tells us that y = mx+b where m is the slope.  So that means 1/2 is the slope of the origonal line.  Now we need the slope of a line perpenduclar to that line.  That means we use the opposite resiprocal which is -2.<br>

Therefore we know that our line will have slope -2.  Now we have the slope and the point it passes through, which is (-3,2).  So we can use the point slope formula of y-y1=m(x-x1) where the point (-3,2) are (x1-y1) respectively, m is the slope and x and y are just dummy variables.<br>

so now we plug in and simplify. <br>

y-y1=m(x-x1)
y-2=-2(x--3)
y-2=-2(x+3)
y-2+2=-2(x+3)+2
y = -2(x+3)+2
y = -2x-6+2
y = -2x-4<br>

so the equation of the line perpendicular to the line given going through the point given is y = -2x-4.  <br>

Now we could also solve this problem another way.  We know that m = -2 and that y = mx+b and we have an x and a y from the point given.  so we can substitute all our knows into the slope-intercept formula and then solve for b.<br>

y = mx+b
2 = -2(-3)+b
2 = 6+b
-4 = b<br>

so again we get y = -2x-4 as our answer.  <br>