Question 885058: Please help.
write the equation of the line in standard form which passes through (-9,3) perpendicular to x-3y=0 Answer by algebrapro18(249) (Show Source):
You can put this solution on YOUR website! So to start we need a point and the slope of a line to determine the equation of the line. The point was given as (-9,3) and we know that equations of perpendicular lines have slopes that are opposite reciprocals of each other. So lets first find the slope of that line. We can do that by solving the given equation for y.
x-3y=0
-3y=-x + 0
y = 1/3x + 0
Since the equation of a line is y = mx+b where m is the slope we see that the above line has slope 1/3. That means that the slope of the line we want going through point (-9,3) would have a slope m = -3 since the reciprocal of 1/3 is 3 and the opposite of positive is negative. So now we know the point and the slope of our line so we can find its equation. There are two ways to do this and I will show you both ways. The first way is to use the point slope formula and the second way is to use the formula y=mx+b to solve for b.
1) POINT-SLOPE FORMULA
The point slope formula is , where x1 = -9 and y1 = 3. Now we can just plug and chug. REMEMBER from above m = -3.
distribute the negative sign distribute the -3 add 3 to both sides to get y by its self add 3x to put the equation into standard form
2) Solving for b.
So we know that x = -9 and y = 3 from the point given and we were able to find that m = -3. So now we just plug everything into y = mx+b and solve for b.
multiplying -3 * -9 subtracting 27 to both sides to get b by its self
-24 = b
So our equation of the line would be y = -3x-24 and putting that in standard form we get 3x+y = -24.
As you can see which ever way you choose to do this you get the same answer of 3x+y = -24.
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