Find the equation of the line through (4, 3) to the line that is perpendicular to the line 3x+y=7
The line equation is:
y-y1 = m(x-x1 ) where m is the slope and (x1,y1) is a point pass through the line:
We have the point (4,3)
Now for the slope:
Since the line is perpendicular to the line 3x+y=7
Then we could determine the slope.
First we will determine the slope for 3x+y=7
==> y = -3x+7
The slope = -3
The slope for the perpendicular line is 1/3 ( flip and change sign)
Then the equation for the line is:
y= (1/3)x -4/3 +3
y= (1/3)x + 5/3
The equation of a line perpendicular to a line ax+by+c is of the form bx-ay +k.
Therefore the line perpendicular to 3x+y = 7 or 3x+y-7 = 0 is of the form: x-3y+k = 0, where k could be determined by the fact that x-3y+k = 0should pass through the given point (4,3): So
4-3(3)+k = 0.
4-9+k = 0
-5+k = 0
Therefore, the required line is x-3y+5 = 0 is the line perpendicular to 3x+y = 7.
If 2 lines are perpendicular then the product of their slopes is:
Let's find the slope of the line 3x+y=7.
We'll put the equation into the standard form: y = mx + n, where m is the slope of the line.
For this reason, we'll isolate y to the left side and we'll subtract 3x both sides:
y = 7-3x
y = -3x + 7
The slope of the line y = -3x + 7 is m1 = -3.
The slope of the line that passes through the point (4,3) is:
We'll divide by -3, both sides:
m2 = 1/3
Now, we'll write the equation of a line which passes through a given point and tyhe valueo of it's slope is known.
y-3 = m2(x-4)
y-3 = (1/3)(x-4)
We'll add 3 both sides:
y = (1/3)x - 4/3 + 3
y = x/3 + 5/3