
Mark W. answered 06/21/14
Tutor
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Writing/Math Instructor for High School, College and Grad School Prep
This is actually a perpendicular lines problem. What the question is really asking you is:
(1) "Find the equation of the line perpendicular to 4x+6y+1=0 that passes through the point (-1,1)” &
(2) “Find the point where this line and the original line intersect”
First you need to find the slope of the line that is given to you here in standard form. To do this, you must solve for "y". Thus:
4x+1=-6y
y=-4x/6 - 1/6
y= -2/3x - 1/6
Your slope here is the coefficient on x: -2/3
The slope of the line perpendicular to this will be the negative reciprocal, or 3/2.
Now, write an equation containing this slope, and the x and y values you were given originally, like so:
1= 3/2 (-1) + b
1= -3/2 + b
b= 5/2
The equation of the line that passes though (-1,1) and is perpendicular to the original line is:
y= 3/2 x + 5/2
To find the point on the original line, you will need to solve the two equations simultaneously. Given that you now have both lines in slope/intercept form (you've solved for "y"), it will be easy to subtract one from the other:
y= 3/2 x + 5/2 MINUS
y= -2/3 x - 1/6
You get:
0= 13/6 x + 16/6, or 0= 13x + 16; thus x= -16/13
Plug this into the original linear equation:
y= -2/3 (-16/13) – 1/6 OR
y= 32/39 – 1/6 OR
y= 64/78 – 13/78
y= 51/78 OR
y= 17/26
The math is ugly but the answer is correct (I think). The point where the two lines intersect (and your answer) is:
(x= -16/13, y= 17/26)