consider the point P(-3, 7) and the line L with equation 2y - x = 3.?

Question

a) give an equation for the line L⊥, perpendicular to L, through P.

b) Where does L⊥ meet L?

Answer 1

QUESTION A:

Let's solve the line in terms of y, so it is in slope-intercept form:

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

So the slope of the line is 1/2. The slope of a perpendicular line will be the *negative reciprocal*, or -2 in this case.

Now your problem reduces to: "Given a slope of -2 and a point P(-3, 7), what is the equation of the line?" Here you can use the point-slope form:

y - y1 = m(x - x1)

Using your values:
m = -2 and (x1, y1) = (-3, 7)

Substitute in and solve for y:
y - 7 = -2(x - (-3))
y - 7 = -2(x + 3)
y - 7 = -2x - 6
y = -2x + 1

QUESTION B:

You now have two equations: the original line (y = (1/2)x + 3/2) and the perpendicular line you figured out in part A (y = -2x + 1). Equate these and solve for x:

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

Equate them:
(1/2)x + 3/2 = -2x + 1

Multiply both sides by 2 to get rid of fractions:
x + 3 = -4x + 2

Add 4x to both sides:
5x + 3 = 2

Subtract 3 from both sides:
5x = -1

Divide both sides by 5:
x = -1/5

Now plug this into either equation to get y:
y = -2x + 1
y = -2(-1/5) + 1
y = 2/5 + 1
y = 7/5

So the point of intersection is (-1/5, 7/5)

Answer 2

The perpendicular line you seek is y=mx+b.
We need to find values for m and b
Lines that are perpendicular to one another have slopes that are negative reciprocals.
The line 2y-x=3, when re-arranged into slope intercept form, is 2y=x+3, or y=(1/2)x+3/2.
Its slope is 1/2.
Therefore the line perpendicular to it has a slope
of -2/1 or simply -2
The equation we seek is thus y=-2x+b.
y=-2x+b passes through (-3,7).
Therefore, X=-3 and y=7,satisfy this equation.
Substituting, 7=-2(-3)+b
7=6+b, from which b=1
Our equation is y=2x+1

These lines meet at the point of intersection.
At this point, the x and y values are the same for
both lines.
Since the y's are the same,
2x+1=(1/2)x+3/2
Multiplying both sides of the equation by 2,
4x+2=x+3
4x-x=3-2
3x=1
x=1/3
Substitute this value back into either equation to get the y value.
2y-x=3
2y-1/3=3
2y=3+1/3 ,
2y=10/3
y=10/6, or 5/3

The lines meet at (1/3, 5/3)

Verify:
2y-x=3
2(5/3)-1/3=3
10/3-1/3=3
9/3=3
3=3


y=2x+1
5/3=2(1/3)+1
5/3=2/3+1
5/3=5/3
Our calculations and answers are correct

Answer 3

a) give an equation for the line L, perpendicular to L, through P.
okay. Line L has the equation 2y-x=3.
This can be rewritten as:
2y=x+3
y=(1/2)x+(3/2)
right?
now we know that a line perpendicular would have the opposite reciprocal slope, so the slopw would be -2

say the equation for the new line is
(y-y1)=m(x-x1)
(y-7)=(-2)(x+3)

You can either leave this as is, or you can convert it to slope intercept form.

y-7=(-2)(x+3)
y-7=-2x-6
y=-2x+1


b) Where does L⊥ meet L?
you have your 2 equations
line L:2y-x=3
line L⊥: y=-2x+1

2y-x=3
2(-2x+1)-x=3
-4x+2-x=3
-5x=1
x=-1/5
y=-2x+1
y=-2(-1/5)+1=2/5+1=7/5

(-1/5, 7/5)