Coordinate Geometry

Question

1. Find k if the following sets of lines are concurrent: y = x – 1, y = 2x + 1, y = kx + 2

2.Triangle formed by the lines y = x + 3, y = 3 – x and y = 2x + 6 is right-angled.

b. Find the area of this triangle

 

3. Find the distance between the parallel lines 3x – 4y = 12 and 3x – 4y = 24

Answer 1

1)  For the 3 lines to be concurrent, they intersect at a single point

If we work with the first two equations, we can determine where they intersect and then use that point to determine k.

y = x-1

y = 2x+1

 

We can use substitution or elimination to solve.

In this case, substitution is pretty straight forward, so we can substitute (x-1) in the second equation for y

x-1 = 2x+1

Using algebra to solve,

-x = 2 and x = -2

 

Using the first equation and the fact that x=-2 we have y = -2-1 = -3

 

So the point of intersection is (-2,-3)

 

Finally, evaluate the 3rd equation, y = kx + 2 at the point (-2,-3) to determine k

-3 = -2k + 2

-5 = -2k

k = 5/2

 

2) The first two equations listed intersect to form the right angle.  We know this because the slops of equation 1 is 1 and the slope of equation 2 is -1.  When two lines are perpendicular, the slope of the second line is the negative inverse of the slope of the first line.

We need to find the 3 vertices and can do so by solving the 3 equations in pairs of two, meaning...

 

Solve (1)                       Solve (2)                      Solve (3)

y = x+3                        y = 3-x                         y = x+3

y = 3-x                         y = 2x+6                      y = 2x+6

 

Solution                        Solution                        Solution

(0,3)                             (-1,4)                           (-3,0)

 

We need to find the length of the base and the length of the altitude

The distance between the first two point is the base.  The distance between points 1 and 3 is the altitude.

The distance formula is

d = √[(x₂-x₁)² + (y₂-y₁)²]

 

For the base

d = √[(0-(-1))² + (3-4)²] = √(1+1) = √2

 

For the altitude (height)

d = √[0-(-3))2 + (3-0)2] = √(9+9) = √18 = 3√2

 

Finally the area of a triangle is 1/2 base • height = (1/2)(√2)(3√2) = 3

 

3)

3x - 4y = 12 can be rewritten as y = (3/4)x - 3

3x - 4y = 24 can be rewritten as y = (3/4)x - 6

Find a line perpendicular to both lines above and note the points that line intersects each of the parallel lines.  I graphed the two lined and then found a line perpendicular to the first line at the y-intercept (0,-3) which would intersect the second line at the point (1,-5).

Using these two points and the distance formula above, will yield the distance between the two parallel lines.

 

d = √[(0-1)² + ((-3)-(-5)²] = √(1+4) = √5

 

Answer 2

1. The 3 lines are concurrent if they all intersect at the same point.

 

    The first two lines intersect when x-1=2x+1

 

                                                    So, x = -2

                                                         y = x-1 = -3

 

    The first two lines intersect at (-2,-3)

 

    So, if the third line also goes through the point (-2,-3), then we           must find k so that -3 = -2k + 2

 

                               -5 = -2k

 

                                k = 5/2

-----------------------------------------------------------------------------------

2. The lines y=x+3 and y=3-x are perpendicular since the slopes of the     two lines are negative reciprocals.  The point of intersection is the       vertex of the right angle of the triangle.

 

    They intersect when x+3=3-x.  So, 2x=0

 

                                                      x = 0 and y = x+3 = 3

 

    Vertex of right angle is (0,3).

 

    The lines y = 2x+6 and y = x+3 intersect when 2x+6=x+3

 

                                                                           x = -3

                                                                           y = x+3 = 0

    So, (-3,0) is also a vertex of the triangle.

 

    Finally, the lines y=3-x and y=2x+6 intersect when 3-x=2x+6

 

                                                                              3x = -3

 

                                                                               x = -1

                                                                               y = 3-x = 4

    The third vertex of the triangle is (-1,4).

 

    Let A = (-1,4), B = (-3,0), C = (0,3)

 

By drawing the triangle, we see that the legs of the right triangle are AC and BC.

 

The area of a right triangle is 1/2 times the product of the lengths of the legs.

 

Length of AC = √[(0-(-1))²+(3-4)²] = √2

Length of BC = √[(0-(-3))²+(3-0)²] = √18

 

Area of triangle = (1/2)(√2)(√18) = 3

 

-----------------------------------------------------------------------------------

 3. Since the lines are parallel, the distance between the lines is the distance between one of the lines and any point on the other line.

 

The distance between the line ax+by+c=0 and the point (x₀,y₀) is given by the formula: l ax₀+by₀+c l/√(a²+b²)

 

So, the distance between (0,-6) and the line 3x-4y=12 is

 

      l 3(0)-4(-6)-12 l / √(3²+(-4)²)  = 12/5 = 2.4