Co-Ordinate Geometry

Question

The equation of the line l is 5+y-2x=0.

(a) Find the co-ordinates of the points where l cuts the axes.

 

(b) Find the slope of the line l

 

The line j goes through the point (11,6) and is perpendicular to the line l.

(c) (i)  Write down the slope of the line j.

 

(ii) Find the equation of the line j

Answer 1

(a) y-intercept (where the line crosses the y-axis): the value of y when x=0, so we plug in 0 for x:

5+y-2(0)=0

5+y=0

y=-5

 

x-intercept: the value of x when y=0:

5+0-2x=0

5-2x=o

2x=5

x=2.5

 

(b) put the equation in slope-intercept form (y=mx+b): y=2x-5

m (coefficient of x) is the slope: 2

 

(c)(i) slope of a perpendicular line is the opposite (m*-1) and reciprocal (1/m) of the original slope:

original slope: m=2, so (1/2)*-1=-1/2

new equation (general form): y=-1/2x+b

(ii) to find b, plug in the coordinates of the point for x and y:

6=-1/2(11)+b

6=-11/2+b

6=-5.5+b

b=6+5.5=11.5

equation: y=-1/2x+11.5 

 

Answer 2

Okay,

a)           5 + y - 2x = 0

 

rewrite in the slope intercept form  y = mx + b

 

           y = 2x - 5

 

Now, y-intercept is when x = 0

  

            y = 2(0) -5

            y = -5

 

x-intercept is when y = 0

 

           0 = 2x - 5

 

           2x = 5

 

            x = 5/2

 

b) Therefore the line cuts the axes at (0,-5) and (5/2,0)

 

Slope of the line, as we have the equation in the slope-intercept form

 

    slope = 2

 

c) Now, to fine equation of line perpendicular to l (y = 2x -5), we first determine the slope

 

    For perpendicular line m2 = -1/m1   the negative reciprical

i)

     So, slope of new line = -1/2

 ii)

   Now to get the equation we use the point-slope form

 

       Point is (11,6)

     

        y - y1 = m(x - x1)

 

       y - 6 = -1/2(x - 11)

 

       y - 6 = -(1/2)x + 11/2

 

        y = -(1/2)x + 23/2

 

        y = -(1/2)x + 11.5   (if I could do arithmetic I could rule this world!)