Homework Help

Determine the equation of the line that passes through the points (1;2) and (3;4)...

user profile pic

lillyan | Student, Undergraduate | (Level 2) Honors

Posted August 22, 2010 at 5:58 PM via web

dislike 2 like

Determine the equation of the line that passes through the points (1;2) and (3;4) .Determine the distance between the point (5;6) and the line.

5 Answers | Add Yours

Top Answer

user profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted August 22, 2010 at 6:02 PM (Answer #1)

dislike 1 like

We have to determine the equation of the line that passes through the given points.

Because we know the coordinates of the points, we'll calculate the equation of the line in this way:

(x-x1)*(y2-y1)=(y-y1)(x2-x1)

(x-1)(4-2)=(y-2)(3-1)

2(x-1)=2(y-2)

We'll divide by 2 both sides:

x-1 = y-2

The equation will become:

l: y-x-1=0

We'll note the point (5,6) as M.

The distance between the M point and the l line is:

d(M,l)=|(-1*5+1*6-1)|/sqrt[(-1)^2+(1)^2]

d(O,l) = |0|/sqrt 2

d(O,l) = 0

Because there is no distance between the M point and the line, the M point is located on the line l.

user profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 22, 2010 at 6:55 PM (Answer #2)

dislike 0 like

The equation for the ine is:

y- y1 = m(x-x1)

m= (y2-y1)/(x2-x1) = (4-2)/(3-1) = 1

==> y-2 = 1(x-1)

==> y= x - 1 + 2

==> y= x+1

==> -x + y -1 = 0

The distance between the line and the point (5,6) is:

d = lax+ by + cl/sqrt(a^2 + b^2)

   = l -1*5 + 1*6 -1 l / sqrt(1+1)

 = 0/sqrt2=0

Then the distance is 0, which means that the point is located on the line.

user profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted August 22, 2010 at 7:01 PM (Answer #3)

dislike 0 like

The line passing through (x1,y1) and (x2,y2) is:

(y-y1 )= [(y2-y1)/(x2-x1)](x-x1)....(1)

(x1,y1) = (1 , 2)  and (x2, y2) = (3,4)

So the line  (1) becomes:

y-2 = [(4-2)/(3-1)](x-1)

y-2 = x-1 .

x-y+1 =0. ....(2)

To find the  distance of the lin (2)  from (5 ,6), we use the formula of the distance d from the point (k,l)  to a line Ax+by +C  given by:

d = |(Ak+By+C) /sqrt(A^2+B^2) |

So (k,l) = (5,6) and  AX+by +C = x-y+1.

So d = |(1*5 -1*6 + 1)/(sqrt[1^2+ (-1)^2]| = 0/ 2 = 0

 

user profile pic

thewriter | College Teacher | (Level 1) Valedictorian

Posted August 22, 2010 at 7:08 PM (Answer #4)

dislike 0 like

The equation of a line that passes through the points (x1,y1) and (x2,y2) is (y-y1)=[(y2-y1)/(x2-x1)]*(x-x1)

Given the points (1,2) and (3,4), substituting the values for x1,x2, y1, y2 we get.

y-2=[(4-2)/(3-1)](x-1)=x-1 therefore the line is y=x+1

Substituting 5,6 into the equation we see that 6=5+1, therefore this means that the point (5,6) lies on the line and hence the required distance is 0.

user profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted August 22, 2010 at 8:19 PM (Answer #5)

dislike 0 like

Equation of a line is given by the equation:

y = mx + c

Where:

m = slope of the line

c = y coordinate of the line where it cuts x-axis.

To find the value of slope of line passing through the two given points we use the formula:

Slope = m = (y2 - y1)/(x2 - x1)

Where the coordinates of the two points are (x1, y1) and (x2, y2)

Substituting the values of x1, y1, x2, and y2 in the equation of slope we get:

m = (4 - 2)/(3 - 1)

= 2/2 = 1

To find value of c, we substitute the values of m, x1and y1 in the general equation of the line. This gives us:

2 = 1*1 + c

==> c = 1

Substituting values of m and c in the general equation of line, the equation of line passing through the two given points becomes:

y = x + 1

To find the distance between this line and the point (5, 6), we not that the y coordinate of this point is 1 more than the x coordinate.

In other words if we represent the coordinates of this point by (x3, y3):

y3 = x3 +1

Thus the points satisfy the equation of the given line.

This means the point lies on the given line. Therefore:

Distance of the point from the line = 0

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes