# The points of A and B have coordinates (1,6) and (5,-2) respectively. The mid-point of AB is M. A straight line passes through M and is...  perpendicular to AB. i) Show that this line has...

The points of A and B have coordinates (1,6) and (5,-2) respectively. The mid-point of AB is M. A straight line passes through M and is...

perpendicular to AB.

i) Show that this line has equation x-2y+1=0

ii) Given that this line passes through the point (k,k+5), find the values of the constant k.

Asked on by hbenson92

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

i) You need to find the equation of line passing through midpoint and orthogonal to AB, thus you should remember that the slope of line AB is the inverse of the slope of orthogonal line passing through M.

You need to find the slope of AB such that:

`m_(AB) = (-2-6)/(5-1) =gt m_(AB) = -8/4`

`m_(AB) = -2`

You need to find coordinates of midpoint such that:

`x_M = (1+5)/2 =gt x_M = 3 ;y_M = (6-2)/2 =gt y_M = 2`

Notice that you know the slope of orthogonal line and you know coordinates of point M that lies on orthogonal line, hence you may write the equation such that:

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

You need to multiply both sides by 2 such that:

`2y - 4 = -x + 3`

You need to move all terms to the left side such that:

`x + 2y - 7 = 0`

Hence, the equation of orthogonal line `x + 2y - 7 = 0`  is not equal to the given equation `x - 2y + 1 = 0` .

ii) You need to remember that the point `(k,k+5)`  lies on line `x + 2y - 7 = 0`  if its coordinates check the equation of line such that:

`k+ 2(k+5) - 7 = 0 =gt k + 2k + 10 - 7 = 0`

Collecting like terms yields:

`3k + 3 = 0 =gt 3k = -3 =gt k = -1`

Hence, evaluating the value of k under given conditions yields k = -1.

ayeshadaw | Student, Undergraduate | eNotes Newbie

Posted on

i) First find the midpoints. x=1+5/2=3 y=+6-2/2=2 and you know that the gradient is perpendicular to AB. so the gradient is 1/2 sub in the numbers into a striaght line equation. y-2=1/2(x-3) muliply both sides by 2. y-4=2(x-3) move the 2 across. 2y-4=x-3 Then move the 3 across. 2y-4+3 -4+3=-1 2y-1=x so then move the 2y-1 across to x. 0=x-2y+1 ii)k-2(k+5)+1=0 k-2k-10+1=0 k-2k-9=0 -k-9=0 move the -k across. -9=k --> k=-9 Hope this helps. Its the right answer, because i checked the mark scheme on the aqa website.

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