The coordinates of the points A, B and C are (−2, 3), (1, 1) and (1, 4) respectively. Find the

equation of the perpendicular from the point B to the line AC and the coordinates of the

foot of that perpendicular.

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The coordinates of the points A, B and C are A(−2, 3), B(1, 1) and

C (1, 4) .

equation of line AC

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

`3(y-3)=(x+2)`

3y-9=x+2 (i)

slope of line AC=1/3

Let BM be the perpendicular from B on to AC.

product of slopes of AC and BM =-1

Thus slope of lineBM=-3

Thus equation of line BM is

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

`y-1=-3x+3`

`3x+y=4` (ii)

The point of intersection of (i) and (ii) are

`3y-x=11`

`3x+y=4`

is the coordinate of foot of perpendicular from B to AC.

solving above equations ,we have

`x=1/10,y=37/8`

Thus coordinate of `M(1/10,37/8)`

We have used formula for line AC is

`y-y_1=((y_2-y_1)/(x_2-x_1))(x-x_1)`

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