# A(-1,1) and B(3,4) are two vertices of triangle ABC.If the area of the triangle is 15 units sq,find the distance of C from AB.

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You should remember that the distance from a point to a line is the perpendicular line dropped from the point to the line.

Notice that the perpendicular from C to AB also represents the height of triangle.

You should remember the formula of area of triangle such that:

A = height * base / 2

You may consider CD as height and AB as base of triangle ABC.

`A = (CD*AB)/2`

You may evaluate the length of the base AB using the following formula such that:

`[AB] = sqrt((xB-xA)^2 + (y_B-y_A)^2)`

`[AB] = sqrt((3+1)^2 + (4-1)^2)`

`[AB] = sqrt(16 + 9)`

`[AB] = sqrt25 =gt [AB] = 5`

The problem provides the value of area of triangle, hence, you should substitute 15 for area and 5 for base such that:

`15 = (CD*5)/2 =gt 30 = CD*5 =gt CD = 6 ` units.

**Hence, evaluating the distance from C to AB yields CD = 6 units.**

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