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.
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.