In right angled triangle ABC angle B=90 degree
and AB= root 34 unit. The coordinates of point B and Care (4,2)and (-1,y) respectively.
If area of triangle ABC = 17 sq.units, then find value of y.
area = (1/2)(BC)(AB)
17 = (1/2)(√(5^2 + (2-y)^2 ) (√34
34/√34 = √(25 + 4 - 4y + y^2)
square both sides
34 = 29 - 4y + y^2
y^2 - 4y - 5 = 0
(y-5)(y+1) = 0
y = 5 or y = -1
To find the value of y, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
In this case, the base of the triangle is AB, which has a length of √34 units.
The height of the triangle can be found by calculating the distance between the line containing points B(4,2) and C(-1,y) and the point A.
We can use the distance formula to find the height:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
For points B(4,2) and A(0,0), the distance is √((4-0)^2 + (2-0)^2) = √20 units.
So the area of the triangle is (1/2) * √34 * √20 = √(34*20)/2 = √680/2 = √170/2 units.
Given that the area of the triangle is 17 sq. units, we set √170/2 = 17 and solve for √170:
√170/2 = 17
√170 = 34
Therefore, the value of y is 34.