If vertices of triangle are A(x, 4); B(2, - 6) and C(5, 4); find x. Area of triangle is 35 sq cm

If vertices of triangle are A(x, 4); B(2, - 6) and C(5, 4); find x. Area of triangle is 35 sq cm

1) Obviously there are two cases of triangles we can form here; (i) ABC is an acute triangle, or (ii) ABC is an Obtuse triangle with
2) In either case, as explained by previous contributor the height from B on to AC = 10 units. Only AC is either (5-x) or (x-5)

3) Taking it as (5-x), x = -2 [Already shown by the previous contributor]

Taking it as (x-5), it is (1/2)*(x-5)*10 = 35

Solving x = 12

Thus A is either (-2,4) or (12, 4)

Alternatively you may apply area of triagle given its 3 vertices as:

(1/2){x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)} sq units.

[Note: In the above formula, we must take only in anticlockwise direction. So it either ABC or ACB]

So it is:

i) Taking ABC anticlockwise,

(1/2){x(-6-4) + 2(4-4) + 5(4+6)} = 35; Solving, x = -2

ii) Taking ACB as anticlock wise,

(1/2){x(4+6) + 5(-6-4) + 2(4-4)} = 35; Solving x = 12

Thus x is either 12 or -2

A and C both have the same y coordinate so let's consider AC to be the base of the triangle. If you compare the y-coordinates of B and C you'll find that the height of the triangle is 4 - -6 = 10.

Area = ½ * base * height
35 = ½ * (5-x) * 10
35 = 5(5-x)
7 = 5-x
x = -2

I hope this helps.

X=-6.