Question

If the area of triangle abc with vertices [x,3] [4,4] and [3,5] is 4 sq units find x

Answer 1

Answer:

x=-3

Step-by-step explanation:

Area of triangle having vertices

$({x_1,y_1}), ({x_2,y_2}),({x_3,y_3})\:is$

$\frac{1}{2}[{x_1}({y_2-y_3})+{x_2}({y_3-y_1})+{x_3}({y_1-y_2})]$

Given:

Area of the triangle = 4 Sq.units

$\frac{1}{2}[x(4-5)+4(5-3)+3(3-4)]= 4$

x(4-5)+4(5-3)+3(3-4)= 8

-x+8-3=8

x=-3

Answer 2

given,
area of triangle = 4 sq unit .

vertices of triangle are : (x, 3), (4, 4) and (3, 5)

we know,
if (x1, y1) , (x2, y2) and (x3, y3) are vertices of triangle then, area of triangle = 1/2 [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]

so, area of given triangle = 1/2 [x(4 - 5) + 4(5 - 3) + 3(3 - 4)]

or, 4 = 1/2 [-x + 4 × 2 + 3 × -1 ]

or, 8 = -x + 8 - 3

or, - x - 3 = 0

or, x = -3

hence, value of x = -3