Question

If the area of a triangle is 5 square units and its two vertices are (2,1) and (3,-2). If the third vertex is (x,y), where y=x+3, find the value of x and y.

Answer 1

SOLUTION

Let A = A(x,y), B=(2,1), C= (3,2)

Area of ∆ABC

=)1/2[x(1+2)+2(-2-y)+3(y-1)]=5{given}

=)1/2|3x+y-7|=5

=) |3x+y-7|= 10

=) 3x+y-7=+-10

=) 3x+y= 17

Or

=) 3x+y= -3

If 3x+y= 17 and y= x+3(given)

Then x= 7/2, y= 13/2

Also if,

3x+y= -3 and y= x+3

Then,

=) x= -3/2 , y= 3/2

hope it helps ✔️

Answer 2

Step-by-step explanation:

Given that,

A(x₁,y₁) = (2,1)

B(x₂,y₂) = (3,-2)

C(x₃,y₃) = (7/2,y)

Area of the triangle = 5 sq. units

y = ?

Solution:

Area of triangle = 1/2 I x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) I

5 = 1/2 I 2(-2-y) + 3(y-1) + 7/2(1+2) I

5 = 1/2 I -4-2y+3y-3+7/2+7 I

5 = 1/2 I -7+7+y+7/2 I

5 = 1/2 I y+7/2 I

10 = I y+7/2 I

10 = y+7/2 , -10 = y+7/2

y = 10 - 7/2 , y = -10-7/2

y = (20-7)/2 , y = (-20-7)/2

y = 13/2 , y = -27/2

These are the required values of y.