Question

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex lies on y = x + 3. Find the third vertex.

Answer 1

GIVEN: The area of triangle is 5. Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y = x + 3

TO FIND: The third vertex.

PROOF: Let the third vertex be (x, y)

We know area of triangle formed by three points (x₁, y₁) (x₂, y₂) (x₃, y₃)is given by 

`Δ =1/2 [x_1y_2 +x_2y_3+x_3y_1) -(x_2y_1+x_3y_2+x_1y_3)` 

NOW 

Taking three points (x, y), (2, 1) and (3, -2) 

`Δ=1/2[(x-4+3y)-(2y+3-2x)` 

`Δ=1/2[3x+y-7]`

`5=1/2[3x+y-7]`

`+-10=3x+y-7` 

`10=3x+y-7 or -10=3x+y-7`  

`0=3x+y-17 ......(1) or 0=3x+y+3 .....(2) ` 

Also it is given the third vertex lies on y = x + 3

Substituting the value in equation (1) and (2) we get   

`+-10=3x+y-7` 

`10=3x+y-7 `  

`0=3x+y-17  ......(1)`

`0=3x+(x+3)-17` 

`x=7/2` 

Again, substituting the value of x in equation 1 we get 

`0=3x+y-17 ......(1)` 

`0= 3(7/2)+y-17` 

`y=13/2` 

Hence `(7/2,13/2) `  

Similirly 

`-10=3x +y-7` 

`0=3x+y+3......(2)` 

`0=3x+(x+3) ` 

`x=3/2` 

Again, substituting the value of x in equation 2 we get 

`0=3x+y+3 ......(2)` 

`0=3 ((-3)/2)+y+3`  

 `y=3/2` 

Hence `((-3)/2,3/2)` 

Hence the coordinate of `(7/2, 13/2) and ((-3)/2, 3/2)`