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+x3_y_2+x_1y_3)` 

NOW 

Taking three point (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 equatin i 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 subsitiuting the value of x in cquation 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)`