Question

Anna drew a triangle which has an area of 4 sq.units. The two vertices of the triangle are (2,6) and (3,−2) and the third vertex with coordinate (x1, y1) lies on the line y = x + 3. Which one is the highest among the possible values of x1 + y1 ?​

Answer 1

Step-by-step explanation:

Given :-

Anna drew a triangle which has an area of 4 sq.units. The two vertices of the triangle are (2,6) and (3,−2) and the third vertex with coordinate (x1, y1) lies on the line y = x + 3.

To find :-

Which one is the highest among the possible values of x1 + y1 ?

Solution :-

Given three vertices of a triangle are (2,6) , (3,-2) and (x1, y1)

Let (x1, y1) = (2,6) => x1 = 2 and y1 = 6

Let (x2, y2) = (3,-2) => x2 = 3 and y2 = -2

Let (x3, y3) = (x1, y1) => x3 = x1 and y3 = y1

We know that

The area of a triangle formed by the three vertices (x1, y1) , (x2, y2) and (x3, y3) is

∆ = (1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units

On substituting these values in the above formula then

=> ∆ = (1/2) | 2(-2-y1)+3(y1-6)+x1(6-(-2)) |

=> ∆ = (1/2) | 2(-2-y1)+3(y1-6)+x1(6+2) |

=> ∆ = (1/2) | -4-2y1+3y1-18+8x1 |

=> ∆ = (1/2) | -22+y1+8x1 | sq.units

According to the given problem

Area of the triangle = 4 sq.units

=> (1/2) | -22+y1+8x1 | = 4

=> | -22+y1+8x1 | = 4×2

=> | -22+y1+8x1 | = 8 -------------(1)

Given that

The third vertex with coordinate (x1, y1) lies on the line y = x + 3

So , It satisfies the given equation

=> y1 = x1+3 --------------------(2)

On substituting the value of y1 in (1) then

=> | -22+x1+3+8x1 | = 8

=> | -19+9x1 | = 8

=> -19+9x1 = 8 or -19+9x1 = -8

=> 9x1 = 8+19 or 9x1 = -8+19

=> 9x1 = 27 or 9x1 = 11

=> x1 = 27/9 or x1 = 11/9

=> x1 = 3 or x1 = 11/9

Now,

If x1 = 3 then y1 = 3+3 = 4

If x1 = 11/9 then y1 = (11/9)+3

=> y1 = (11+27)/9

=> y1 = 38/9

We have ,

x1 = 3 and y1 = 4

x1 = 11/9 and y1 = 38/9

Now,

x1+y1 = 3+4 = 7

x1+y1 = (11/9)+(38/9)

=> x1+y1 = (11+38)/9

=> x1+y1 = 49/9 or 5.44

7 > 5.44

The heighest value of x1+y1 = 7

Answer:-

The heighest possible value of x1+y1 is 7

Used formulae:-

→ The area of a triangle formed by the three vertices (x1, y1) , (x2, y2) and (x3, y3) is ∆ = (1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units

→ If | x | = a then x = a or x = -a