Question

Anna drew a triangle which has an area of 10 sq.units. The two vertices of the triangle are (4,7) and (5,-4) and the third vertex with coordinate (x1, y1) lies on the line y=x+1. 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 10 sq.units. The two vertices of the triangle are (4,7) and (5,-4) and the third vertex with coordinate

(x1, y1) lies on the line y=x+1.

To find :-

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

Solution :-

Given that

The vertices of a triangle are (4,7) , (5,-4) and (x1, y1)

Let (x1, y1) = (4,7) => x1 = 4 and y1 = 7

Let (x2, y2) = (5,-4) => x2 = 5 and y2 = -4

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

We know that

Area of a triangle formed by the given vertices 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) | 4(-4-y1)+5(y1-7)+x1(7-(-4)) |

=> ∆ = (1/2) | 4(-4-y1)+5(y1-7)+x1(7+4) |

=> ∆ = (1/2) | -16-4y1+5y1-35+11x1 |

=> ∆ = (1/2) | -51+y1+11x1 | sq.units

According to the given problem

Area of the given triangle is 10 sq.units

=> (1/2) | -51+y1+11x1 | = 10

=> | -51+y1+11x1 | = 10×2

=> | -51+y1+11x1 | = 20-----------(1)

And Given that

the third vertex with coordinate (x1, y1) lies on the line y=x+1.

So, it satisfies the given equation

=> y1 = x1 + 1 -------------------(2)

On substituting the value of y1 from (2) in (1) then

=> | -51+x1+1+11x1 | = 20

=> | -50+12x1 | = 20

=> -50+12x1 = 20 or -50+12x1 = -20

=> 12x1 = 20+50 or 12x1 = -20+50

=> 12x1 = 70 or 12x1 = 30

=> x1 = 70/12 or 30/12

=> x1 = 35/6 or 5/2

On substituting the value of x1 in (2) then

=> y1 = (35/6)+1 or (5/2)+1

=> y1 = (35+6)/6 or (5+2)/2

=> y1 = 41/6 or 7/2

Now

The value of x1+y1

=> (35/6)+(41/6) or (5/2)+(7/2)

=> (35+41)/6 or (5+7)/2

=> 76/6 or 12/2

=> 38/3 or 6

The heighest value = 38/3

Answer :-

The required possible value of x1+y1 is 38/3

Used formulae:-

→ Area of a triangle formed by the given vertices is

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

→ | x | = a => x = a or -a