Question

Anna drew a triangle which has an area of 6 sq.units. The two vertices of the triangle are (3,7) and (5,−4) and the third vertex with coordinate (x1, y1) lies on the line y = x + 4. Which one is the highest among the possible values of x1 + y1 ?

Answer 1

Answer:

Step-by-step explanation:

C lies on y=x+4

Let v be x

Then=(v,v+4)

Now apply area of triangle formula

6=1/2[X1(y2-y3)+X2(y3-y1)+x3(y1-y2)

After solving

(V-3)11+(v-3)2=12

Then c=(51/13,103/13)

Now X1+y1=51/13+51/13+4=11.8

Answer 2

Given  : Anna drew a triangle which has an area of 6 sq. units.

The two vertices of the triangle are (3,7) and (5,−4) and the third vertex with coordinate (x₁, y₁) lies on the line y = x + 4.

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

Solution:

(x₁, y₁) lies on the line y = x + 4

=> y₁ = x₁ + 4

(x₁, x₁ + 4)

x₁ + y₁  = x₁ +x₁ + 4 = 2x₁ + 4

(3,7) , (5,−4)  , (x₁, x₁ + 4)

Area of triangle = (1/2) | 3 ( -4 -  x₁  -  4) + 5(x₁ + 4 - 7) + x₁(7 - (-4))|

= (1/2) | -24 - 3x₁ + 5x₁ - 15 + 11x₁ |

=  (1/2) |13x₁ -39|

(1/2) |13x₁ -39| = 6

=> |13x₁ -39| = 12

=> 13x₁ -39 = ±12

=> x₁  = 51/13  , 27/13

x₁ + y₁  =  2x₁ + 4  = 2(27/13) + 4  = 106/13

or 2 ( 51/13) + 4  = 154/13

highest among the possible values of x₁ + y₁   = 154/13

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