Question

Anna drew a triangle which has an area of 6 sq.units. The two vertices of the triangle are
(
3
,
12
)
and
(
7
,
−
2
)
and the third vertex with coordinate
(
x
1
,
y
1
)
lies on the line
y
=
x
+
7
. Which one is the highest among the possible values of
x
1
+
y
1
?

Answer 1

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

Solution :-

given that, (x1,y1) lies on the line y = x + 4 .

So, putting both values,

→ y1 = x1 + 4

Then,

→ Coordinates of third vertex are (x1, y1) = (x1, x1 + 4)

Now, we know that, area of ∆ with coordinates of three vertex as (x1,y1) , (x2,y2) and (x3,y3) is given by :-

• (1/2) * | x1(y1 - y2) + x2(y2 - y3) + x3(y1 - y2) |

Let,

• (x1, y1) = (3, 7)
• (x2, y2) = (5, -4)
• (x3, y3) = (x1, x1 + 4)

then,

→ Area of ∆ = (1/2) * | 3{(-4) - (x1 + 4)} + 5(x1 + 4 - 7) + x1{7 - (-4)} |

→ (1/2) * | (-24) - 3x1 + 5x1 - 15 + 11x1 | = 6

→ | 16x1 - 3x1 - 15 - 24 | = 6 * 2

→ | 13x1 - 39 | = 12

→ 13x1 - 39 = ± 12

taking 12,

→ 13x1 - 39 = 12

→ 13x1 = 12 + 39

→ x1 = (51/13)

taking (-12) ,

→ 13x1 - 39 = (-12)

→ 13x1 = (-12) + 39

→ x1 = (27/13)

therefore, at x1 = (51/13) :-

→ x1 + y1 = x1 + (x1 + 4) = 2x1 + 4 = 2(51/13) + 4 = (102 + 52/13) = (154/13)

and, at x1 = (27/13) :-

→ x1 + y1 = x1 + (x1 + 4) = 2x1 + 4 = 2(27/13) + 4 = (54/13) + 4 = (54 + 52)/13 = (106/13) .

comparing both possible values of (x1 + y1) now,

→ (154/13) > (106/13) .

Hence, we can conclude that, the highest among the possible values of x1 + y1 is equal to (154/13) .

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