Question

S) Anna drew a triangle which has an area of 10 sq units. The two vertices of the triangle are (4,11) and (8-2) and the thurd vertex with coordinate (2013) fies on the line y | 2 Which one is the highest among the possible values of *+91?​

Answer 1

Answer:

Transcribed image text: 5) Anna drew a triangle which has an area of 9 sq.units. The two vertices of the triangle are (4,9) and (6,-3) and the third vertex with coordinate (21, Yı) lies on the line y = 1 + 7. Which one is the highes among the possible values of a1 + yı?

Answer 2

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

 $\[\left( {{x_1},{\rm{ }}{y_1}} \right)\]$lies on the line $y=x+1$.

To find :- Which one is the highest among the possible values of $\[{x_1} + {y_1}\]$?

Solution :-

Know that from the question,

The vertices of a triangle are $(4,7) , (5,-4)$ and $\[\left( {{x_1},{y_1}} \right)\]$.

Let $\[\left( {{x_1},{y_1}} \right)\]=\ (4,7)$.

$\[ \Rightarrow {x_1} = 4,{y_1} = 7\]$

Let $\[\left( {{x_2},{y_2}} \right) = \left( {5, - 4} \right)\]$

$\[ \Rightarrow {x_2} = 5,{y_2} = - 4\]$

Let $\[\left( {{x_3},{y_3}} \right) = \left( {{x_1},{y_1}} \right)\]$

$\[ \Rightarrow {x_3} = {x_1},{y_3} = {y_1}\]$

Know that

Area of a triangle formed by the given vertices is $\[\Delta = \frac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|\]$

Substitute these values in the above formula.

$\[ \Rightarrow \Delta = \frac{1}{2}\left| {4\left( { - 4 - {y_1}} \right) + 5\left( {{y_1} - 7} \right) + {x_1}\left( {7 - \left( { - 4} \right)} \right)} \right|\]$

$\[ \Rightarrow \Delta = \frac{1}{2}\left| { - 16 - 4{y_1} + 5{y_1} - 35 + 11{x_1}} \right|\]$

$\[ \Rightarrow \Delta = \frac{1}{2}\left| { - 51 + {y_1} + 11{x_1}} \right|sq.\,units\]$  

Put the Area of the given triangle as 10 sq.units

$\[ \Rightarrow 10 = \frac{1}{2}\left| { - 51 + {y_1} + 11{x_1}} \right|\]$

$\[ \Rightarrow 20 = \left| { - 51 + {y_1} + 11{x_1}} \right|\]------(1)$

Observe that from the question, the third vertex with coordinate $\[\left( {{x_1},{y_1}} \right)\]$lies on the line$y=x+1.$

Understand that, it satisfies the given equation

$\[ \Rightarrow {y_1} = {x_1} + 1\]-------(2)$

Substitute the value of $\[{{y_1}}\]$ from equation (2) in equation (1).

$\[\begin{array}{l}20 = \left| { - 51 + {y_1} + 11{x_1}} \right|\\ \Rightarrow 20 = \left| { - 51 + {x_1} + 1 + 11{x_1}} \right|\\ \Rightarrow 20 = \left| { - 50 + 12{x_1}} \right|\end{array}\]$

$\[\begin{array}{l} \Rightarrow \pm 20 = \left( { - 50 + 12{x_1}} \right)\\ \Rightarrow 20 = \left( { - 50 + 12{x_1}} \right)\,\,or\,\, - 20 = \left( { - 50 + 12{x_1}} \right)\\ \Rightarrow 12{x_1} = 50 + 20\,\,or\,12{x_1} = 50 - 20\\ \Rightarrow 12{x_1} = 70\,\,or\,\,12{x_1} = 30\end{array}\]$

$\[\begin{array}{l} \Rightarrow {x_1} = \frac{{70}}{{12}}\,\,or\,\,{x_1} = \frac{{30}}{{12}}\\ \Rightarrow {x_1} = \frac{{35}}{6}\,\,or\,\,{x_1} = \frac{5}{2}\end{array}\]$

Substitute the value of $\[{x_1}\]$ in equation (2).

$\[\begin{array}{l}{y_1} = {x_1} + 1\\ \Rightarrow {y_1} = \frac{{35}}{6} + 1\,\,or\,\,{y_1} = \frac{5}{2} + 1\end{array}\]$

$\[ \Rightarrow {y_1} = \frac{{35 + 6}}{6}\,\,or\,\,{y_1} = \frac{{5 + 2}}{2}\]$

$\[ \Rightarrow {y_1} = \frac{{41}}{6}\,\,or\,\,{y_1} = \frac{7}{2}\]$

Find the  value of $\[{x_1} + {y_1}\]$.

$\[{x_1} + {y_1} = \frac{{35}}{6} + \frac{{41}}{6}\,\,\,or\frac{5}{2} + \frac{7}{2}\]$

$\[ \Rightarrow {x_1} + {y_1} = \frac{{35 + 41}}{6}\,\,or\,\,\frac{{5 + 7}}{2}\]$

$\[ \Rightarrow {x_1} + {y_1} = \frac{{76}}{6}\,\,or\,\,\frac{{12}}{2}\]$

$\[ \Rightarrow {x_1} + {y_1} = \frac{{38}}{6}\,\,or\,\,6\]$

Therefore, The highest value $=\frac{38}{3}$

Hence, The required possible value of $\[{x_1} + {y_1}\]$ is $\frac{38}{3}$.