Question

The area of the triangle is 5 sq. units two of its vertices are \[(2,1),(3, - 2)\]. The third vertex lies on \[y = x + 3\]. The third vertex is
A. \[\left( {\dfrac{7}{2},\dfrac{3}{2}} \right)\]
B. \[\left( {\dfrac{{ - 3}}{2},\dfrac{3}{2}} \right)\]
C. \[\left( {\dfrac{{ - 3}}{2},\dfrac{{13}}{2}} \right)\]
D. \[\left( {\dfrac{7}{2},\dfrac{5}{2}} \right)\]

Answer 1

Hint: As we know, two vertices are \[(2,1),(3, - 2)\] and let the third vertices be \[(x,y)\]. We can substitute these in the equation to find the area of the triangle using determinant method which is given by \[Area = \left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|} \right|\]. We can substitute the equation of the line to eliminate one variable. Then we can equate the determinant to the given area. Then we can solve for the variable. Then we can find the other coordinate by substituting it in the equation of the line.

Complete step-by-step answer:
We know that area of the triangle of the triangle with vertices $\left( {{x_1},{y_1}} \right)$, $\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ is given by \[Area = \left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|} \right|\]
We are provided information that the area of the triangle is 5 sq. units, two of its vertices are \[(2,1),(3, - 2)\]. Let the third vertices be \[(x,y)\]. So, we get the area as
 \[ \Rightarrow \left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  2&1&1 \\
  3&{ - 2}&1 \\
  x&y&1
\end{array}} \right|} \right| = 5\]
Now, open the modulus and write the possible values at R.H.S.
 \[ \Rightarrow \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  2&1&1 \\
  3&{ - 2}&1 \\
  x&y&1
\end{array}} \right| = \pm 5\]
As the \[{3^{rd}}\] vertex lies on the line \[y = x + 3\], it will satisfy the equation. So, we can substitute for y.
 \[ \Rightarrow \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  2&1&1 \\
  3&{ - 2}&1 \\
  x&{x + 3}&1
\end{array}} \right| = \pm 5\]
Now, cross-multiply the digit on R.H.S and expand the given determinant, we get,
 \[ \Rightarrow 2( - 2 - x - 3) - 1(3 - x) + 1(3x + 9 + 2x) = \pm 10\]
On simplification, we get
 \[ \Rightarrow 2( - 5 - x) - 3 + x + 5x + 9 = \pm 10\]
On opening bracket, we get
 \[ \Rightarrow - 10 - 2x - 3 + x + 5x + 9 = \pm 10\]
On adding like terms, we get
 \[ \Rightarrow 4x - 4 = \pm 10\]
Now we can solve for x in both the cases.
When \[4x - 4 = 10\] , we get
 \[ \Rightarrow 4x = 14\]
Dividing the equation by 4 we get
 \[ \Rightarrow x = \dfrac{7}{2}\]
When \[4x - 4 = - 10\]
 \[ \Rightarrow 4x = - 6\]
Dividing the equation by 4 we get
 \[ \Rightarrow x = - \dfrac{3}{2}\]
As now using the equation of line given in the question which is \[y = x + 3\] So the values of y are
For \[x = - \dfrac{3}{2}\],
 \[y = x + 3\]
On substituting the value of x, we get
 \[ \Rightarrow y = - \dfrac{3}{2} + 3\]
On simplification we get
  \[ \Rightarrow y = \dfrac{3}{2}\]
For \[x = \dfrac{7}{2}\],
 \[y = x + 3\]
On substituting the value of x, we get
 \[ \Rightarrow y = \dfrac{7}{2} + 3\]
On simplification we get
  \[ \Rightarrow y = \dfrac{{13}}{2}\]
Hence, the coordinates are \[\left( { - \dfrac{3}{2},\dfrac{3}{2}} \right)\] and \[\left( {\dfrac{7}{2},\dfrac{{13}}{2}} \right)\].
Out of them only one is given in the option. Therefore, the required coordinate is \[\left( { - \dfrac{3}{2},\dfrac{3}{2}} \right)\]
Hence, option B is the required answer.

Note: Alternatively, we are given that the third vertex lies on \[y = x + 3\].
So, any point on the line will satisfy the equation \[x - y + 3 = 0\].
So, we can check whether the given options satisfy the equation or not.
Consider option A \[\left( {\dfrac{7}{2},\dfrac{3}{2}} \right)\].
On substituting the values on the equation \[x - y + 3 = 0\], we get
 \[ \Rightarrow \dfrac{7}{2} - \dfrac{3}{2} + 3 = 2 + 3 = 5 \ne 0\]
So, this won’t lie in the given line. Thus, option A cannot be the answer.
Consider option B \[\left( { - \dfrac{3}{2},\dfrac{3}{2}} \right)\].
On substituting the values on the equation \[x - y + 3 = 0\] , we get
 \[ \Rightarrow \dfrac{{ - 3}}{2} - \dfrac{3}{2} + 3 = - 3 + 3 = 0\]
So, this point lies in the given line. Thus, option B is an answer.
Consider option C \[\left( {\dfrac{{ - 3}}{2},\dfrac{{13}}{2}} \right)\].
On substituting the values on the equation \[x - y + 3 = 0\], we get,
 \[ \Rightarrow \dfrac{{ - 3}}{2} - \dfrac{{13}}{2} + 3 = 8 + 3 = 11 \ne 0\]
So, this won’t lie in the given line. Thus, option C cannot be the answer.
Consider option D \[\left( {\dfrac{7}{2},\dfrac{5}{2}} \right)\].
On substituting the values on the equation \[x - y + 3 = 0\], we get
 \[ \Rightarrow \dfrac{7}{2} - \dfrac{5}{2} + 3 = 1 + 3 = 4 \ne 0\]
So, this won’t lie in the given line. Thus, option D also cannot be the answer.
Therefore, the required coordinate is \[\left( { - \dfrac{3}{2},\dfrac{3}{2}} \right)\]
Hence, option B is the required answer.