Question

The area of the triangle ABC is \[{\text{20sq}}{\text{.unit}}\]. The coordinate of vertex A are \[( - 5,0)\]and B are \[(3,0)\]. The vertex C lies on the line \[x - y = 2\], the coordinate of C are
A. \[(5,3)\]
B. \[( - 3, - 5)\]
C. \[( - 5, - 7)\]
D. \[(7,5)\]

Answer 1

Hint: As we know, two vertices are \[( - 5,0),(3,0)\] and let the third vertices be \[(x,y)\], so we use this data and apply the determinant method of finding the area of the triangle which is\[\left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|} \right|\], and solve for the unknown coordinates.

Complete step by step answer:

So we can use the above provided information that the area of the triangle is \[20\] sq. units and two of its vertices \[({x_1},{y_1}),({x_2},{y_2})\] are\[( - 5,0),(3,0)\], proceeding with the given, we have,
The formula of area of the triangle which is\[\left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|} \right|\],
On substituting values, we get,
\[\left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  { - 5}&0&1 \\
  3&0&1 \\
  x&y&1
\end{array}} \right|} \right| = 20\]
Now, open the mode and write the possible values at R.H.S also substitute \[x = 2 + y\]
\[\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  { - 5}&0&1 \\
  3&0&1 \\
  {y + 2}&y&1
\end{array}} \right| = \pm 20\]
Now , cross-multiply the digit on R.H.S and expand the given determinant
\[
   \Rightarrow - 5( - y) - 0 + 1(3y) = \pm 40 \\
   \Rightarrow 8y = \pm 40 \\
   \Rightarrow y = \pm 5 \\
 \]
On solving both the equation we can obtain
\[
  for,y = 5 \\
  x = y + 2 = 5 + 2 \\
  x = 7 \\
  for,y = - 5 \\
  x = y + 2 = - 5 + 2 \\
  x = - 3 \\
 \]
Hence , the coordinates are \[(7,5)\] and \[( - 3, - 5)\]
Hence, option (b) and (d) are required answers.

Note: Triangle : A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted.
The formula for finding area of triangle when points in 2D plane are given is \[\left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|} \right|\], whereas if points are given in 3D plane then the formula for area of triangle will be \[\left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&{{z_1}} \\
  {{x_2}}&{{y_2}}&{{z_2}} \\
  {{x_3}}&{{y_3}}&{{z_3}}
\end{array}} \right|} \right|\].
Keep in mind that the area of the triangle is always positive so it cannot be negative and use modulus and open modulus carefully.