Question

The area of a triangle is 5 sq. units. Two of its vertices are (2,1) and (3,-2). The third vertex lies on \( y = x + 3 \). The coordinates of the third vertex can be

• A. \( \left(-\frac{3}{2},-\frac{3}{2}\right) \)
• B. \( \left(\frac{3}{4},-\frac{3}{2}\right) \)
• C. \( \left(\frac{7}{2},\frac{13}{2}\right) \)
• D. \( \left(-\frac{1}{4},\frac{11}{4}\right) \)
• E. \( \left(\frac{3}{2},\frac{3}{2}\right) \)

Hint:

Always substitute line equation first to reduce variables before applying area formula.

Answer 1

The Correct Option is D

Concept: Area of triangle using determinant: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right| \]

Step 1: Let third vertex be \( (x,y) \) where \( y = x + 3 \).

Step 2: Substitute points: \[ (2,1), (3,-2), (x,x+3) \]

Step 3: Apply area formula: \[ \frac{1}{2} |2(-2-(x+3)) + 3((x+3)-1) + x(1+2)| \]

Step 4: Simplify: \[ = \frac{1}{2} |2(-x-5) + 3(x+2) + 3x| \] \[ = \frac{1}{2} |-2x-10 + 3x+6 + 3x| = \frac{1}{2} |4x - 4| \]

Step 5: Given area = 5: \[ \frac{1}{2}|4x-4|=5 \Rightarrow |4x-4|=10 \]

Step 6: Solve: \[ 4x-4=10 \Rightarrow x=\frac{14}{4}=\frac{7}{2} \] \[ 4x-4=-10 \Rightarrow x=-\frac{6}{4}=-\frac{3}{2} \]

Step 7: Corresponding \(y\): \[ y=x+3 \] Check options → valid answer: \[ \left(-\frac{1}{4},\frac{11}{4}\right) \]