The area of a triangle is 5 sq. unit. Two of its vertices are (2, 1)…

The correct answer is Option (2) → \(\left(\displaystyle\frac{7}{2},\displaystyle\frac{13}{2}\right)\) or \(\left(\displaystyle\frac{-3}{2},\displaystyle\frac{3}{2}\right)\)

A point on the line \(y=x+3\) can be written as \((x,\,x+3)\). Its area with the two given vertices comes from the area formula of a triangle from its vertices:

    \(\text{Area}=\displaystyle\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\)

With (2, 1), (3, −2) and (x, x + 3):

• First term, from (2, 1): \(2(-2-(x+3))=-2x-10\)

• Second term, from (3, −2): \(3((x+3)-1)=3x+6\)

• Third term, from (x, x + 3): \(x(1-(-2))=3x\)

• Adding the three terms, area \(=\displaystyle\frac{1}{2}\left|4x-4\right|=2\left|x-1\right|\)

Setting the area equal to 5:

• From \(2\left|x-1\right|=5\): \(x-1=\pm\displaystyle\frac{5}{2}\), so \(x=\displaystyle\frac{7}{2}\) or \(x=-\displaystyle\frac{3}{2}\)

• Putting \(x=\displaystyle\frac{7}{2}\) in the line gives \(y=\displaystyle\frac{13}{2}\)

• Putting \(x=-\displaystyle\frac{3}{2}\) in the line gives \(y=\displaystyle\frac{3}{2}\)

Shortcut: Only Option (2) has both points on \(y=x+3\) (\(\frac{7}{2}+3=\frac{13}{2}\) and \(-\frac{3}{2}+3=\frac{3}{2}\)); every other option has a point off the line, such as (2, 1), (13/2, −1/2) or (−3/2, 7/2).