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rectangular cartensian co-ordinates — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSrectangular cartensian co-ordinates1 Mark
MCQ
The area of a triangle is $5$. If two of its vertices are $(2,1), (3,-2)$ and the third vertex lies on the line $y = x + 3$, then the third vertex is
  • A
    $\left( { - \frac{7}{2}, - \frac{{13}}{2}} \right)$
  • B
    $\left( { - \frac{7}{2},\frac{{13}}{2}} \right)$
  • C
    $\left( {\frac{7}{2}, - \frac{{13}}{2}} \right)$
  • ✓
    $\left( {\frac{7}{2},\frac{{13}}{2}} \right)$
✓

Answer

Correct option: D.
$\left( {\frac{7}{2},\frac{{13}}{2}} \right)$
d
(d) Let the third vertex be $(p, q).$ Since it lies on the line $y = x + 3,$ therefore $q = p + 3$ .....$(i)$
Also area of the triangle is $5,$
$\therefore$  $\frac{1}{2}\,[2\,( - 2 - q) + 3\,(q - 1) + p\,(1 + 2)] = \pm \,5$
$ \Rightarrow \,\,q + 3p - 7 = \pm \,10$.....$(ii)$
Thus on solving $(i)$ and $(ii),$ we get $p = \frac{7}{2},\,\, - \frac{3}{2}$ and $q = \frac{{13}}{2},\,\frac{3}{2}.$ Hence the third vertex is either $\left( {\frac{7}{2},\,\frac{{13}}{2}} \right)$ or $\left( { - \frac{3}{2},\,\,\frac{3}{2}} \right)$.

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