The area of a triangle is 5. Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y = x + 3. Find the third vertex.
GIVEN: The area of triangle is 5.Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y = x+3
TO FIND: The third vertex.
PROOF: Let the third vertex be (x, y)
We know area of triangle formed by three points $\left(x_{1,} y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$
Now
Taking three points(x, y), (2, 1) and (3, −2)
$\Delta=\frac{1}{2}|(x-4+3 y)-(2 y+3-2 x)|$
$\Delta=\frac{1}{2}|3 x+y-7|$
$5=\frac{1}{2}|3 x+y-7|$
$\pm 10=3 x+y-7$
$10=3 x+y-7$ or $-10=3 x+y-7$
$0=3 x+y-17 \quad \ldots \ldots .(1)$ or $0=3 x+y+3 \quad \ldots \ldots(2)$
Also it is given the third vertex lies on y = x+3
Substituting the value in equation (1) and (2) we get
$\pm 10=3 x+y-7$
$10=3 x+y-7$
$0=3 x+y-17$......(1)
$0=3 x+(x+3)-17$
$x=\frac{7}{2}$
Again subsituting the value of $x$ in equation 1 we get
$0=3 x+y-17 \quad \ldots \ldots(1)$
$0=3\left(\frac{7}{2}\right)+y-17$
$y=\frac{13}{2}$
$\operatorname{Hence}\left(\frac{7}{2}, \frac{13}{2}\right)$
Similiarly
$-10=3 x+y-7$
$0=3 x+y+3$ .....(2)
$0=3 x+(x+3)$
$x=\frac{-3}{2}$
Again subsituting the value of $x$ in cquation 2 we get
$0=3 x+y+3$ .............(2)
$0=3\left(\frac{-3}{2}\right)+y+3$
$y=\frac{3}{2}$
Hence $\left(\frac{-3}{2}, \frac{3}{2}\right)$
Hence the coordinates of $\left(\frac{7}{2}, \frac{13}{2}\right)$ and $\left(\frac{-3}{2}, \frac{3}{2}\right)$
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