 # The area of a triangle is 5. Two of its vertices are (2, 1) Question:

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.

Solution:

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 (xy)

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(xy), (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 $\left(\frac{-3}{2}, \frac{3}{2}\right)$  