Question

Find the vertex of a triangle if its two vertices are $(-1,4)$ and $(5,2)$ and mid-point of one side is $(0,3)$.

Answer 1

Given 2 vertices A (-1,4) B(5,2)

And mid point of one side is (0,3)

Let $3^{rd}$ vertex be $(x_1,y_1)$

case I : Join vertex $\left(x_1{y}_1\right)$ with (-1,4) & M.P is (0,3)

Eq of line $=\frac{y-4}{3-4}=\frac{x+1}{0+1}\Rightarrow -(y-4)=x+1$

$x=-y+3\Rightarrow x_1=y_1+3$

By distance formula = (0,3) is M.P of (-1,4) & $\left(x_1{y}_1\right)$

$\sqrt{(0+1{)}^2+(3-4{)}^2}=\sqrt{(x_1-0{)}^2+(y_1-3{)}^2}$

$1+1=x_1^2+y_1^2-6y_1+9$

$x_1^2+y_1^2-6y_1+7=0[\because x_1=-y_1+3]$

$(-y_1+3{)}^2+y_1^2-6y_1+7=0$

$y_1^2+y_1^2+9-6y_1-6y_1+7=0$

$2y_1^2-12y_1+16=0\to y_1^2-6y_1+8=0$

$y_1^2-2y_1-4y_1+8=0$

$\therefore y_1=4,y_1=2$

$(y_1-4)(y_1-2)=0$

$y_1=4$ then $x_1=-4+3=-1$

$y_1=2$ then $x_1=-2+3=1$

so vertex is (1,2) since (-1,4) is already existing

case II :- Join vertex $\left(x_1{y}_1\right)$ with (5,2) & M.P is(0,3)

eq of line $\frac{y-2}{3-2}=\frac{x-5}{0-5}\Rightarrow -5(y-2)=(x-5)$

$x=-5y+10+5\Rightarrow x_1=-5y_1+15$

By distance formula : (0,3) is M.P of(5,2) &$\left(x_1{y}_1\right)$

$\sqrt{(0-5{)}^2+(3-2{)}^2}=\sqrt{(x_1-0{)}^2+(y_1-3{)}^2}$

$25+1=x_1^2+y_1^2-6y_1+9\Rightarrow x_1^2+y_1^2-6y_1-17=0[\because x_1=5y_1+15]$

$(-5y_1+15{)}^2+y_1^2-6y_1-17=0\Rightarrow 25y_1^2+225-150y_1+y_1^2-6y_1-17=0$

$26y_1^2-156y_1+208=0\Rightarrow y_1^2-6y_1+8=0$

$y_1^2-4y_1-2y_1+8=0\Rightarrow (y_1-4)(y_1-2)=0$

$y_1=4$ then $x_1=-5$ $y_1=2$ then $x_1=5$

So vertex is $(-5,4)$ since $(5,2)$ is already existing

$\therefore$ The third vertex can be either $(1,2)or(-5,4)$