Question

If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex.

OR

Find the value of k, if the points P (5, 4), Q (7, k) and R (9, -2) are collinear.

Answer 1

Let the two given vertices be A (3, 0) and B (6, 0).
Let the coordinates of the third vertex be C (x, y).
It is given that the triangle ABC is equilateral.
Therefore, AB = BC = CA (Sides of an equilateral triangle)
$\Rightarrow \sqrt{(6-3{)}^2+(0-0{)}^2}=\sqrt{(x-6{)}^2+(y-0{)}^2}=\sqrt{(x-3{)}^2+(y-0{)}^2}$
$\Rightarrow 9=(x-6{)}^2+y^2=(x-3{)}^2+y^2$
$\therefore (x-6{)}^2+y^2=(x-3{)}^2+y^2$
$\Rightarrow -12x+36=-6x+9$
$\Rightarrow -6x=-27$
$\Rightarrow x=\frac{9}{2}$
$Now,y^2+(x-6{)}^2=9$
$\Rightarrow y^2+{(\frac{9}{2}-6)}^2=9(\therefore x=\frac{9}{2})$
$\Rightarrow y^2=9-\frac{9}{4}$
$\Rightarrow y^2=\frac{27}{4}$
$\Rightarrow y=\pm \sqrt{\frac{27}{4}}=\pm \frac{3\sqrt{3}}{2}$
Thus, the coordinates' of the third vertex are $(\frac{9}{2},\frac{3\sqrt{3}}{2})or(\frac{9}{2}-\frac{3\sqrt{3}}{2})$
OR
Let Q(7,k) divide the line segment joining P(5,4) and (9,-2) in the ratio $\lambda$ : 1
$\therefore$ Coordinates of the Point $Q=(\frac{9\lambda +5}{\lambda +1},\frac{-2\lambda +4}{\lambda +1})$
$\therefore \frac{9\lambda +5}{\lambda +1}=7andk=\frac{-2\lambda +4}{\lambda +1}$
$\Rightarrow 9\lambda +5=7\lambda +7$
$\Rightarrow 2\lambda =2$
$\Rightarrow =\lambda =1$
$Now,k=\frac{-2\lambda +4}{\lambda +1}$
$\Rightarrow k=\frac{-2\times 1+4}{1+1}$
$\Rightarrow k=\frac{-2+4}{2}$
$\Rightarrow k=1$
Thus, the value of k is 1