The point A divides the join of P (−5, 1) and Q(3, 5) in the ratio k:1.


The point A divides the join of P (−5, 1)  and Q(3, 5) in the ratio k:1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C(7, −2) is equal to 2 units.


GIVEN: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio k: 1

Coordinates of point B (1, 5) and C (7, −2)

TO FIND: The value of k

PROOF: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio k: 1

So the coordinates of $A$ are $\left[\frac{3 k+1}{k+1}, \frac{5 k+1}{k+1}\right]$

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 Area of ΔABC= 2 sq units.

Taking three points A $\left[\frac{3 k+1}{k+1}, \frac{5 k+1}{k+1}\right], B(1,5)$ and $C(7,-2)$

$2=\frac{1}{2}\left\{\left(\frac{3 k-5}{k+1}\right) 5-2+7\left(\frac{5 k+1}{k+1}\right)\right\}-\left(\frac{5 k+1}{k+1}\right)+35-2\left(\frac{3 k-5}{k+1}\right)$

$2=\frac{1}{2}\left\{\left(\frac{15 k-25}{k+1}\right)-2+\left(\frac{35 k+7}{k+1}\right)\right\}-\left(\frac{5 k+1}{k+1}\right)+35-\left(\frac{6 k-10}{k+1}\right)$

$2=\frac{1}{2}\left(\frac{15 k-25-2 k+35 k+7}{k+1}\right)-\left(\frac{5 k+1+35 k+35-6 k+10}{k+1}\right)$

$2=\frac{1}{2}\left(\frac{(48 k-20)-(34 k+46)}{k+1}\right)$

$2=\frac{1}{2}\left(\frac{14 k-66}{k+1}\right)$

$2=\left(\frac{7 k-33}{k+1}\right)$

$\pm 2=\left(\frac{7 k-33}{k+1}\right)$

$\Rightarrow 7 k-33=\pm 2(k+1)$

$\Rightarrow 7 k-33=2(k+1), 7 k-33=-2(k+1)$

$\Rightarrow 7 k-33=2 k+2,7 k-33=-2 k-2$

$\Rightarrow 7 k-2 k=33+2,7 k+2 k=+33-2$


$\Rightarrow 5 k=35,9 k=31$

$\Rightarrow k=\frac{35}{5}, k=\frac{31}{9}$

$\Rightarrow k=7, k=\frac{31}{9}$

Hence $k=7$ or $\frac{31}{9}$

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