Question

If the points $A(6,1),B(8,2),C(9,4)$ and $D(k,p)$ are the vertices of a parallelogram taken in order, then find the values of $k$ and $p$.

Answer 1

ABCD is a parallelogram.

So by distance formula we have,

Distance between two points = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{(8-6{)}^2+(2-1{)}^2}=\sqrt{4+1}=\sqrt{5}$

$BC=\sqrt{(9-8{)}^2+(4-2{)}^2}=\sqrt{1+4}=\sqrt{5}$

$AB=CD$ and $BC=AD$ (opposite sides of a parallelogram)

$A{B}^2=C{D}^2$

$\Rightarrow 5=(k-9{)}^2+(p-4{)}^2$

$\Rightarrow 5=k^2+81-18k+p^2+16-8p$ --------- (1)

Also, $B{C}^2=A{D}^2$

$\Rightarrow 5=(K-6{)}^2+(P-1{)}^2$

$\Rightarrow 5=k^2+36-12k+p^2+1-2p$

$\Rightarrow k^2+p^2-12k-2p+32=0$ ------------(2)

from (1) and(2), We get;

$k^2+p^2-18k-8p+92=k^2+p^2-12k-2p+32$

$\Rightarrow 18k-8p-92=12k+2p-32$

$\Rightarrow 18k-12k=2p+8p-32+92$

$\Rightarrow 6k=10p+60$

$\Rightarrow 3k=5p+30$

$\Rightarrow k=\frac{5p+30}{3}$ --------(3)

From (1) and (3), we get;

$5=(k-9{)}^2+(p-4{)}^2$

$\Rightarrow 5={(\frac{5p+30}{3}-9)}^2+(p-4{)}^2$

$\Rightarrow 5={\left(\frac{5p+30-27}{3}\right)}^2+(p-4{)}^2$

$\Rightarrow 5={\left(\frac{5p+3}{3}\right)}^2+(p-4{)}^2$

$\Rightarrow 5=\frac{25p^2+9+30p}{9}+p^2+16-8p$

$\Rightarrow 5=\frac{25p^2+9+30p+9p^2+144-72p}{9}$

$\Rightarrow 45=34p^2-42p+153$

$\Rightarrow 34p^2-42p+108=0$

$p=3$

$thenk=7\to from\left(3\right)$