Rectangular and Cartesian products Questions and answers

Let \( \Large \left(h,\ k\right) \) be any point on the given line, then

\( \Large \left(h-a_{1}\right)^{2}+ \left(k-b_{1}\right)^{2}= \left(h-a_{2}\right)^{2}+ \left(k-b_{2}\right)^{2} \)

=> \( \Large 2 \left(a_{1}-a_{2}\right)h+2 \left(b_{1}-b_{2}\right)k = a^{2}_{1}+b^{2}_{1}-a^{2}_{2}-b^{2}_{2} \)

=> \( \Large \left(a_{1}-a_{2}\right)h+ \left(b_{1}-b_{2}\right)k=\frac{1}{2} \left(a^{2}_{1}+b^{2}_{1}-a^{2}_{2}-b^{2}_{2}\right) \)

Since (h, k) lies on the given line

\( \Large \left(a_{1}-a_{2}\right)h+ \left(b_{1}-b_{2}\right)k=c \)

On comparing equations (i) and (ii), we get

\( \Large c = \left(\frac{1}{2}\right) \left(a^{2}_{1}+b^{2}_{1}\right)-a^{2}_{2}-b^{2}_{2} \)

Let \( \Large O \left(0,\ 0\right) \) be the arthocentre \( \Large A \left(h,\ k\right) \) be the third vertex and \( \Large \beta \left(-2,\ 3\right)\ and\ C \left(5,\ -1\right) \) the other two vertices. Then the slope of the line through

A and O is \( \Large \frac{k}{h} \) while the line through B and C has the slope \( \Large \frac{\{ -1-3 \}}{ \left(5+2\right) }=\frac{-4}{7} \). By the property of the orthocentre, these two lines must be Perpendicular, so we have

\( \Large \left(\frac{k}{h}\right) \left(-\frac{4}{7}\right)=-1 => \frac{k}{h}=\frac{7}{4} \) ...)i)

Also, \( \Large \frac{5-2+h}{3}+\frac{-1+3+k}{3}=7 \)

=> h +k = 16 ...(ii)

Which is not satisfied by the points given in the options (a), (b) or (c).

Let \( \Large P \left(h,\ k\right) \) be the required point then

\( \Large 4PA^{2} = 9PB^{2} \)

=> \( \Large 4 \left(h^{2}+k^{2}\right)=9 \left(h-4\right)^{2}+9 \left(k+3\right)^{2} \)

=> \( \Large 4h^{2}+4k^{2}=9 \left(h^{2}+16-8h\right) + 9 \left(k^{2}+9+6k\right) \)

=> \( \Large 5h^{2}+5k^{2}-72h+54k+225=0 \)

Therefore, Required locus of \( \Large P \left(h,\ k\right)\ is \)

\( \Large 5x^{2}+5y^{2}-72x+54y+225=0 \)