Find the locus of the image of the point (2,3) on the line (2x-3y+4) +

1.	Find the locus of the image of the point (2,3) on the line (2x-3y+4) + lambda (x-2y+3). Here lambda is any constant.
Answer» Given: A point P(2,3) A line (2x-3y+4)+λ(x-2y+3)=0 To FIND: The locus of the image of the point P(2,3) on the line (2x-3y+4)+λ(x-2y+3)=0 Diagram: $\setlength{\unitlength}{1cm}\thicklines\begin{picture}(30,15)\put(0,0){\line(2,1){5}}\put(0,2.5){\line(2,-1){5}}\thinlines\multiput(0,1.23)(0.4,0){14}{\line(1,0){0.2}}\multiput(2.49,2.95)(0,-0.4){9}{\line(0,-1){0.2}}\put(3.39,2.6){\circle{0.1}}\put(3.7,2.7){$P(2,3)$}\put(1.48,-0.24){\circle{0.1}}\put(0.5,-0.7){$R(h,k)$}\put(2.48,1.24){\circle{0.2}}\put(1.6,1.7){$Q(1,2)$}\put(1.48,-0.24){\line(2,3){2}}\put(6.5,2){$(2x-3y+4=0)$}\put(6.67,0.4){$x-2y+3=0$}\put(6.5,2.1){\vector(-1,0){2}}\put(6.6,0.5){\vector(-1,0){2.1}}\end{picture}$ Here, dotted lines and line PQR are some of those lines which passes through point Q. Solution:* Here, the equation (2x-3y+4)+λ(x-2y+3)=0 represents the equations of all those lines which passes through the intersection of line 2x-3y+4=0 and x-2y+3=0 LET the intersection point of lines 2x-3y+4=0 and x-2y+3=0 be Q Now, here $\rm{2x-3y+4=0}------(1)$ $\rm{x-2y+3=0}-------(2)$ On multiplying (2) by 2, we get $\rm{2x-4y+6=0}-------(3)$ On subtracting (2) from (1), we get $\rm{2x-3y+4-(2x-4y+6)=0}$ $\rm{2x-3y+4-2x+4y-6=0}$ $\rm{y-2=0}$ $\rm{y=2}$ On putting y=2 in (1), we get $\rm{2x-3(2)+4=0}$ $\rm{2x-6+4=0}$ $\rm{2x-2=0}$ $\rm{2x=2}$ $\rm{x=\dfrac{2}{2}}$ $\rm{x=1}$ So, the coordinates of point Q is (1,2) $\rule{190}{1}$ Now, let the image of point P be R(h,k) Since, point R is image of P, So,by geometry $\longrightarrow\rm{PQ=QR}$ Using Distance formula, $\sf{\sqrt{(1-2)^{2}+(2-3)^{2}}=\sqrt{(h-1)^{2}+(k-2)^{2}}}$ On SQUARING both the sides, we get $\rm{(1-2)^{2}+(2-3)^{2}=(h-1)^{2}+(k-2)^{2}}$ $\sf{(-1)^{2}+(-1)^{2}=h^{2}+(1)^{2}-2h+k^{2}+(2)^{2}-4k}$ $\rm{1+1=h^{2}+1-2h+k^{2}+4-4k}$ $\rm{2=h^{2}-2h+k^{2}-4k+5}$ $\rm{h^{2}-2h+k^{2}-4k+5=2}$ $\rm{h^{2}+k^{2}-2h-4k+5-2=0}$ $\rm{h^{2}+k^{2}-2h-4k+3=0}$ On substituting (h,k) with (x,y), we get $\longrightarrow\rm\green{x^{2}+y^{2}-2x-4y+3=0}$ Hence, the locus of the image of the point P(2,3) on the line (2x-3y+4)+λ(x-2y+3)=0 is x²+y²-2x-4y+3=0.

Find the locus of the image of the point (2,3) on the line (2x-3y+4) + lambda (x-2y+3). Here lambda is any constant.

Answer»

Given:

A point P(2,3)

A line (2x-3y+4)+λ(x-2y+3)=0

To FIND:

The locus of the image of the point P(2,3) on the line (2x-3y+4)+λ(x-2y+3)=0

Diagram:

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(30,15)\put(0,0){\line(2,1){5}}\put(0,2.5){\line(2,-1){5}}\thinlines\multiput(0,1.23)(0.4,0){14}{\line(1,0){0.2}}\multiput(2.49,2.95)(0,-0.4){9}{\line(0,-1){0.2}}\put(3.39,2.6){\circle*{0.1}}\put(3.7,2.7){$P(2,3)$}\put(1.48,-0.24){\circle*{0.1}}\put(0.5,-0.7){$R(h,k)$}\put(2.48,1.24){\circle*{0.2}}\put(1.6,1.7){$Q(1,2)$}\put(1.48,-0.24){\line(2,3){2}}\put(6.5,2){$(2x-3y+4=0)$}\put(6.67,0.4){$x-2y+3=0$}\put(6.5,2.1){\vector(-1,0){2}}\put(6.6,0.5){\vector(-1,0){2.1}}\end{picture}$

Here, dotted lines and line PQR are some of those lines which passes through point Q.

Solution:

Here, the equation (2x-3y+4)+λ(x-2y+3)=0 represents the equations of all those lines which passes through the intersection of line 2x-3y+4=0 and x-2y+3=0

LET the intersection point of lines 2x-3y+4=0 and x-2y+3=0 be Q

Now, here

$\rm{2x-3y+4=0}------(1)$