Question

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.

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Answer 1

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.

Answer 2

✍️Given,

(x−2y+3)+λ(2x−3y+4)=0

image of point (2,3)

✍️we have,

x−2y+3=0

2x−3y+4=0

solving the above equations, we get,

$x = 1$

$y = 2$

$\sqrt{( {2 - 1)}^{2} + {(3 - 2)}^{2} } = \sqrt{ {(h - 1)}^{2} + {(k - 2)}^{2} }$

$\sqrt{2} = \sqrt{ {h}^{2} + {k}^{2} - 2h - 4k + 1 + 4}$

$\sqrt{2} = \sqrt{ {h}^{2} + {k}^{2} - 2h - 4k + 5 }$

${h}^{2} + {k}^{2} - 2h - 4k + 5 = 2$

${h}^{2} + {k}^{2} - 2h - 4k + 3 = 0$

✍️ Therefore, the required locus is x²+y²-2x-4y+3=0