Question

Find the image of the point P(-3,1) in theline 2x-3y=1.

Answer 1

there's formula  image of (h,k) in line ax+ by + c =0

x-h/ a = y - k/ b = -2(ah + bk +c)/a² + b²

so here h= -3  k = 1 a= 2 b = -3  c = -1

x- (-3) / 2 = y - 1/ -3 =  -2( 2(-3) -3(1) -1) / 4 + 9

solving gives x = -3/13  and  y = -41/13

hope my ans is correct

Answer 2

$\underline{ \underline {{ \huge{ \red{ \tt{S}}}}{ \textbf{\textsf{OLUTION :-}}}}}$

$\blue{ \sf \: Given \: line \: is \: 2x - 3y = 4 \: \: \: \: \: \: ...(i)} \\ : \longrightarrow\blue{ \sf - 3y = - 2x + 4} \\ : \longrightarrow \blue{ \sf \: y = \frac{2}{3}x - \frac{4}{3} } \\ \\ \therefore \: \blue{ \sf the \: slope \: of \: the \: line \:(i) = \frac{2}{3}.}$

From P, draw a perpendicular PM on line (i) produced to point P' so that P'M = MP, then P' will be the image of P on line (i) and the line (I) is the right bisector of the segment PP'.

$\blue{ \sf \: Let \: P' \: be \: ( \alpha , \beta ). } \\ \therefore \blue{ \sf{ \:Then \: slope \: of \: PP' = \frac{ \beta - 1}{ \alpha + 3} }}$

$\blue{ \sf \: As \: (i) \: is \: perpendicular \: to \: PP' , we \: get} \\ : \longrightarrow \blue{ \sf \: \frac{ \beta - 1}{ \alpha + 3} \: . \: \frac{2}{3} = - 1 } \\ \\ : \longrightarrow \blue{ \sf \: \frac{2 \beta - 2}{3 \alpha + 9} = - 1} \\ \\ : \longrightarrow \blue{ \sf \:2 \beta - 2 = - 3 \alpha - 9} \\ \\ : \longrightarrow \blue{ \sf \:3 \alpha + 2 \beta + 7 = 0 \: \: \: \: \: ...(ii) }$

Since,

$\boxed{ \blue{ \sf{Mid-point \: of \: PP' \: = \: M( \frac{ \alpha - 3}{2} ,\frac{ \beta + 1}{2} )}} }$

$: \longrightarrow \blue {\sf \:2. \frac{ \alpha - 3}{2} - 3. \frac{ \beta + 1}{2} = 4 } \\ \\ : \longrightarrow \blue{ \sf \: \frac{2( \alpha - 3) - 3( \beta + 1)}{2} = 4 } \\ \\ : \longrightarrow \blue {\sf \: 2 \alpha - 6 - 3 \beta - 3 = 4 \times 2} \\ \\ : \longrightarrow \blue{ \sf \: 2 \alpha - 3 \beta - 9= 8} \\ \\ : \longrightarrow \blue{ \sf \: 2 \alpha - 3 \beta - 9 - 8 = 0} \\ \\ : \longrightarrow \blue{ \sf \: 2 \alpha - 3 \beta - 17 = 0 \: \: \: \: \: \: \: ...(iii)}$

Now,

$\blue { \sf \: From \: eq. (ii) \: and \: (iii),} \\ : \longrightarrow \blue{ \sf \: 3 \alpha + 2 \beta + 7 = 0 \: \: \: \: } \times 3 \\ \blue{ \sf \:2 \alpha - 3 \beta - 17 = 0 \: \: } \times 2$

⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀9α + 6β + 21 = 0

⠀⠀⠀⠀⠀⠀4α - 6β - 34 = 0

⠀⠀⠀⠀⠀(+)⠀(+)⠀(+)

⠀⠀⠀⠀⠀_______________

⠀⠀⠀⠀⠀⠀13α⠀⠀⠀- 13 = 0

⠀⠀⠀⠀⠀⠀

$: \longrightarrow \blue { \sf \: 13 \alpha - 13 = 0} \\ : \longrightarrow \blue{ \sf \:13 \alpha = 13 } \\ : \longrightarrow \blue{ \sf \: \alpha = \frac{ \cancel{13}}{ \cancel{13}} } \\ \\ : \longrightarrow \boxed{\red{ \sf \: \alpha = 1 \: .}}$

$\blue{ \sf \:On \: putting \: the \: value \: of \: \alpha \: in \: eq. (ii), } \\ \\ : \longrightarrow \blue{ \sf \: 3 \alpha + 2 \beta + 7 = 0} \\ : \longrightarrow \blue{ \sf \: 3 \times 1 + 2 \beta + 7 = 0} \\ : \longrightarrow \blue{ \sf \:2 \beta + 10 = 0 } \\ : \longrightarrow \blue{ \sf \: 2 \beta = - 10} \\ : \longrightarrow \blue{ \sf \: \beta = - \frac{ \cancel{10} {}^{5} }{ \cancel{2}} } \\ \\ : \longrightarrow \boxed{ \red{ \sf \: \beta = - 5 .} }$

therefore,

The image of P = P'(1 , -5).

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@MissSolitary✌️

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