Question

Find the image of (1.-2) wr.t the straight line 2x – 3y + 5 = 0.​

Answer 1

Given Question :-

• Find the image of (1, -2) w.r.t the straight line 2x – 3y + 5 = 0.

Answer

$\begin{gathered}\begin{gathered}\bf \: Given \: - \begin{cases} &\sf{a \: point \: (1,-2)} \\ &\sf{a \: line \: 2x - 3y + 5 = 0} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find \: - \begin{cases} &\sf{image \: of \: point \: (1,-2) \: in \: 2x - 3y + 5 = 0}  \end{cases}\end{gathered}\end{gathered}$

Identities and concept Used :-

Slope of line

Let us consider a line ax + by + c = 0, then slope(m) is

$\bigstar \: \: \boxed{ \pink{ \rm \: slope \: (m) = - \dfrac{coefficient \: of \: x}{coefficient \: of \: y} = - \dfrac{a}{b} }}$

Condition for perpendicular lines

Let us consider a line l having slope m and let other ljne having slope M, then two lines are perpendicular iff

$\bigstar \: \boxed{ \pink{ \rm \: m \: \times \: M \: = \: - 1 \: \: or \: \: M \: = - \dfrac{1}{m} }}$

Equation of line :

Let us consider a line which passes thr ough the point (a, b) and having slope 'm' is given by

$\bigstar \: \: \boxed{ \pink{ \rm \: y - b \: = m(x - a) \: }}$

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the midpoint of AB, then coordinates of C(x, y) is given by

$\bigstar \: \: {\underline{\boxed{ \pink{\rm \: (x, y) \: = \: {\quad \dfrac{x_1 + x_2}{2} \; or\; \dfrac{y_1 + y_2}{2} \quad}}}}}$

$\sf \: where \: coordinates \: are \: A(x_1, \: y_1) \: \: and \: \: B(x_2, \: y_2)$

Solution :-

Let AB represents the equation of line 2x - 3y + 5 = 0 and let P represents the coordinate (1, - 2).

Let image of the point P(1, - 2) be Q(a, b).

Now, Let further join Line segment PQ, intersects AB at N.

As we know, distance of object = distance of image

That is, PN = NQ

So, N is the midpoint of PQ

Also, PN is perpendicular to AB.

Step : - 1. Slope of PN

Given

Equation of line AB, 2x - 3y + 5 = 0 ----(i)

$\rm \: Slope \: of \: A B \: = - \dfrac{2}{ - 3} = \dfrac{2}{3}$

Since, PN is perpendicular to AB.

$\rm : \implies \:Slope \: of \: PN \: = \: - \dfrac{3}{2}$

Step :- 2 Equation of PN

Now, equation of line passes throu gh P(1, - 2) having slope - 3/2 is given by

$\rm : \implies \:y - ( - 2) = - \dfrac{3}{2} (x - 1)$

$\rm : \implies \:2y + 4 = - 3x + 3$

$\rm : \implies \:3x + 2y + 1 = 0 - - (ii)$

Step : 3 To find coordinates of N

Now,

We have

$\rm : \implies \:2x - 3y + 5 = 0 - - (i)$

and

$\rm : \implies \:3x + 2y + 1 = 0 - - (ii)$

Now, multiply equation (i) by 2 and equation (ii) by 3 and add, we get

$\rm : \implies \:4x - 6y + 10 \: + 9x + 6y + 3 = 0$

$\rm : \implies \:13x \: + \: 13 \: = \: 0$

$\rm : \implies \: \boxed{ \pink{ \rm \: \: x \: = \: - \: 1 \: }}$

On Substituting x = - 1 in equation (ii) we get

$\rm : \implies \: - 3 + 2y + 1 = 0$

$\rm : \implies \:2y - 2 = 0$

$\rm : \implies \: \boxed{ \pink{ \rm \: y \: = \: 1 \: }}$

$\rm : \implies \:Coordinates \: of \: N \: are \: (-1, 1)$

Step : -4 To find coordinates of Q

Now

We have

• Coordinates of P (1, - 2)

• Coordinates of N (- 1, 1)

• Coordinates of Q (a, b)

Since, N is the midpoint of PQ.

So, by using midpoint Formula,

$\rm : \implies \:(1,-2) \: = \: (\dfrac{a + 1}{2} , \dfrac{b - 2}{2} )$

On comparing we get,

$\rm : \implies \:\dfrac{a + 1}{2} = - 1\: \: and \: \: \dfrac{b - 2}{2} = 1$

$\rm : \implies \:a \: = - 3 \: \: and \: \: b \: = \: 4$

Hence,

$\rm : \implies \:Coordinates \: of \: Q \: are \: (-3, 4)$

Hence,

$\boxed{ \pink{ \rm \: Image \: of P(1, - 2) \:w.r.t. \: 2x - 3y + 5 = 0 \: is \: Q(- 3, 4)}}$