Question

[Maths]

Find the image of the point $(1, 2)$ in the line $x-2y-7=0$.
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Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of line is

$\rm :\longmapsto\:x - 2y - 7 = 0 - - - (1)$

↝ Let the coordinates (1, 2) is represented as P.

↝ So, we have the find the image of the point P in the line x - 2y - 7 = 0.

↝ Let assume that the image of the point P in the line x - 2y - 7 = 0 be Q (a, b).

➢ We know, given line x - 2y - 7 = 0 act as a line mirror.

↝ So, Distance of the object = Distance of the image.

↝ So, line x - 2y - 7 = 0 is perpendicular bisector of PQ.

Let assume that N be midpoint of PQ.

Now,

We know that,

The slope of line ax + by + c = 0, is given by

$\red{\boxed{ \rm{ \: m \: = \: - \: \frac{coefficient \: of \: x}{coefficient \: of \: y}}}}$

Thus,

Slope of line x - 2y - 7 = 0 is

$\rm :\longmapsto\:m = - \: \dfrac{1}{( - 2)} = \dfrac{1}{2}$

Further, we know that,

Two lines having slope m and M are perpendicular iff Mm = - 1.

Since, PN is perpendicular to x - 2y - 7 = 0

So,

Slope of PN = - 2

We know,

Equation of line passing through the point (a, b) having slope m is given by y - b = m ( x - a ).

➢ So, Equation of PN passing through the point (1, 2) having slope, - 2 is

$\rm :\longmapsto\:y - 2 = - 2(x - 1)$

$\rm :\longmapsto\:y - 2 = - 2x + 2$

$\bf\implies \:2x + y = 4 - - - - (2)$

Now, Solving equation (1) and (2), to get the coordinates of N.

$\rm :\longmapsto\:x - 2y - 7 = 0 - - - (1)$

and

$\rm :\longmapsto\:2x + y = 4 - - - (2)$

On multiply equation (2) by 2, we get

$\rm :\longmapsto\:4x + 2y = 8 - - - (3)$

On adding equation (1) and (3), we get

$\rm :\longmapsto\:5x = 15$

$\bf\implies \:x = 3$

On substituting x = 3, in equation (2), we get

$\rm :\longmapsto\:6 + y = 4$

$\rm :\longmapsto\:y = 4 - 6$

$\bf\implies \:y \: = \: - \: 2$

Hence,

• Coordinates of N is ( 3, - 2 ).

Now, We know

➢ The midpoint M of line segment joining the points (a, b) and (c, d) is given by

$\red{\boxed{ \rm{ \: Coordinates \: of \: M = \bigg(\dfrac{a + c}{2}, \dfrac{b + d}{2}\bigg)}}}$

So, here we have

↝ N (3, - 2) is the midpoint of line segment joining the points P (1, 2) and Q(a, b).

➢ So, using Midpoint Formula,

$\rm :\longmapsto\:(3, - 2)= \bigg(\dfrac{a + 1}{2}, \dfrac{b + 2}{2}\bigg)$

So, on comparing we get

$\rm :\longmapsto\:3= \dfrac{a + 1}{2} \: \: \: and \: \: \: \dfrac{b + 2}{2} = - \: 2$

$\rm :\longmapsto\:a + 1 = 6 \: \: \: and \: \: \: b + 2 = - 2$

$\rm :\longmapsto\:a= 6 - 1 \: \: \: and \: \: \: b = - 2 - 2$

$\rm :\longmapsto\:a= 5 \: \: \: and \: \: \: b = -4$

Hence, Coordinates of Q is (5, - 4)

So,

• The image of the point P(1, 2) in the line x - 2y - 7 = 0 be Q (5, - 4).

Answer 2

Answer:

The image of the point is (5, -4)