Question

Find the image of the point P(1, 2) in the line x – 3y + 4 = 0.

Answer 1

Let line AB be x – 3y + 4 = 0 and point P be (1, 2)

Let the image of the point P(1, 2) in the line mirror AB be Q(h, k).

Since line AB is a mirror. Then PQ is perpendicularly bisected at R.

Since R is the midpoint of PQ.

We know that,

Midpoint of a line joining (x₁, y₁) & (x₂, y₂)  = \(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\)

So, Midpoint of the line joining (1, 2) & (h, k) = \(\frac{1+h}{2},\frac{2+k}{2}\)

Since point R lies on the line AB. So, it will satisfy the equation of line AB x – 3y + 4 = 0

Substitution the x = 1+h/2 & y = 2+k/2 in the above equation, we get

⇒ 3 + h – 3k = 0

⇒ h – 3k = -3 …(i)

Also, PQ is perpendicular to AB

We know that, if two lines are perpendicular then the product of their slope is equal to -1

∴ Slope of AB × Slope of PQ = -1

⇒ Slope of PQ = -1/Slope of AB

Now, we find the slope of line AB i.e. x – 3y + 4 = 0

We know that, the slope of an equation is

= - 3

Now, Equation of line PQ formed by joining the points P(1, 2) and Q(h, k) and having the slope – 3 is

y₂ – y₁ = m(x₂ – x₁)

⇒ k – 2 = (-3)(h – 1)

⇒ k – 2 = -3h + 3

⇒ 3h + k = 5 …(ii)

Now, we will solve the eq. (i) and (ii) to find the value of h and k

h – 3k = -3 …(i)

and 3h + k = 5 …(ii)

From eq. (i), we get

h = -3 + 3k

Putting the value of h in eq. (ii), we get

3(- 3 + 3k) + k = 5

⇒ - 9 + 9k + k = 5

⇒ - 9 + 10k = 5

⇒ 10k = 5 + 9

⇒ 10k = 14

⇒ k = 14/10 = 7/5

Putting the value of k in eq. (i), we get

h - 3(7/5) = -3

⇒ 5h – 21 = -3 × 5

⇒ 5h – 21 = -15

⇒ 5h = -15 + 21

⇒ 5h = 6

⇒ h = 6/5