First slide

Equation of line in Straight lines

Question

The image of the point (–8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is

Moderate

A

(16, –2)
B

(–16, 2)
C

(16, 2)
D

(–16, –2)

Solution

Equation of the given line is
4x + 7y + 13 = 0 (1)
Let Q (a, b) be the image of the point P(–8, 12) w.r.t. line (1).
Then, PQ $⊥$ line (1) and PC = CQ.
Equation of the line PC is
$(y - 12) = \frac{7}{4} (x + 8)$
[PC is $⊥$ to the line (1) and passes through (–8, 12)]
or 7x – 4y + 104 = 0 (2)
Solving Eq. (1) and (2), we get
x = –12 and y = 5. $∴$ C ≡ (–12, 5)
Since C is mid-point of PQ,
$∴ - 12 = \frac{α - 8}{2} a n d 5 = \frac{β + 12}{2}$
$\Rightarrow α = - 16 a n d β = - 2$
$∴ Q = (- 16, - 2)$

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