Question

The image of the point p(3,8) in the line x+ 3y -7 = 0  A) (-1,-4)
B) (-1,4)
C)(1,-4)
D) (1,4)​

Answer 1

Answer:

The image of the point p(3,8) in the line x+ 3y -7 = 0  A) (-1,-4)

B) (-1,4)

C)(1,-4)

D) (1,4)

Ans = (-1,4)

Answer 2

Let line AB be x+3y=7 and point P be (3,8)

Let Q(h,k) be the image of point P(3,8) in the line x+3y=7

Since line AB is a mirror,

1) Point P and Q are at equal distance from line AB, i.e., PR=QR, i.e., R is the mid-point of PQ

2) Image is formed perpendicular to mirror i.e., line PQ is perpendicular to line AB

Since R is the midpoint of PQ

Mid point of PQ joining (3,8) and (h,k) is

$\\ \\ \\ \\ ( \frac{h + 3}{2} \: \: , \: \: \frac{k + 8}{2} ) \\ \\ \\Co o rdinate \: \: of \: \: point \: \: R \\ = ( \frac{h + 3}{2} \: , \: \frac{k + 8}{2} )$

Since point R lies on line AB

Therefore,

$\\ \\ \\ \\ \\ ( \frac{3 + h}{2} ) + 3( \frac{8 + k}{2} ) = 7 \\ \\ \\ \\ h + 3k = - 13 \: - - - (i)$

Also, PQ is perpendicular to AB

THEREFORE,

Slope of PQ × Slope of AB = - 1

$</em></strong></p><p></p><p><strong><em>[tex] \\ \\ \\ \\ \\ Slope \: \: \: of \: \: \: AB \: \: = - \frac{1}{3}$

Therefore,

$\\ \\ \\ \\ \\ Slope \: \: of \: \: PQ = 3 \: = \frac{k - 8}{h - 3} \\ \\ \\ 3h - k = 1 \: - - - (ii)$

SOLVING EQUATIONS (i) and (ii)

h = -1 , k = -4

Hence image of Q is ( -1 , -4 )