Question

Question 18 Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Class X1 - Maths -Straight Lines Page 234

Answer 1

concept : - first of all we have to find the equation of line perpendicular to given line , then find their point of intersection. finally use mid-point concept.

solution :-
see attachment , equation of line AB is, x+3y=7-----------(1)
so, the slope of line AB (m1) = -1/3

Let P(3,8) be the given point for which image to be found.

Here, PQ ⊥ AB
so, slope of PQ × slope of AB = -1
Let slope of PQ = m
m × (-1/3) = -1
m = 3

Now, equation of line PQ by using formula
y - y1 = m(x - x1)
Where , (x1, y1) = (3, 8). and m = 3

y-8 = 3(x - 3)
y - 8 = 3x - 9
y - 3x +1 = 0 --------------(2)

To determine point Q , solve equations (1) and (2).

doing. eqn (1) × 3 + eqn(2)
3x + 9y -21 + y - 3x +1 = 0
10y -20 = 0
y = 2 , put it in equation (1)
x + 3×2 = 7
x = 1
so, co-ordinate of point Q = (1,2)

now, mid-point of P(3,8) and R(h,k) is
{(x1 + x2)/2 , (y1 + y2)/2 } = { (3+h)/2, (8+k)/2}

compare with co-ordinate Q(1,2)
1 = (3+h)/2
2 = 3 + h
h = -1

2 = (8+k)/2
4 = 8 + k
k = -4

hence, image of point P is R (-1,-4)