Question

Find the image of P(-3, 5) about the line x - y + 2 = 0

• A. (0, 2)
• B. (3, -1)
• C. (2, 0)
• D. (1, 3)

Answer 1

The correct option is B

(3, -1)

We can use the formula to find the image directly. We will first solve it without using the formula.

PR = QR $\Rightarrow$ R is the mid point of P and Q. We will first find R, foot of the perpendicular , because it is easy to find.

Let the co-ordinates of R be (h,k)

Slope of PR = $\frac{-1}{slopeoftheline}$

= $\frac{-1}{1}$

= - 1

Same slope can be calculated using P(-3,5) and R(h,k)

$\Rightarrow$ $\frac{k-5}{h+3}=-1$

$\Rightarrow$ k - 5 = -h - 3

k + h = 2 - (1)

R(h,k) is also a point on x - y + 2 = 0

$\Rightarrow$ h - k + 2 = 0

$\Rightarrow$ h - k = -2 -(2)

h + k = 2 -(1)

(1) + (2) $\Rightarrow$ 2h = 0

$\Rightarrow$ h = 0 and k = 2

Now that we got the co-ordinates of R, we can calculater the co-ordinates of PQ. Let it be (x,y)

$\Rightarrow$ $\frac{x_1-3}{2}=0and\frac{y_1+5}{2}=2$

$\Rightarrow$ $x_1=3and{y}_1=-1$

$\Rightarrow$ Q $\equiv$ (3,-1)

We will calculate the same using formula

The image (x,y) of a point (x,y) about a line ax + by + c = 0 is given by

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-2\frac{(a{x}_1+b{y}_1+c)}{a^2+b^2}$

$\Rightarrow$ $\frac{x+3}{1}=\frac{y-5}{-1}=-2\frac{(-3-5+2)}{1^2+1^2}$

$\Rightarrow$ x + 3 = $\frac{y-5}{-1}$ = -2 (-3) = 6

$\Rightarrow$ x = +3 and y = -1