The image of the point (3,5) in the line x-y+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the image of the point $P(3,5)$ in the line $x-y+1=0$ be $P'(x,y)$.Using the formula for the image of a point $(x_1, y_1)$ in the line $ax+by+

Vedclass · Accepted Answer

$(x-2)^{2}+(y-4)^{2}=4$

Vedclass · Answer

(D) Let the image of the point $P(3,5)$ in the line $x-y+1=0$ be $P'(x,y)$.Using the formula for the image of a point $(x_1, y_1)$ in the line $ax+by+c=0$:$\frac{x-x_1}{a} = \frac{y-y_1}{b} = -2 \left( \frac{ax_1+by_1+c}{a^2+b^2} \right)$Substituting the values $x_1=3, y_1=5, a=1, b=-1, c=1$:$\frac{x-3}{1} = \frac{y-5}{-1} = -2 \left( \frac{3-5+1}{1^2+(-1)^2} \right)$$\frac{x-3}{1} = \frac{y-5}{-1} = -2 \left( \frac{-1}{2} \right) = 1$So,$x-3=1 \implies x=4$ and $y-5=-1 \implies y=4$.The image point is $(4,4)$.Now,check which option is satisfied by the point $(4,4)$:For option $D$: $(4-2)^2 + (4-4)^2 = 2^2 + 0^2 = 4$.Thus,the point $(4,4)$ lies on the circle $(x-2)^2 + (y-4)^2 = 4$.

The image of the point $(3,5)$ in the line $x-y+1=0$ lies on:

Similar Questions

The unit normal vector to the line joining $i - j$ and $2i + 3j$ and pointing towards the origin is

The equation of the line perpendicular to the line $ax + by + c = 0$ and passing through $(a, b)$ is:

The coordinates of the foot of the perpendicular from $(x_1, y_1)$ to the line $ax + by + c = 0$ are

The foot of the perpendicular from the point $(2,4)$ upon the line $x+y=4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module