Let Q be the image of a point P(1, 2) with re | vedclass

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

Question

(A) The image $Q(x_1, y_1)$ of point $P(1, 2)$ with respect to $x + y + 1 = 0$ is given by $\frac{x_1 - 1}{1} = \frac{y_1 - 2}{1} = -2 \frac{1(1) + 1(

Vedclass · Accepted Answer

$\sqrt{10}$

Vedclass · Answer

(A) The image $Q(x_1, y_1)$ of point $P(1, 2)$ with respect to $x + y + 1 = 0$ is given by $\frac{x_1 - 1}{1} = \frac{y_1 - 2}{1} = -2 \frac{1(1) + 1(2) + 1}{1^2 + 1^2} = -2 \frac{4}{2} = -4$. Thus,$x_1 - 1 = -4 \implies x_1 = -3$ and $y_1 - 2 = -4 \implies y_1 = -2$. So,$Q = (-3, -2)$. $M$ is the midpoint of $PQ$,so $M = (\frac{1-3}{2}, \frac{2-2}{2}) = (-1, 0)$. The image $R(x_2, y_2)$ of point $Q(-3, -2)$ with respect to $x - y - 1 = 0$ is given by $\frac{x_2 + 3}{1} = \frac{y_2 + 2}{-1} = -2 \frac{1(-3) - 1(-2) - 1}{1^2 + (-1)^2} = -2 \frac{-3 + 2 - 1}{2} = -2 \frac{-2}{2} = 2$. Thus,$x_2 + 3 = 2 \implies x_2 = -1$ and $y_2 + 2 = -2 \implies y_2 = -4$. So,$R = (-1, -4)$. $N$ is the midpoint of $QR$,so $N = (\frac{-3-1}{2}, \frac{-2-4}{2}) = (-2, -3)$. The distance $MN = \sqrt{(-2 - (-1))^2 + (-3 - 0)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.

Let $Q$ be the image of a point $P(1, 2)$ with respect to the line $x + y + 1 = 0$ and $R$ be the image of $Q$ with respect to the line $x - y - 1 = 0$. If $M$ and $N$ are the midpoints of $PQ$ and $QR$ respectively,then $MN =$

Similar Questions

If the point $P(4,1)$ undergoes a reflection in the line $x-y=0$,then a translation through a distance of $2$ units along the positive $X$-axis and finally projected on the $X$-axis,then the coordinates of $P$ in the final position are

Let $ABC$ be a triangle and $A=(1,2)$. If $x-3y-5=0$ and $x+5y-9=0$ are the perpendicular bisectors of the sides $AB$ and $BC$ respectively,then the length of the side $AC$ is

The image of a point $A(3, 8)$ in the line $x + 3y - 7 = 0$ is

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

$A$ ray of light is incident along a line which meets another line,$7x - y + 1 = 0$,at the point $(0, 1)$. The ray is then reflected from this point along the line,$y + 2x = 1$. Then the equation of the line of incidence of the ray of light is

Vedclass Test Series

Exam Paper Generator

Online Exam Module