Let L be the line y = 2x in the two-dimension | vedclass

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

Question

(C) Let $P(0, 1)$ be the given point and $L: 2x - y = 0$ be the line.The formula for the image $(x_1, y_1)$ of a point $(x_0, y_0)$ in the line $ax +

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is true,Statement $2$ is a correct explanation for Statement $1$.

Vedclass · Answer

(C) Let $P(0, 1)$ be the given point and $L: 2x - y = 0$ be the line.The formula for the image $(x_1, y_1)$ of a point $(x_0, y_0)$ in the line $ax + by + c = 0$ is given by $\frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.Substituting the values: $\frac{x_1 - 0}{2} = \frac{y_1 - 1}{-1} = -2 \frac{2(0) - 1(1)}{2^2 + (-1)^2} = -2 \frac{-1}{5} = \frac{2}{5}$.Thus,$x_1 = 2 	imes \frac{2}{5} = \frac{4}{5}$ and $y_1 = 1 - \frac{2}{5} = \frac{3}{5}$.So,Statement $1$ is true.By definition,the image of a point in a line is the point such that the line is the perpendicular bisector of the segment joining the point and its image. This implies the points lie on opposite sides and are equidistant from the line. Thus,Statement $2$ is true and is the correct explanation for Statement $1$.

Similar Questions

If the foot of the perpendicular drawn from the origin to a straight line is $(3, -4)$,then the equation of the line is:

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 $MN =$

The perpendicular from the origin to the line $y=mx+c$ meets it at the point $(-1, 2)$. Find the values of $m$ and $c$.

If the image of point $P(2, 3)$ in a line $L$ is $Q(4, 5)$,then the image of point $R(0, 0)$ in the same line is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module