The image of the point (2, 4) with respect to | vedclass

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

Question

(A) Let the image of the point $A(2, 4)$ in the line mirror $DE$ be $C(\alpha, \beta)$. Then, $AC$ is perpendicular to $DE$.The midpoint $B$ of $AC$ i

Vedclass · Accepted Answer

$\left(-\frac{14}{13}, -\frac{8}{13}\right)$

Vedclass · Answer

(A) Let the image of the point $A(2, 4)$ in the line mirror $DE$ be $C(\alpha, \beta)$. Then, $AC$ is perpendicular to $DE$.The midpoint $B$ of $AC$ is $\left(\frac{\alpha + 2}{2}, \frac{\beta + 4}{2}\right)$.Since point $B$ lies on the line $2x + 3y - 6 = 0$, we have:$2\left(\frac{\alpha + 2}{2}\right) + 3\left(\frac{\beta + 4}{2}\right) - 6 = 0$$\Rightarrow 2\alpha + 4 + 3\beta + 12 - 12 = 0$$\Rightarrow 2\alpha + 3\beta + 4 = 0$  --- $(i)$Since $AC \perp DE$, the product of their slopes is $-1$. The slope of $DE$ is $-\frac{2}{3}$, so the slope of $AC$ must be $\frac{3}{2}$.$\frac{\beta - 4}{\alpha - 2} = \frac{3}{2}$$\Rightarrow 2\beta - 8 = 3\alpha - 6$$\Rightarrow 3\alpha - 2\beta + 2 = 0$  --- $(ii)$Solving equations $(i)$ and $(ii)$:Multiply $(i)$ by $2$ and $(ii)$ by $3$:$4\alpha + 6\beta + 8 = 0$$9\alpha - 6\beta + 6 = 0$Adding these, $13\alpha + 14 = 0 \Rightarrow \alpha = -\frac{14}{13}$.Substituting $\alpha$ in $(i)$:$2(-\frac{14}{13}) + 3\beta + 4 = 0$$-\frac{28}{13} + 3\beta + \frac{52}{13} = 0$$3\beta = -\frac{24}{13} \Rightarrow \beta = -\frac{8}{13}$.Thus, the image of the point $(2, 4)$ is $\left(-\frac{14}{13}, -\frac{8}{13}\right)$.

The image of the point $(2, 4)$ with respect to the straight line $2x + 3y - 6 = 0$ is

Similar Questions

The point $Q$ is the image of the point $P(1,5)$ about the line $y=x$ and $R$ is the image of the point $Q$ about the line $y=-x$. The circumcentre of the $\Delta PQR$ 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 coordinates of the foot of the perpendicular from $(0,0)$ upon the line $x+y=2$ are

$A$ ray of light coming from the point $(1, 2)$ is reflected at a point $A$ on the $x$-axis and then passes through the point $(5, 3)$. The coordinates of the point $A$ are:

The area (in square units) of the quadrilateral formed by the point of intersection of the lines $x+y-1=0$ and $x-y+1=0$,the point $P(1,1)$,and the feet of the perpendiculars from this point onto the lines is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module