The image of a point A(3, 8) in the line x + | vedclass

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

Question

(A) Let the image of point $A(3, 8)$ be $A'(x_1, y_1)$.The line passing through $A$ and perpendicular to $x + 3y - 7 = 0$ has the form $3x - y + k = 0

Vedclass · Accepted Answer

$(-1, -4)$

Vedclass · Answer

(A) Let the image of point $A(3, 8)$ be $A'(x_1, y_1)$.The line passing through $A$ and perpendicular to $x + 3y - 7 = 0$ has the form $3x - y + k = 0$.Since it passes through $(3, 8)$,we have $3(3) - 8 + k = 0$,which gives $k = -1$. So,the line is $3x - y - 1 = 0$.To find the intersection point of $x + 3y = 7$ and $3x - y = 1$,we solve the system:$x + 3y = 7$ $(i)$$9x - 3y = 3$ (ii)Adding $(i)$ and (ii),we get $10x = 10$,so $x = 1$. Substituting $x=1$ into $(i)$,$1 + 3y = 7$,so $3y = 6$,$y = 2$. The intersection point is $(1, 2)$.Since $(1, 2)$ is the midpoint of $AA'$,we have:$\frac{x_1 + 3}{2} = 1 \Rightarrow x_1 = -1$$\frac{y_1 + 8}{2} = 2 \Rightarrow y_1 = -4$Thus,the image is $(-1, -4)$.

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

Similar Questions

The coordinates of the foot of the perpendicular from the point $(1, 2)$ on the line $x - 3y + 7 = 0$ are

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

Find the image of the point $(-1, 3)$ with respect to the line $x - y + 1 = 0$.

Let $P$ be a point inside a $\triangle ABC$ with $\angle ABC = 90^{\circ}$. Let $P_1$ and $P_2$ be the images of $P$ under reflection in $AB$ and $BC$ respectively. The distance between the circumcenters of $\triangle ABC$ and $\triangle P_1PP_2$ is

Consider the lines $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$,where $\lambda$ is a parameter. All these lines pass through a fixed point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$,then the value of $d^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module