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

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(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

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