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The image of the point $\left (3, 8 \right)$ in the line $x+3y=7$ is :

  1. $\left (1, 4 \right)$
  2. $\left (4, 1 \right)$
  3. $\left (-1, -4 \right)$
  4. $\left (-4, -1 \right)$
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Answer is C

Let (x,y) be the image of given point (3,8). Line joining point (3, 8) and image (x, y) is perpendicular to given line x + 3y = 7

So, $m_{1}*m_{2} =-1$  -----(1)

For the line x + 3y = 7,  3y = -x + 7 → y = $\frac{-x+7}{3}$, so [ y = mx + c; where m is the slope]

slope $m_{1}=\frac{-1}{3}$  ------(2)

Slope of line joining the point (3, 8) and the image (x, y) is $m_{2}$ =$\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{y-8}{x-3}$  ------(3)

Putting (2) and (3) in (1), we will get 3x – y – 1 = 0

On solving both equations of lines x + 3y = 7 and 3x – y = -1, we will get the midpoint (x,y) = (1, 2)

In order to find the image (x, y), $\frac{3+x}{2}=1; \frac{8+y}{2}=2;$

On solving, we will get image as (-1, -4)

 

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