Question

The distance from the origin to the image
ot (1,-1) with respect to the line soon
x+y-5=0
​

Answer 1

Answer:

We want to determine the distance from the origin of the image of (1,1) with respect to the line x+y+5=0.

Let the coordinates of the image be (a,b).

The distance of (1,1) from the line x+y+5=0 is |1+1+512+12√|=|72√|.

The distance of (a,b) from the line x+y+5=0 is |a+b+512+12√|=|a+b+52√|.

The point and its reflection are both equidistant from the line but lying on opposite sides.

⇒a+b+52√=−72√⇒a+b+5=−7.

⇒a+b=−12.

The slope of the line joining the point and its image is the negative reciprocal of the line with respect to which the point is reflected.

The slope of the line x+y+5=0 is −1.

⇒b−1a−1=1⇒a=b.

But we have already shown that a+b=−12.

⇒a=b=−6.

⇒ The coordinates of the image is (−6,−6).

⇒ The distance from the origin of the image of (1,1) with respect to the line x+y+5=0 is (−6)2+(−6)2−−−−−−−−−−−√=62–√.