Question -
Answer -
Given:
The line:┬а3x тАУ 5y+ 7 = 0
Comparing ax + by + c= 0 and 3x┬атИТ┬а5y + 7 = 0, we get:
a = 3, b=┬атИТ┬а5 and c = 7
So, the distance ofthe point (4, 5) from the straight line 3x┬атИТ┬а5y + 7 = 0 is
тИ┤ The required distanceis 6/тИЪ34