Question -
Answer -
Given:
The equation isparallel to 3x┬атИТ┬а4y + 5 = 0 and pass through (2, 3)
The equation of theline parallel to 3x┬атИТ┬а4y + 5 = 0 is
3x тАУ 4y+┬а╬╗┬а= 0,
Where,┬а╬╗┬аisa constant.
It passes through (2,3).
Substitute the valuesin above equation, we get
3 (2) тАУ 4 (3) + ╬╗ = 0
6 тАУ 12 +┬а╬╗┬а=0
╬╗┬а= 6
Now, substitute thevalue of ╬╗┬а=┬а6 in 3x тАУ 4y +┬а╬╗┬а= 0, we get
3x┬атИТ┬а4y + 6
тИ┤ The required line is3x┬атИТ┬а4y + 6 = 0.