Answer: Putting y=0 in 3x−4y+9=0, we get x=−3. Thus, (−3,0) is a point on the line 3x−4y+9=0. Length of the perpendicular from (−3,0) to 6x−8y−15=0 is given by d=62+(−8)2∣−3×6−8×0−15∣=1033 Reply
Answer:
Putting y=0 in 3x−4y+9=0, we get x=−3.
Thus, (−3,0) is a point on the line 3x−4y+9=0.
Length of the perpendicular from (−3,0) to 6x−8y−15=0 is given by
d=62+(−8)2∣−3×6−8×0−15∣=1033