Question

Find the pt. on x-asis, whose distances from the line x\3 +y\4 =1 are 4 unit​

Answer 1

Answer:

The given equation of line is

3

x

+

4

y

=1 or 4x+3y−12=0

Let (a,0) be the point on the x-axis whose distance from the given line is 4 units.

Thus, using distance formula,

4=

4

2

+3

2

∣4a+3⋅0−12∣

⇒4=

5

∣4a−12∣

⇒∣4a−12∣=20

⇒±(4a−12)=20

⇒(4a−12)=20or−(4a−12)=20

⇒4a=20+12or4a=−20+12

⇒a=8or−2

Thus, the required points on the x-axis are (−2,0) and (8,0)