Find the points on the x-axis whose distance from the line equation (x/3) + (y/4) = 1 is given as 4units.

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Given that,

The equation of a line = (x/3) + (y/4) = 1

It can be written as:

4x + 3y -12 = 0 …(1)

Compare the equation (1) with general line equation Ax + By + C = 0,

we get the values A = 4, B = 3, and C = -12.

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

we know that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

D = |Ax1+By1+ C|/ √A2+ B2

Now, substitute the values in the above formula, we get:

4 = |4a+0 + -12|/ √42+ 32

⇒4 = |4a-12|/5

⇒|4a-12| = 20

⇒± (4a-12)= 20

⇒ (4a-12)= 20 or -(4a-12) =20

Therefore, it can be written as:

(4a-12)= 20

4a = 20+12

4a = 32

a = 8

(or)

-(4a-12) =20

-4a +12 =20

-4a = 20-12

-4a= 8

a= -2

⇒ a= 8 or -2

Hence, the required points on x axis are (-2, 0) and (8, 0).

Thanks