SJ
Last Activity: 4 Years ago
Welcome to Askiitians
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