Question

Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

Answer 1

I think the answer is (4,0)
Hope it helps u

Answer 2

$\mathtt{ \huge{ \fbox{SOLUTION :}}}$

Let ,

(a,0) be the point on the x axis

Given ,

The distance of a point from a line is 4 units

The equation of line is x/3 + y/4 = 1

it can be written as ,

4x + 3y = 12

4x + 3y - 12 = 0 -------- (1)

On comparing eq (1) with general equation of a line Ax + By + C = 0 , we obtain

A = 4 , B = 3 and C = -12

We know that , the distance from a line is given by

$\sf \large{\fbox{D = \frac{ |Ax + By + C| }{ \sqrt{{(A)}^{2} + {(B)}^{2}} } } }$

Substitute the known values , we get

$\sf \mapsto 4 = \frac{ |4a + 3(0) - 12| }{ \sqrt{ {(4)}^{2} + {(3)}^{2} } } \\ \\\sf \mapsto 4 = \frac{ |4a - 12| }{5} \\ \\\sf \mapsto 20 = |4a - 12| \\ \\\sf \mapsto 20 = ±(4a - 12)$

When , (4a - 12) is positive

$\sf \mapsto 20 = (4a - 12) \\ \\\sf \mapsto 4a = 32 \\ \\\sf \mapsto a = 8$

When , (4a - 12) is negative

$\sf \mapsto 20 = - (4a - 12) \\ \\\sf \mapsto - 4a = 8 \\ \\\sf \mapsto a = - 2$

Hence , The required point on the x axis is (8,0) or (-2,0)