Question

Show that the distance of the point (6, -2) from
the line 4x + 3y = 12 is half the distance of the
point (3, 4) from the line 4x - 3y = 12.​

Answer 1

refer to the attachment:)

Answer 2

Answer:

The distance from point (6,-2) on line 4x + 3y=12 is 6/5 units

The distance from point (3,4) on line 4x - 3y=12 is 12/5 units

Step-by-step explanation:

The perpendicular distance from any point (m,n) on any line having equation Ax + By + C=0 is given by,

d= | Am + Bn + C | / √(A^2 + B^2)

We are taking modulus of numerator because distance is always positive but there can be values of A,B,C, m,n such that Am+Bn+C can be negative. That is the reason for taking modulus of the numerator i.e, | Am + Bn + C |.

Now The distance from point (6,-2) on line 4x + 3y=12 is given by,

D1 = | 4*6 + 3*(-2) + (-12)| / √(4^2 + 3^2)

= | 24 - 6 -12 | / √ ( 16+9)

= | 6| / √ (25)

= 6/5 units

Now The distance from point (3,4) on line 4x - 3y=12 is given by,

D2 = | 4*3 + (-3)*4 + (-12)| / √(4^2 + (-3)^2)

= | 12 - 12 -12 | / √ ( 16+9)

= | -12| / √ (25)

= 12/5 units

Thus D1 is half of D2.