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 3x-4y =12

Answer 1

Given that,

Point = (6,-2)

The equation of line is

$4x+3y=12$

Another point = (6,-2)

The equation of line is

$3x-4y=12$

Let d₁ be the distance between point and line.

We need to calculate the distance d₁

Using formula of distance

$d_{1}=|\dfrac{4x+3y-12}{\sqrt{4^2+3^2}}|$

Put the value of x and y

$d_{1}=|\dfrac{4\times6+3\times(-2)-12}{\sqrt{4^2+3^2}}|$

$d_{1}=\dfrac{6}{5}\ unit$

Let d₂ be the distance between point and line.

We need to calculate the distance d₂

Using formula of distance

$d_{2}=|\dfrac{4x-3y-12}{\sqrt{4^2+3^2}}|$

Put the value of x and y

$d_{2}=|\dfrac{4\times3-3\times(4)-12}{\sqrt{4^2+3^2}}|$

$d_{2}=|\dfrac{-12}{5}\ unit|$

$d_{2}=\dfrac{12}{5}\ unit$

$d_{2}=2\times\dfrac{6}{5}$

Here, $\dfrac{6}{5}=d_{1}$

$d_{2}=2d_{1}$

$d_{1}=\dfrac{d_{2}}{2}$

So, 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.

Hence, This is proved.

Answer 2

Answer:

D1=1/2d2

Step-by-step explanation:

see the attachment for the answer.