Question

Show that the distance of the point (6-2)
)
Soom the lire 4x+3y=12 is half of the
point distance of the point (34) foom the
line 4x-3y=12​

Answer 1

Answer:

Step-by-step explanation:

1). The shorter distance between point and line is the length of the perpendicular from the point to the line.

2). For two lines "m" and "n", if m⊥n , then their slopes are opposite reciprocals.

3) (a). y = mx + b ("m" is a slope)

(b). $y - y_{1}=m(x - x_{1})$ point-slope form of a line.

4). $d=\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1^2})}$

~~~~~~~~~~~~

Consider point (6, - 2) and line 4x + 3y = 12

Rewrite standard form in slope-intercept form (shortly - solve for "y")

$y= - \frac{4}{3} x + 4$ .... (1), m = - 4/3 , the slope of perpendicular line is 3/4

y + 2 = (3/4)(x - 6) ⇒ y = (3/4)x - (13/2) .... (2)

Lines (1) and (2) intersect at (5.04, - 2.72) {to find this point we have to solve system of equation, I did it graphically}

$d_{1}$ = √[(5.04 - 6)^2 + (-2.72 + 2)^2] = 1.2 unit

Same steps we have to take for point (3, 4) and line 4x - 3y = 12 .... (3)

y = (4/3)x - 4 , then slope of ⊥ line is (- 3/4)  and equation is

y - 4 = (-3/4)(x - 3) ⇒ y = (-3/4)x + 25/4 .... (4)

Lines (3) and (4) intersect at (4.92, 2.56)

$d_{2}$ = √[(4.92 - 3)^2 + (2.56 - 4)^2] = 2.4 units.

As we can see, $d_{2}$ = 2$d_{1}$