Question 216692
find the distance between P(7,-4) and the line with equation x-3y+5=0.
round to the nearest tenth
:
Find the slope of the given equation
x - 3y + 5 = 0
-3y = -x - 5
Multiply eq by -1 to get rid of all those negatives
3y = x + 5
y = {{{1/3}}}x + {{{5/3}}}
:
The slope: m1 = {{{1/3}}}
Find the slope of a line perpendicular (m2) to this equation
m1 * m2 = -1; the relationship of slopes of perpendicular lines
{{{1/3}}}m2 = -1
m2 = -1*3
m2 = -3
:
Find the equation of the perpendicular line that passes thru P(7,-4)
Use the point/slope formula y - y1 = m(x - x1)
y -(-4) = -3(x - 7)
y + 4 = -3x + 21
y = -3x + 21 - 4
y = -3x + 17 is the equation of the perpendicular line
:
Find the value of x common to both equations
{{{1/3}}}x + {{{5/3}}} = -3x + 17
multiply both sides by 3
x + 5 = -9x + 51
x + 9x = 51 - 5
10x = 46
x = {{{46/10}}}
x = 4.6
:
Find the value for y:
y = -3(4.6) + 17
y = -13.8 + 17
y = +3.2
:
Find the distance from points x1 = 7, y1 =-4 and x2 = 4.6, y2 =3.2
The distance formula d = {{{sqrt((x2-x1)^2 + (y2-y1)^2)}}}
Using our point values
d = {{{sqrt((4.6-7)^2 + (3.2-(-4))^2)}}}
d = {{{sqrt((-2.4)^2 + (7.2)^2)}}}
d = {{{sqrt(5.76 + 51.84)}}}
d = {{{sqrt(57.6)}}}
d = 7.589 ~ 7.6 dist between P(7,-4) and Eq x - 3y + 5 = 0.