Question

I NEED HELP PLEASE, THANK YOU!

Answer 1

Answer:

$\sqrt{37}$

OR

≈ 6.08

Step-by-step explanation:

Turn the line given into slope-intercept form:

6x - y = -3

-y = -6x - 3

y = 6x + 3

We need to set this equal to the perpendicular version, and to do that we take the slope 6x and use the opposite reciprocal, and plug into the point-slope form:

$y-y_{1} = m(x - x_{1} )$

$y-2 = \frac{-1}{6} (x - 6 )$

y - 2 = $\frac{-1}{6} x + 1$

y = $\frac{-1}{6} x + 3$

6x + 3 = $\frac{-1}{6} x + 3$

x = 0

y = 0 + 3

y = 3

$d = \sqrt{(x_2 - x_1)^2+(y_2-y_1)^2}$

$d =\sqrt{37}$

Answer 2

Answer:

$\sqrt{37}$

Step-by-step explanation:

To find the distance between a point (m, n ) and the line

Ax + By + C = 0

d = $\frac{|Am +Bn+C|}{\sqrt{A^2+B^2} }$

Here (m, n) = (6, 2) and rearranging the line

6x - y + 3 = 0 ← in general form

with A = 6, B = - 1 and C = 3 , then

d = $\frac{|6(6)-1(2)+3|}{\sqrt{6^2+(-1)^2} }$

   = $\frac{|36-2+3|}{\sqrt{36+1} }$

   = $\frac{37}{\sqrt{37} }$  

Rationalise the denominator by multiplying numerator/ denominator by $\sqrt{37}$

= $\frac{37}{\sqrt{37} }$ × $\frac{\sqrt{37} }{\sqrt{37} }$

=$\frac{37\sqrt{37} }{37}$ ← cancel 37 on numerator/ denominator

= $\sqrt{37}$