Find the distance between the point and the line.Round your answer to four decimal places. Point Line
$( 4,5 ) \quad 7 x + y = 1$
A) $d \approx 4.7068$
B) $d \approx 5.7068$
C) $d \approx 6.7068$
D) $d \approx 3.7068$
E) $d \approx 7.7068$

Question

Find the distance between the point and the line.Round your answer to four decimal places. Point Line
$( 4,5 ) \quad 7 x + y = 1$ 
A) $d \approx 4.7068$
B) $d \approx 5.7068$
C) $d \approx 6.7068$
D) $d \approx 3.7068$
E) $d \approx 7.7068$

Quizplus · Accepted Answer

The distance $d$ between a point $(x_0, y_0)$ and a line $Ax + By + C = 0$ is given by the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. For the point $(4,5)$ and the line $7x + y = 1$ (which can be rewritten as $7x + y - 1 = 0$), we have $A = 7$, $B = 1$, $C = -1$, $x_0 = 4$, and $y_0 = 5$. Plugging these values into the formula gives $d = \frac{|7(4) + 1(5) - 1|}{\sqrt{7^2 + 1^2}} = \frac{|28 + 5 - 1|}{\sqrt{49 + 1}} = \frac{32}{\sqrt{50}} = \frac{32}{5\sqrt{2}} = \frac{32\sqrt{2}}{10} \approx 4.7068$.

Find the Distance Between the Point and the Line (4,5)7x+y=1( 4,5 ) \quad 7 x + y = 1(4,5)7x+y=1

Find the Distance Between the Point and the Line $( 4,5 ) \quad 7 x + y = 1$