Question

if (13,6) is one endpoint and (-3,k) is the other endpoint, what is k if the distance is √260? k=4 coach omega was admiring clouds while sitting by the pond. k=160 the victim was crushed by a fallen object. k=16 the victim was not mauled by a cougar. k=6 dr. alpha was cleaning up popcorn in the movie theater.

Explanation:

Step1: Recall distance formula

The distance \( d \) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \( d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \). Here, \((x_1,y_1)=(13,6)\), \((x_2,y_2)=(-3,k)\), and \( d=\sqrt{260} \). Substitute into the formula:
\( \sqrt{260}=\sqrt{(-3 - 13)^2+(k - 6)^2} \)

Step2: Square both sides

Squaring both sides to eliminate the square roots:
\( 260=(-3 - 13)^2+(k - 6)^2 \)

Step3: Simplify the x - difference

Calculate \((-3 - 13)^2=(-16)^2 = 256\). So the equation becomes:
\( 260=256+(k - 6)^2 \)

Step4: Solve for \((k - 6)^2\)

Subtract 256 from both sides:
\( (k - 6)^2=260 - 256=4 \)

Step5: Solve for k

Take square roots of both sides: \( k - 6=\pm2 \)

• Case 1: \( k - 6 = 2\Rightarrow k=2 + 6=8 \)? Wait, no, wait, maybe I made a mistake. Wait, wait, let's check again. Wait, the options have k = 4, 160, 16, 6. Wait, maybe I miscalculated. Wait, let's re - do step 3. Wait, \((-3-13)=-16\), squared is 256. Then \( 260 = 256+(k - 6)^2\), so \((k - 6)^2=4\), so \( k - 6 = 2\) or \( k - 6=-2 \). So \( k=8\) or \( k = 4\). Oh, \( k = 4\) is one of the options. So when \( k - 6=-2\), \( k=6 - 2 = 4\).

Answer:

\( k = 4 \) (corresponding to "Coach Omega was admiring clouds while sitting by the pond.")