Question

3. Find the value of k, if the distance between the points P(4,-5) and Q(-3,k) is 5 units​

Answer 1

$\underline{\bf{GIVEN:-}}$

• Coordinates of point P = (4,-5)
• Coordinates of point Q = (-3,k)
• Distance between P and Q is 5 units.

$\\ \underline{\bf{TO \: FIND:-}}$

• The distance between points P and Q.

$\\ \underline{\bf{SOLUTION :-}}$

Here,

$\sf x_{1} = 4 \: and \: x_{2} = -3 \\ \sf y_{1} = -5 \: and \: y_{2} = k$

By using Distance Formula,

$\sf \: Distance \: between \: P \: and \: Q = \sqrt{( x_{2} - x_{1}) + (y_{2} - y_{1})} \\ \\ \sf 5 = \sqrt{( - 3 - 4) + (k + 5)} \\ \\ \sf \: 5 = \sqrt{ - 7 + k + 5} \\ \\ \sf5 = \sqrt{k - 2} \\ \\ \sf \: (5) {}^{2} = ( \sqrt{k - 2} ) {}^{2} \\ \\ \sf 25 = k - 2 \\ \\ \boxed{\boxed{ \bf{ \purple{k = 27}}}}$

Therefore, the value of k is 27.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

Find the value of k, if the distance between the points P(4,-5) and Q(-3,k) is 5 unit.

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• Point P's coordinate = (4,-5)
• Point Q's coordinate = (-3,k)
• Distance between P and Q is 5 unit.

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• The distance between points P and Q.

${\bigstar}\large{\boxed{\sf{\pink{Full \: solution}}}}$

$\large\purple{\texttt{Here we have}}$

$\large\green{\texttt{x¹ = 4 and x² = 3}}$

$\large\green{\texttt{y¹ = -5 and y² = k}}$

$\bold{\purple{\fbox{\orange{By using formula of Distance}}}}$

$\large\green{\texttt{Distance between p and q -}}$

• √ (x² - x¹ ) + (y² - y¹ )
• 5 = √ -7 + k + 5
• 5 = √ k - 2
• (5)² = ( √k - 2 )²
• 25 = k - 2
• k = 25 + 2
• k = 27

$\large\red{\texttt{27 is the answer}}$

@Itzbeautyqueen23

Hope it's helpful

Thank you :)