Question

Question 11 *
If the distance between the points (4, p) and
(1, 0) is 5 units, then the value of p is
(a) 4 only
(6) 4
(C) – 4 only (d) 0
O
a​

Answer 1

Given:

•The distance between the points (4,p) and (1,0) is

5 units .

To find :

• The value of p .

solution:

• As we know that :

$\star \boxed{ \red{\bold{\: d = \sqrt{( {x_2 - x_1)}^{2} + ({y_2 - y_1)}^{2} } }}}$

Here ,

• x1 = 4 and y1= p

•x2= 1 and y2= 0

• d = 5

$\implies \: 5 = \sqrt{ ({1 -4)}^{2} + ({0 - p)}^{2} } \\ \\ \implies \: 25 = 9 + {p}^{2} \\ \\ \implies \: {p}^{2} = 16 \\ \\ \implies \: p = \sqrt{16} \\ \\ \implies \: p = \pm \: 4$

Answer 2

Answer:

• 1st Point = (4 , p) = (x₁ , y₁)
• 2nd Point = (1 , 0) = (x₂ , y₂ )
• Distance b/w points = 5

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf (Distance)^2=(x_2-x_1)^2+(y_2-y_1)^2\\\\\\:\implies\sf (5)^2=(1-4)^2+(p-0)^2\\\\\\:\implies\sf 25 = (-\:3)^2+(p)^2\\\\\\:\implies\sf 25 = 9 + (p)^2\\\\\\:\implies\sf 25 - 9 = (p)^2\\\\\\:\implies\sf 16 = (p)^2\\\\\\:\implies\sf \sqrt{16} = p\\\\\\:\implies\underline{\boxed{\sf p = \pm \:4}}$

⠀

$\therefore\:\underline{\textsf{Hence, Required value of p is \textbf{ \pm 4}}}.$

$\rule{180}{1.5}$

Verification:

• when value of p = 4

$\dashrightarrow\sf 16=(p)^2\\\\\\\dashrightarrow\sf 16=(4)^2\\\\\\\dashrightarrow\sf 16=16$

⠀

• when value of p = - 4

$\dashrightarrow\sf 16=(p)^2\\\\\\\dashrightarrow\sf 16=(-\:4)^2\\\\\\\dashrightarrow\sf 16=16$