Question

If the distance between the
points (3.p) (-1,0) is 6
units, then value of p is​

Answer 1

Answer: 2√5

Step-by-step explanation: Simple method of distance formula.

Answer 2

Answer:-

Given:

Distance between (3 , p) and ( - 1 , 0) = 6 units.

We know that,

Distance between two points with coordinates $\sf (x_1 , y_1)$ and $\sf (x_2 , y_2)$ is given by:

$\sf \sqrt{ {(x _{2} - x _{1}) }^{2} + (y _{2} - y _{1} ) ^{2} }$

Let ,

• x2 = - 1

• x1 = 3

• y2 = 0

• y1 = p

Hence,

$\implies \sf \: \sqrt{ {( - 1 - 3)}^{2} + {(0 - p)}^{2} } = 6 \\ \\ \sf \: on \: squaring \: both \: sides \: we \: get \\ \\ \implies \sf { \bigg( \sqrt{ {( - 4)}^{2} + ( - p) ^{2} } \bigg)}^{2} = {6}^{2} \\ \\ \sf \implies \: 16 + {p}^{2} = 36 \\ \\ \implies \sf \: {p}^{2} = 36 - 16 \\ \\ \implies \sf \: {p}^{2} = 20 \\ \\ \implies \sf \: p = \sqrt{20} \\ \\ \implies \sf \large{p = 2 \sqrt{5} }$

Hence, the value of p is 2√5.

Verification:-

$\sf\sqrt{( - 4) ^{2} + ( - 2 \sqrt{5} ) ^{2} } = 6 \\ \\ \implies \sf \sqrt{16 + 20} = 6 \\ \\ \implies \sf \: \sqrt{36} = 6 \\ \\ \sf \implies \large 6 = 6$

Hence, Verified.