Question

find the value of p if the distance between points A (2,p) and B(-1,2) is 5​

Answer 1

Solution:

To solve this problem, we need to know Distance Formula.

Distance Formula:

Let P(x₁, y₁) and Q(x₂, y₂) be two points on the Cartesian Plane. Then the distance between the two points is given as:-

$\rm\longrightarrow Distance = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Here, we are given with the following information:

→ Distance between A(2, p) and B(-1, 2) is 5 unit.

By distance formula:

$\rm\longrightarrow 5 = \sqrt{(-1-2)^{2}+(2-p)^{2}}$

$\rm\longrightarrow 5 = \sqrt{(-3)^{2}+(2-p)^{2}}$

$\rm\longrightarrow 5 = \sqrt{9+(2-p)^{2}}$

Squaring both sides, we get:

$\rm\longrightarrow 5^{2}= 9+(2-p)^{2}$

$\rm\longrightarrow 25= 9+(2-p)^{2}$

$\rm\longrightarrow 25-9=(2-p)^{2}$

$\rm\longrightarrow (2-p)^{2}=16$

$\rm\longrightarrow (2-p)=\sqrt{16}$

$\rm\longrightarrow (2-p)=\pm 4$

$\rm\longrightarrow 2=\pm 4+p$

$\rm\longrightarrow p=\pm4+2$

$\rm\longrightarrow p=4+2, -4+2$

$\rm\longrightarrow p=6,-2$

→ So, the possible values of p are 6 and -2.

Answer:

• The values of p are 6 and -2.

Learn More:

1. Section formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\rm\longrightarrow R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\rm\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle.

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\rm\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$

Answer 2

$\bigstar{\underline{ \boxed{\pmb{\red{\sf{ \; Given \; :-}}}}}}$

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• Find the Value of p if the distance between points A (2,p) and B(-1,2) is 5.

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$\bigstar \; \underline{\boxed{\pmb{\green{\sf{ \; To \; Find \; :-}}}}}$

• Value of P = ?.

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${\underline{\boxed{\pmb{\pink{\sf{\; Solution \; :-}}}}}}$

Formula Used :

${ \boxed{ \underline{ \sf{ Distance = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}}}$

${ \qquad{\rule{150pt}{2pt}}}$

${ \dashrightarrow{ \qquad{ \sf{ 5 = \sqrt{(-1-2)^{2}+(2-p)^{2}}}}}} \\ \\ { \dashrightarrow{ \qquad{ \sf{5 = \sqrt{(-3)^{2}+(2-p)^{2}}}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{5 = \sqrt{9+(2-p)^{2}} }}}} \\ \\ { \dashrightarrow{ \qquad{ \sf{ 5^{2}= 9+(2-p)^{2}}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{ 25= 9+(2-p)^{2}}}}} \\ \\ { \dashrightarrow{ \qquad{ \sf{25-9=(2-p)^{2}}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{(2-p)^{2}=16}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{(2-p)=\sqrt{16}}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{(2-p)=\pm 4}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{2=\pm 4+p}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{ p=\pm4+2}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{ p=4+2, -4+2}}}} \\ \\{ \dashrightarrow{ \qquad{ \sf{ p=6,-2}}}}$

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Therefore

The values of p are 6 and -2.

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