Question

find the positive value of m for which the distance between the point A(5,- 3) and b (13, M) is 10 units​

Answer 1

${\huge {\mathfrak {Answer :-}}}$

We have,

A(5, - 3) and B(13, m)

By condition,

➡️ [(x2 - x1)^2 + (y2 - y1)^2]^1/2 = 10

Or, [(13 - 5)^2 + (m + 3)^2] = 10^2

Or, (8)^2 + (m + 3)^2 = 100

Or, (m + 3)^2 = 100 - 64 = 36

Or, m + 3 = 6

Or, m = 6 - 3 = 3.

➡️ $<b>$ Hence, the positive value of m is 3.

${\huge {\red {Thanks}}}$

Answer 2

Answer:

Step-by-step explanation:

Location of point A = (5,-3)

Location of point B = (13,-3)

Location of point C = (13,M)

The straight distance between point A and B =(13-5)=8

The The straight distance between point A and C = 10

Therefore

The straight distance between point B and C

$BC=\sqrt{AC^{2}}-AB^{2}\\\\\Rightarrow{BC}=\sqrt{{10^{2}-8^{2}}$

$\\\\\Rightarrow{BC}=\sqrt{100-64}\\\\\Rightarrow{BC}=\sqrt{36}\\\\\Rightarrow{BC}=6$

$CD=BC-BD=6-3=3$

Therefore the value of M = 3