Question

find the positive value for m for which the distance between two points A(5,-3) and B(13,m) is 10 units

Answer 1

using distance formula

that is, AB²=(x2-x1)²+(y2-y1)²

(10)²=(13-5)²+(m-(-3))²

100=64+m²+9+6m

100-(64+9)=m²+6m

m²+6m -27=0

m²+(9-3)m -27=0

m(m+9)-3(m+9)=0

m=3,-9.

therefore positive value of m is 3.

Answer 2

Answer =

Step-by-step explanation:

${ab}^{2} = \sqrt{ {(x2 - x1)}^{2} } + {(y2 - y1)}^{2} \\ {(10) }^{2} = {(13 - 5)}^{2} + {(m - ( - 3))}^{2} \\ 100 = 64 + {m}^{2} + 9 + 6m \\ 100 - (64 + 9) = {m}^{2} + 6m \\ {m}^{2} + 6m - 27 = 0 \\ {m}^{2} + (9 - 3)m - 27 = 0 \\ m(m + 9) - 3(m + 9) = 0 \\ m = 3. - 9$

Therefore the positive value of m is 3