Question

Find the value of ‘p’ for which the distance between A(2,-3) and B(1,p)is 10 units.

Answer 1

Step-by-step explanation:

using distance formula - ✓(2-1)^2 + (-3-p)^2 = 10

1 + 9 + p^2 + 6p =100

p^2 + 6p= 90

using quadratic formula- (-6 +- ✓36+360)/2

Answer 2

Given that ,

The distance b/w A(2,-3) and B(1,p) is 10 units

We know that , the distance between two points is given by

$\sf\large\fbox{D = \sqrt{ {( x_{2} -x_{1} )}^{2} + {y_{2} -y_{1} )}^{2} } \: \: }$

Thus ,

10 = √{(1 - 2)² + (p + 3)²}

Squaring on both sides , we get

100 = (p)² + 9 + 6p

(p)² + 6p - 91 = 0

(p)² + 13p - 7p - 91 = 0

p(p + 13) - 7(p + 13) = 0

(p - 7)(p + 13) = 0

p = 7 or p = -13

$\therefore\sf \underline{The \: value \: of \: p \: will \: be \: 7 \: or -13}$