Question

if the point p(k-12) is equidistant from A(3,k) B(k,5) find the value of k​

Answer 1

$\huge \tt \red{★ \: Solution \: ★}$

$\begin{gathered}\begin{gathered}\underline{\bigstar\:\boldsymbol{Using\:section\:formula\::}}\\ \\\end{gathered} \end{gathered} </p><p>$

$\begin{gathered}\begin{gathered}\star\;{\boxed{\sf{\pink{(x,y) = \bigg( \dfrac{m x_2 + n x_1}{m + n}\;,\; \dfrac{m y_2 + n y_1}{m + n} \bigg)}}}}\\ \\\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\sf Here \begin{cases} \sf{x_1 , y_1 = -6,10} \\ \sf{x_2 , y_2 = 3,-8} \end{cases}\\\end{gathered}\end{gathered} </p><p>$

Therefore,⠀⠀

$\begin{gathered}\begin{gathered}:\implies\sf \dfrac{m \times 3 + n \times -6}{m + n} = -4 \\\\\\:\implies\sf m \times 3 + n \times -6 = - 4m -4n\\\\\\:\implies\sf 3m - 6n = -4m - 4n\\\\\\:\implies\sf 3m + 4m = 6n - 4n \\\\\\:\implies\sf 7m = 2n\\\\\\:\implies\sf \dfrac{m}{n} = \dfrac{ 2}{7}\\\\\\:\implies{\underline{\boxed{\sf{\purple{m : n = 2 : 7}}}}}\;\bigstar\\ \\\end{gathered} \end{gathered} </p><p>$

$\therefore\;{\underline{\sf{The\;ratio\; in \;which \;(-4,6)\; divides\; the \;line\; segment\; is\; {\textsf{\textbf{2 : 7}}}.}}}$

Answer 2

Solution :-

We have, AP = BP

So by distance formula

Distance In points = (x₂- x₁)² + (y₂- y₁)²

AP² - BP²

➼ (k - 4)² + (2 - k)²

➼ 1 + 9k² + 16 - 8k + 4 + k² - 4k = 10

➼ 2k² - 12k + 20 = 10

➼ k² - 6k + 5 = 0

➼ (k - 5) (k - 1) = 0

k = 1 , 5