Question

5. Find the point on x-axis which is equidistant from the following pair of points :

(i) (3, 2) and (-5. - 2)​

Answer 1

Answer:

• The points are (-1, 0).

Step by Step Explanation:

${\underline{\bf Let\;P(x,0)\;is\;the\;point\;which\;is\;equidistant\;from\;given\;points.}}\\ \\ \\ {\underline{\bf Given:}}\\ \\ \\ \longrightarrow \sf Two\;points\;A(3,2)\;and\;B(-5,-2)\\ \\ \\ {\underline{\bf To\;Find:}}\\ \\ \\ \longrightarrow \sf Find\;the\;point\;on\;x-axis\;which\;is\;equidistant\;from\;given\;points.\\ \\ \\ \sf So, PA = PB\\ \\ \\ \longrightarrow \sf \sqrt{(x-3)^{2}+(0-2)^{2}}=\sqrt{[(x-(-5)]^{2}+[0-(-2)}]^{2}}\\ \\ \\ {\underline{\bf Now,\;squaring\;the\;both\;sides,\;we\;get}}$

$\longrightarrow \sf (x-3)^{2}+(-2)^{2}=(x+5)^{2}+(2)^{2}\\ \\ \\ \longrightarrow \sf x^{2}-6x+9+4=x^{2}+10x+25+4\\ \\ \\ \longrightarrow \sf x^{2}-6x+13=x^{2}+10x+29\\ \\ \\ \longrightarrow \sf x^{2}-6x-10x-x^{2}=29-13\\ \\ \\ \longrightarrow \sf -16x=16\\ \\ \\ \longrightarrow \sf x=-\dfrac{16}{16}\\ \\ \\ \longrightarrow \bf x=-1$

$\large{\boxed{\bf {\therefore Hence,\;the\;points\;are\;(-1,0).}}}$

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