Question

Q. Find the point on the y-axis which is
equidistant from the points A (6,5) and
B (-4,3)​

Answer 1

Answer:

a lies in 2 qandrant b lies in 1 quadrant

Answer 2

$\underline{\underline{\blue{\bf Answer:-}}}$

$\tt\therefore The \:point \:will\:be\:(0,9).$

$\underline{\underline{\orange{\bf Step-by-step\: Explanation:-}}}$

$\underline{\green{\bf Given:-}}$

$\tt:\implies Two \:points\:(6,5)\:and\:(-4,3).$

$\underline{\red{\bf To\:Find:-}}$

$\tt:\implies The\:point\:on\:y-axis\:which\:is\: \\\tt equidistant\: from\; them.$

• As we know that ,

$\tt If \: the\: point\:is\:on\:y\:axis\: then\:its\:x\\\tt coordinate\: will\:be\:0 . Let\: the\: point\:be \: C(0,y).$

$\underline{\color{blue}{\sf Distance\: between\:A\;and\:C :}}$

$\pink{\tt :\implies Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\\\\\tt:\implies Distance=\sqrt{ (6-0)^2+(5-y)^2}\\\\\tt :\implies Distance = \sqrt{ 6^2+5^2+y^2-10y}\\\\\tt:\implies Distance=\sqrt{ 36 + 25 + y^2 - 10y }\\\\\boxed{\red{\hookrightarrow}\bf Distance =\sqrt{ y^2-10y+61}}$

$\underline{\color{blue}{\sf Distance\: between\:B\;and\:C :}}$

$\pink{\tt :\implies Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\\\\\tt:\implies Distance=\sqrt{ (-4-0)^2+(3-y)^2}\\\\\tt :\implies Distance = \sqrt{ 4^2+3^2+y^2-6y}\\\\\tt:\implies Distance=\sqrt{ 16 + 9 + y^2 - 6y }\\\\\boxed{\red{\hookrightarrow}\bf Distance =\sqrt{ y^2-6y+25}}$

$\underline{\color{blue}{\sf Since\: Distance\:is\:equal :}}$

$\tt:\implies \sqrt{y^2-10y+61}=\sqrt{y^2-6y+25}\\\\\tt :\implies y^2-10y+61=y^2-6y+25\\\\\tt:\implies -10y +6y = 25-61 \\\\\tt:\implies -4y = -36 \\\\\tt:\implies y =\cancel{\dfrac{-36}{-4}}\\\\\underline{\boxed{\red{\tt\longmapsto y = 9 }}}$