Question

The equation of locus of point P which is equidistant from point A(-5.4) and B(3-2)​

Answer 1

Answer:

4x - 3y + 7 = 0

Step-by-step explanation:

Let the coordinates of that point P be (x , y).

 Since P  is equidistant from point A(-5.4) and B(3-2)​, using distance formula,

⇒ √(x - (-5))² + (y - 4)² = √(3 - x)² + (-2 -y)²

⇒ (x + 5)² + (y - 4)² = (3 - x)² + (-2 - y)²

⇒ x²+25+10x +y²+16-8y = 9+x²-6x+4+y²+4y

⇒ 10x - 8y + 41 = 4y - 6x + 13

⇒ 16x  - 12y + 28 = 0

⇒ 4x - 3y + 7 = 0

Answer 2

${\large{\sf{\pmb{\underline{Given \; that}}}}}$

★ The equation of pocus of point P which is equidistant from point A(-5.4) and B(3-2)

${\large{\sf{\pmb{\underline{Solution}}}}}$

★ The equation of pocus of point P which is equidistant from point A(-5.4) and B(3-2) = 4x - 3y + 7 = 0

${\large{\sf{\pmb{\underline{Using \; concept}}}}}$

★ Distance formula.

${\large{\sf{\pmb{\underline{Using \; formula}}}}}$

${\underline{\boxed{\sf{\quad \rightarrow Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$ ⠀

${\large{\sf{\pmb{\underline{Full \; Solution}}}}}$

$:\implies \sf Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ :\implies \sf Distance = \sqrt{[x-(-5)]^{2}+(y-4)^{2}} = \sqrt{(3-x)^{2}+(-2-y)^{2}} \\ \\ \bf Now \: let's \: cancel \: square \: root \: by \: each \: other \\ \\ \bf And \: (- -) \: = + \\ \\ :\implies \sf Distance = (x+5^{2} + (y-4)^{2} = (3-x)^{2} + (-2-y)^{2} \\ \\ \bf Now \: let's \: open \: brackets \: and \: solve \\ \\ :\implies \sf Distance = x^{2} + 25 + 10x + y^{2} + 16 - 8y = 9 + x^{2} - 6x +4 + y^{2} + 4y \\ \\ :\implies \sf Distance = 10x - 8y + 41 = 4y - 6x + 13 \\ \\ \bf Combining \: like \: terms \: and \: carry \: on \\ \\ :\implies \sf Distance = 10x + 6x - 8y - 4y + 41 - 13 = 0 \\ \\ :\implies \sf Distance = 16x - 12y + 28 = 0 \\ \\ :\implies \sf Distance = 4x - 3y + 7 = 0$

${\large{\sf{\pmb{\underline{Additional \; knowledge}}}}}$

$\underline{\bigstar\:\textsf{Distance Formula\; :}}$

Distance formula is used to find the distance between two given points.

${\underline{\boxed{\frak{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$ ⠀

$\underline{\bigstar\:\textsf{Section Formula\; :}}$

Section Formula is used to find the co ordinates of the point(Q) Which divides the line segment joining the points (B) and (C) internally or externally.

${\underline{\boxed{\frak{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$ ⠀

$\underline{\bigstar\:\textsf{Mid Point Formula\; :}}$

Mid Point formula is used to find the Mid points on any line.

${\underline{\boxed{\frak{\quad \Bigg(\dfrac{x_1 + x_2}{2} \; or\; \dfrac{y_1 + y_2}{2} \Bigg)\quad}}}}$