Question

Find the values of y for which the distance between the points P (4, -5) and Q (12, y) is 10 units.​

Answer 1

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

Given that,

Distance between the points P (4, -5) and Q (12, y) is 10 units.

We know,

Distance Formula

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂), then distance between A and B is denoted as AB and given by

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

Here,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x_1 = 4} \\ \\ &\sf{x_2 = 12}\\ \\ &\sf{y_1 = - 5}\\ \\ &\sf{y_2 = y}\\ \\ &\sf{AB = 10} \end{cases}\end{gathered}\end{gathered}$

So, on substituting the values, we get

$\rm :\longmapsto\:10 = \sqrt{ {(12 -4 )}^{2} + {(y + 5)}^{2} }$

$\rm :\longmapsto\:10 = \sqrt{ {8}^{2} + {(y + 5)}^{2} }$

$\rm :\longmapsto\:10 = \sqrt{ 64 + {(y + 5)}^{2} }$

On squaring both sides, we get

$\rm :\longmapsto\:100 = 64 + {(y + 5)}^{2}$

$\rm :\longmapsto\:100 - 64 = {(y + 5)}^{2}$

$\rm :\longmapsto\:36 = {(y + 5)}^{2}$

$\rm :\longmapsto\: {6}^{2} = {(y + 5)}^{2}$

$\rm \implies\:y + 5 \: = \: \pm \: 6$

$\rm \implies\:y \: = \: \pm \: 6 - 5$

$\rm \implies\:y \: = \: 6 - 5 \: \: or \: \: - 6 - 5$

$\rm \implies\:y \: = \: 1 \: \: or \: \: - 11$

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Additional Information :-

Section Formula :-

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂) and Let C (x, y) be any point on AB which divides AB internally in the ratio m : n, then coordinates of C is

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

Midpoint Formula :-

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂) and Let C (x, y) be mid point on AB, then coordinates of C is

${\underline{\boxed{\sf{\quad (x,y) \: = \: \bigg(\dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2}\bigg) \quad}}}}$