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## What are the coordinates of B?

Shwetank Mauria
Featured 4 months ago

Coordinates of $B$ are $\left(- 1.6 , - 1.2\right)$

#### Explanation:

Let the coordinates of B be $B \left(p , q\right)$

and coordinates of $A$ being $\left(- 1 , - 3\right)$ and slope of $B A$ is $\frac{q + 3}{p + 1}$ and mid point of $B A$ is $\left(\frac{p - 1}{2} , \frac{q - 3}{2}\right)$. Note that this will be point of intersection of $B A$ with its perpendicular bisector.

as $B A$ is perpendicular to its perpendicular bisector $x - 3 y - 5 = 0$ and hence product of their slope should be $- 1$

as $x - 3 y - 5 = 0$ in slope-intercept form is $y = \frac{1}{3} x - \frac{5}{3}$ (1), its slope is $\frac{1}{3}$ and hence slope of $B A$ is $\frac{- 1}{\frac{1}{3}} = - 3$

and if $\left(x , y\right)$ is a point on $B A$ then $\frac{y + 3}{x + 1} = - 3$ i.e. $- 3 x - 3 = y + 3$ i.e. $3 x + y + 6 = 0$ (2)

and its intersection with $x - 3 y - 5 - 0$ will be given by solving these equations. Putting value from (1) into (B), we have

$3 x + \frac{1}{3} x - \frac{5}{3} + 6 = 0$ or $\frac{10}{3} x = - \frac{13}{3}$ and $x = - \frac{13}{10} = - 1.3$

and $y = - \frac{13}{10} \times \frac{1}{3} - \frac{5}{3} = - \frac{63}{30} = - 2.1$

Hence, coordinates of point of intersection are $\left(- 1.3 , - 2.1\right)$

and hence $\frac{p - 1}{2} = - 1.3$ i.e. $p = - 1.6$ and

$\frac{q - 3}{2} = - 2.1$ i.e $q = - 1.2$

and coordinates of $B$ are $\left(- 1.6 , - 1.2\right)$

graph{(3x+y+6)(x-3y-5)((x+1)^2+(y+3)^2-0.01)((x+1.6)^2+(y+1.2)^2-0.01)=0 [-9.92, 10.08, -5.88, 4.12]}

## A circle cuts the parabola y^2=4ax in the points (at_i^2, 2at_i) for i=1, 2, 3, 4. Prove that t_1+t_2+t_3+t_4=0?

dk_ch
Featured 3 months ago

Let the equation of the circle which cuts the given parabola ${y}^{2} = 4 a x$ be ${\left(x - b\right)}^{2} + {\left(y - c\right)}^{2} = {r}^{2.}$,where $\left(b , c\right)$ is the coordinates of the center(D) of the circle and $r$ is its radius.

Replacing $x = {y}^{2} / \left(4 a\right)$ in the equation of the circle we get the following equation of y of degree 4

${\left(x - b\right)}^{2} + {\left(y - c\right)}^{2} = {r}^{2.}$

$\implies {x}^{2} - 2 b x + {b}^{2} + {y}^{2} - 2 c y + {c}^{2} - {r}^{2} = 0$

$\implies {\left({y}^{2} / \left(4 a\right)\right)}^{2} - 2 b \left({y}^{2} / \left(4 a\right)\right) + {b}^{2} + {y}^{2} - 2 c y + {c}^{2} - {r}^{2} = 0$

$\implies {y}^{4} / \left(16 {a}^{2}\right) + \left(1 - \frac{2 b}{4 a}\right) {y}^{2} - 2 c y + {b}^{2} + {c}^{2} - {r}^{2} = 0 \textcolor{red}{\ldots . . \left[1\right]}$

Here $f \left(y\right) = 0$ is an equation of y of degree 4. So it will have 4 roots of y and those roots will be given by $2 a {t}_{i} \text{ for } i = 1 , 2 , 3 , 4$

Since the coefficient of ${y}^{3}$ of equation [1] is zero , the sum of all 4 roots of y must be zero.

Hence ${\sum}_{i = 1}^{i = 4} \left(2 a {t}_{1}\right) = 0$

$\implies {\sum}_{i = 1}^{i = 4} {t}_{1} = 0$

$\implies {t}_{1} + {t}_{2} + {t}_{3} + {t}_{4} = 0$

## Find the area of a regular octagon with an apothem of 8.5.?

marfre
Featured 3 months ago

$A = 238 \text{ } u n i t {s}^{2}$

#### Explanation:

Area of a regular polygon $= \frac{1}{2} \times p e r i m e t e r \times a p o t h e m$

$A = \frac{P a}{2}$

Perimeter $= \text{number of sides" xx "side length}$

$P = n s = 8 s$

$A = \frac{8 s \cdot 8.5}{2} = 34 s$

Since we know that the octagon is regular and we know the apothem (height from a side to the center of the polygon, we can calculate the side length):

Each angle of the central angle: $\frac{{360}^{\circ}}{8} = {45}^{\circ}$

and each side of the polygon forms an isoceles triangle when joined to center. Further, a perpendicular from center to the side of the polygon divides the side in two equal parts and also cuts this angle at the center (${45}^{\circ}$) in half i.e. ${22.5}^{\circ}$.

Using trigonometry, we can find half the side length ($x$):

$\tan {22.5}^{\circ} = \frac{x}{a} = \frac{x}{8.5}$

$x = 8.5 \cdot \tan {22.5}^{\circ} \approx 3.52$

side length: $s = 2 x \approx 7$

$A = 34 \cdot s \approx 34 \cdot 7 \approx 238 {\text{ units}}^{2}$

## What is the length between PO, PQ, OQ have to be when the points form a equilateral triangle?

CW
Featured 2 months ago

length$= \frac{2}{\sqrt{15}}$ unit

#### Explanation:

We know that the perpendicular distance $d$ between two parallel lines $y = m x + c 1 , \mathmr{and} y = m x + c 2$ is given by:
$d = | c 1 - c 2 \frac{|}{\sqrt{1 + {m}^{2}}}$

Given that the two parallel lines are $y = 2 x + 1 , \mathmr{and} y = 2 x + 2$, as shown in the figure.
$\implies c 1 = 1 , c 2 = 2$ and slope of the two lines $m = 2$
$\implies d = P M = | 1 - 2 \frac{|}{\sqrt{1 + {2}^{2}}} = \frac{1}{\sqrt{5}}$

Given that $\Delta O P Q$ is an equilateral triangle,
$\implies O P = P Q = O Q , \mathmr{and} \angle O P Q = \angle O Q P = \angle Q O P = {60}^{\circ}$
$\implies P O \sin 60 = P M$
$\implies P O \cdot \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{5}}$
$\implies P O = \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{15}}$ unit

Hence, length of each side of the equilateral triangle $= \frac{2}{\sqrt{15}}$ unit

## What is the radius of the incircle of a triangle whose sides are 5, 12 and 13 units?

CW
Featured 3 weeks ago

$r = 2$ units

#### Explanation:

As ${13}^{2} = {5}^{2} + {12}^{2}$, the triangle is a right triangle.
Formula for the inradius ($r$) of a right triangle :
$r = \frac{a \cdot b}{a + b + c}$, or $r = \frac{a + b - c}{2}$
where $a \mathmr{and} b$ are the legs of the right traingle and $c$ is the hypotenuse.
$\implies r = \frac{a \cdot b}{a + b + c} = \frac{5 \cdot 12}{5 + 12 + 13} = \frac{60}{30} = 2$ units

or $r = \frac{a + b - c}{2} = \frac{5 + 12 - 13}{2} = \frac{4}{2} = 2$ units

Proof 1 : $r = \frac{a \cdot b}{a + b + c}$

Area $\Delta A B C = \frac{1}{2} \cdot \left(B C\right) \cdot \left(A C\right) = \frac{1}{2} \cdot r \cdot \left(B C + A C + A B\right)$
$\implies \frac{r \cdot \left(a + b + c\right)}{2} = \frac{a \cdot b}{2}$
$\implies r = \frac{a \cdot b}{a + b + c}$

Proof 2 : $r = \frac{a + b - c}{2}$

Recall that the tangents to a cicle from an external point are equal,
$\implies C D = C F = r , B E = B F = a - r$,
and $A D = A E = b - r$
$\implies c = a - r + b - r$
$\implies c = a + b - 2 r$
$\implies 2 r = a + b - c$
$\implies r = \frac{a + b - c}{2}$

## In triangle ABC AM is median prove that AB^2+AC^2=2×AM^2+1/2+BC^2 solve by apollonius theorem?

Shwetank Mauria
Featured 6 days ago

#### Explanation:

Let us consider the following figure, where we have the median $A M$ and $A P \bot B C$.

Now in $\Delta A B M$, we have

$A {B}^{2} = A {P}^{2} + P {B}^{2} = A {P}^{2} + {\left(B M - P M\right)}^{2}$

= $\textcolor{red}{A {P}^{2}} + B {M}^{2} + \textcolor{red}{P {M}^{2}} - 2 B M \times P M$

or $A {B}^{2} = \textcolor{red}{A {M}^{2}} + B {M}^{2} - 2 B M \times P M$ .....................(1)

and in $\Delta A C M$, we have

$A {C}^{2} = A {P}^{2} + P {C}^{2} = A {P}^{2} + {\left(M C + P M\right)}^{2}$

= $\textcolor{red}{A {P}^{2}} + M {C}^{2} + \textcolor{red}{P {M}^{2}} - 2 M C \times P M$

or $A {C}^{2} = \textcolor{red}{A {M}^{2}} + M {C}^{2} - 2 M C \times P M$

But as $B M = M C$, we can write it as

$A {C}^{2} = \textcolor{red}{A {M}^{2}} + B {M}^{2} - 2 B M \times P M$ .....................(2)

Adding (1) and (2), we get

$A {B}^{2} + A {C}^{2} = 2 A {M}^{2} + 2 B {M}^{2}$ and as $B M = \frac{B C}{2}$

$A {B}^{2} + A {C}^{2} = 2 A {M}^{2} + \frac{2 B {C}^{2}}{4}$

or $A {B}^{2} + A {C}^{2} = 2 A {M}^{2} + \frac{1}{2} B {C}^{2}$

This is Apollonius theorem.

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