8
Active contributors today

## Consider the family of lines #(4a+3)x - (a+1)y - (2a+1) = 0# where #ainR#. Minimum area of the triangle which a member of this family with negative gradient can make with the positive semi axes?

dk_ch
Featured 1 month ago

Option (C)

#### Explanation:

Given family of lines $\left(4 a + 3\right) x - \left(a + 1\right) y - \left(2 a + 1\right) = 0$,where $a \in R$

Rearranging we get

$4 a x + 3 x - a y - y - 2 a - 1 = 0$

$\mathmr{and} , \left(4 x - y - 2\right) a + \left(3 x - y - 1\right) = 0$

Obviously the all lines of the family mus pass through the point of intersection of two lines represented by the following two equations

$4 x - y - 2 = 0. \ldots . \left[1\right]$

and

#3x-y-1=0........[2]

Subtracting [2] from [1] we get $x = 1$

Inserting $x = 1$ in [1] we get $y = 2$

So the coordinates of common point of intersection of all lines of the family will be $\left(1 , 2\right)$

Any line passing through this common point $\left(1 , 2\right)$ may be represented as $y - 2 = m \left(x - 1\right) \ldots . \left[3\right]$,where m is the gradient of the line , a variable one.
Now rearranging [3] in intercept form we get

$y - 2 = m \left(x - 1\right)$

$\implies m x - y = m - 2$

$\implies \frac{x}{\frac{m - 2}{m}} + \frac{y}{2 - m} = 1$

So area of the triangle made by this line with the positive semi axes will be given by

$A = \frac{1}{2} \times \frac{m - 2}{m} \times \left(2 - m\right) = \frac{- {m}^{2} + 4 m - 4}{2 m}$

$\implies A = - \frac{m}{2} + 2 - \frac{2}{m}$

Differentiating w r, to m we get

$\frac{\mathrm{dA}}{\mathrm{dm}} = - \frac{1}{2} + 0 + \frac{2}{m} ^ 2$

Imposing the condition of minimization of $A$i.e . $\frac{\mathrm{dA}}{\mathrm{dm}} = 0$ we get

$- \frac{1}{2} + \frac{2}{m} ^ 2 = 0$

$\implies {m}^{2} = 4$

$\implies m = - 2$, as per given condition the sraight line which forms minimum area with the positive semi axes must have negative gradient.

Hence minimum area of the the triangle should be for m=-2##

${A}_{\text{min}} = \frac{- {\left(- 2\right)}^{2} + 4 \left(- 2\right) - 4}{2 \left(- 2\right)} = 4$

[It matches with Option (C)]

## A circle has a radius of 5 meters. What the length of an arc of that circle that is captured by a central angle that measures 120°?

Steve M
Featured 1 month ago

$\text{Arc Length} = \frac{10 \pi}{3} \approx 10.47 \setminus m$

#### Explanation:

If the circle has radius $5 \setminus m$ then the total arc length of the entire circle (ie the perimeter) is given by the standard formula:

${P}_{\text{Total}} = 2 \pi r$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 \times \pi \times 5$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 10 \pi$

We require the length of the arc of a sector of ${120}^{o}$, we can represent this as a fraction of that of the entire circle:

${P}_{\text{Sector" = 120^0/360^0 xx P_"Total}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{3} \times {P}_{\text{Total}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{3} \times 10 \pi$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{10 \pi}{3}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \approx 10.47 \setminus m$

## An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(1 ,4 )# to #(5 ,1 )# and the triangle's area is #15 #, what are the possible coordinates of the triangle's third corner?

dk_ch
Featured 1 month ago

Let $P R = A$ be the side of the isosceles triangle having coordinates of its end points as follows

$P \to \left(1 , 4\right)$ and $R \to \left(5 , 1\right)$

Let the coordinates of the third point of the triangle be $\left(x , y\right)$.

As $\left(x , y\right)$ is equidistant from P and R we can write

${\left(x - 1\right)}^{2} + {\left(y - 4\right)}^{2} = {\left(x - 5\right)}^{2} + {\left(y - 1\right)}^{2}$

$\implies {x}^{2} - 2 x + 1 + {y}^{2} - 8 y + 16 = {x}^{2} - 10 x + 25 + {y}^{2} - 2 y + 1$

$\implies 8 x - 6 y = 9$

$\implies x = \frac{9 + 6 y}{8.} \ldots . . \left[1\right]$

Again $\left(x , y\right)$ being equidistant from P and R, the perpendicular dropped from $\left(x , y\right)$ to $P R$ must bisect it, Let this foot of the perpendicular or mid point of $P R$ be $T$

So coordinates of $T \to \left(3 , 2.5\right)$

Now height of the isosceles triangle

$H = \sqrt{{\left(x - 3\right)}^{2} + {\left(y - 2.5\right)}^{2}}$

And the base of the isosceles triangle

$P R = A = \sqrt{{\left(1 - 5\right)}^{2} + {\left(4 - 1\right)}^{2}} = 5$

So by the problem its area

$\frac{1}{2} \times A \times H = 15$

$\implies H = \frac{30}{A} = \frac{30}{5} = 6$

$\sqrt{{\left(x - 3\right)}^{2} + {\left(y - 2.5\right)}^{2}} = 6$

$\implies {\left(x - 3\right)}^{2} + {\left(y - 2.5\right)}^{2} = 36. \ldots \left[2\right]$

By [2] and [1] we get

${\left(\frac{9 + 6 y}{8} - 3\right)}^{2} + {\left(y - 2.5\right)}^{2} = 36$

$\implies \frac{1}{64} {\left(6 y - 15\right)}^{2} + {\left(y - 2.5\right)}^{2} = 36$

$\implies {\left(6 y - 15\right)}^{2} + 64 {\left(y - 2.5\right)}^{2} = 36 \times 64$

$\implies 36 {y}^{2} - 180 y + 225 + 64 {y}^{2} - 320 y + 400 = {48}^{2}$

$\implies 100 {y}^{2} - 500 y + 625 = {48}^{2}$

$\implies {y}^{2} - 5 y + 6.25 = {4.8}^{2}$

$\implies {\left(y - 2.5\right)}^{2} = {4.8}^{2}$

$\implies y = 2.5 \pm 4.8$

So $y = 7.3 \mathmr{and} y = - 2.3$

when $y = 7.3$

$x = \frac{9 + 6 \times 7.3}{8} = 6.6$

when $y = - 2.3$

$x = \frac{9 + 6 \times \left(- 2.3\right)}{8} = - 0.6$

So the coordinates of third point will be

$\left(6.6 , 7.3\right) \to \text{Q in figure}$

OR

$\left(- 0.6 , - 2.3\right) \to \text{S in figure}$

## The sun's angular diameter is measured to be 1920". The distance D of the sun from the earth is 1.496×10power11meter. Find the diameter of the sun? {Hint:d=aD}

Dean R.
Featured 4 weeks ago

# d = 1.496 times 10^{11} cdot 1920/3600 cdot {2 pi}/360 = 1.39254×10^9 # meters

#### Explanation:

Well now I have to look up what the double quote means. Probably minutes or seconds.

OK, Wikipedia says For example, 40.1875° = 40° 11' 15" .

That's 40 degrees, 11 minutes, 15 seconds, a minute being 1/60th of a degree and a second being 1/60th of a minute. (Thank you Babylonians.)

So we're to understand 1920" as

$\frac{1920}{3600}$ degrees $= \frac{1920}{3600} \cdot \frac{2 \pi}{360}$ radians

So we're looking at a diameter

$d = 1.496 \times {10}^{11} \cdot \frac{1920}{3600} \cdot \frac{2 \pi}{360}$ meters

# d = 1.39254×10^9 # meters

Check: Google #quad sqrt#

dk_ch
Featured 4 days ago

option (1) $6 \sqrt{2}$

#### Explanation:

Q92

Given that G is the centroid of $\Delta A B C$

$G A = 2 \sqrt{3} , G B = 2 \sqrt{2} \mathmr{and} G C = 2$

AP median is produced such that $G P = P O$

$B , O \mathmr{and} C , O$ are joined. The quadrilateral CBCO is a parallelogram,

As $A G : G P = 2 : 1$ we get

$\Delta B G P = \frac{1}{3} \Delta A B P = \frac{1}{3} \cdot \frac{1}{2} \Delta A B C$

Again $\Delta B G O = 2 \Delta B G P$

So $\Delta A B C = 3 \Delta B G O$

Now $G O = G A = 2 \sqrt{3}$

$B O = G C = 2$

and $G B = 2 \sqrt{2}$

So $G {O}^{2} - {\left(2 \sqrt{3}\right)}^{2} = 12$

$B {O}^{2} + B {G}^{2} = {2}^{2} + {\left(2 \sqrt{2}\right)}^{2} = 12$

So $\Delta B G O$ is rigtangled at B

So #DeltaBGO=1/2BO×BG#
Area of

$\Delta A B C = 3 \Delta B G O = \frac{3}{2} \times B O \times B G = \frac{3}{2} \times 2 \times 2 \sqrt{2} = 6 \sqrt{2}$

## When a rectangle is rotated about a line it becomes a rectangular prism. Do you agree? Im not sure. It might be a cylinder.

Somebody N.
Featured 2 days ago

See below.

#### Explanation:

This question is a little ambiguous. I'm guessing you mean a volume of revolution. If this is the case then a rectangle rotated about a line could be a solid cylinder or a hollow cylinder ( Tube) if the line of rotation is parallel or perpendicular to the sides of the rectangle. If the line of rotation is neither parallel nor perpendicular then a vast number of different solids could result. A big problem here is on the definition. Strictly a prism has a base that is a polygon and faces that are flat, but you could say a cylinder has a polygonal base with an in finite number of sides and an infinite number of faces.

Here are two rotations of a rectangle around different lines.

$y = 0$

$x = - 2$

##### Questions
• · 12 minutes ago
• · 19 minutes ago
• · 23 minutes ago
• · 50 minutes ago
• · An hour ago
• · 2 hours ago
• · 2 hours ago
• · 6 hours ago
• · 10 hours ago
• · 10 hours ago
• · 10 hours ago · in Quadrilaterals
• · 12 hours ago
• 15 hours ago
• · 15 hours ago
• · 15 hours ago
• · 15 hours ago
• · 15 hours ago
• · 15 hours ago