Question

8. The base of an equilateral triangle is x-axis and the vertex is (0,4). the equation of one of the
other two sides is​

Answer 1

Answer:

Step-by-step explanation:

The altitude is perpendicular to the given side, through the given vertex. The perpendicular family to  x+y=2  is gotten by swapping the coefficients on  x  and  y , negating one of them, so is  x−y=constant  and the constant is given by the vertex,

x−y=2−−1=3  

We intersect with

x+y=2  

Adding,

2x=5  

x=5/2  

y=2−x=2−5/2=−1/2  

So our altitude, between  (52,−12)  and  (2,−1),  has length

h:(5/2−2)2+(−1/2−−1)2−−−−−−−−−−−−−−−−−−−−−√=1/2–√  

The sides of the 30/60/90 triangle are in ratio  1:3–√:2.  That’s half an equilateral triangle i.e.  3–√  is the altitude when  2  is the side, so in our case

s=23–√12–√=26–√=6–√3  

Answer:  6–√/3  

Let’s try a different way. Slopes are tangents, so we can compute the slopes of the other two sides through the tangent sum angle formula.  y=−x+2  has a slope of -1,  tanA=−1.  We add  60∘  to that,  tanB=tan60∘=3–√  

m=tan(A+B)=tanA+tanB1−tanAtanB=−1+3–√1+3–√=2−3–√  

The other other slope is  2+3–√  but we won’t need it.

Now we have a point  (p,q)=(2,−1)  and a slope  m=2−3–√  that intersects a line  x+y=2  or generally  ax+by=c.  Let’s do the general case.

Find the squared distance  d2  between  (p,q)  and  (x,y) , the intersection of the two lines  ax+by=c  and  y−q=m(x−p) .

y=m(x−p)+q  

ax+b(mx+q−mp)=c  

ax+bmx=c+bmp−bq  

x=c+bmp−bqa+bm  

We don’t need to solve for   y=cm−amp+aqa+bm   because

d2=(x−p)2+(y−q)2=(x−p)2+(m(x−p))2=(1+m2)(x−p)2  

d2=(1+m2)(c+bmp−bq−ap−bmpa+bm)2  

d2=(c−ap−bq)2(1+m2)(a+bm)2  

Any chance this is right? Let’s try it. We have  (p,q)=(2,−1)  and  m=2−3–√,  and  x+y=2  so  a=1,b=1,c=2.  

m2=(2−3–√)2=7−43–√  

d2=(2−2−(−1))2(1+7−43–√)(3−3–√)2=8−43–√12−63–√=23  

d=23−−√=6–√3✓

Answer 2

Answer:

The equation of the one side other than the base is √3x+y-4 = 0

Step-by-step explanation:

Given,

The base of an equilateral triangle is X-axis

The vertex of the equilateral triangle is (0,4)

To find,

Equation of one side other than the base.

Solution:

Recall the concepts:

The perpendicular drawn from the vertex to the base of an equilateral triangle passes through the midpoint of the base of the triangle.

Mid-point formula,

The mid-point of the line joining (x₁,y₁) and (x₂,y₂) is given by

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

Distance formula,

The distance between the line joining  (x₁,y₁) and (x₂,y₂) is given by

$\sqrt{(x_1 - x_2)^2 + (y_1-y_2)^2}$

The equation of the line joining  (x₁,y₁) and (x₂,y₂) is given by

$\frac{x - x_1}{x_1 - x_2} = \frac{y - y_1}{y_1 - y_2}$

Let ABC be the equilateral triangle, with A as the vertex and BC be the base, and let D be the midpoint of the base BC.

Since the base of the equilateral triangle is the x-axis, the y coordinate of B, C and D is '0'

Again, since vertex A of the equilateral triangle is (0,4), vertex A lies on the y axis.

Since the perpendicular drawn from the vertex to the base of an equilateral triangle passes through the midpoint of the base of the triangle, the midpoint of the base D lies on the y-axis.

That is, x co-ordinate of D = 0

Hence, we have the co-ordinates of D are (0,0)

Since the y coordinate of the vertices B and C are zero, let us take the co-ordinate of B and C as (a,0) and (b,0).

Then by mid-point formula

(0,0) = $(\frac{a+b}{2},0)$

$\frac{a+b}{2} = 0$

a+b = 0 -----------------(1)

Again we have,

2BD = AB --------------(2)

By distance formula, B(a,0),  D(0,0), A(0,4)

$BD = \sqrt{(a-0)^2 +(0-0)^2} = \sqrt{a^2} = a$

$AB = \sqrt{(a-0)^2+ (0-4)^2} = \sqrt{a^2+16}$

Then, equation (2) becomes

2a = $\sqrt{a^2+16}$

Squaring on both sides,

4a² = a²+16

4a²- a² = 16

3a² = 16

a² = $\frac{16}{3}$

a =$\frac{4}{\sqrt{3} }$, or $-\frac{4}{\sqrt{3} }$

Hence the coordinates of the point B is $(-\frac{4}{\sqrt{3} } ,0)$and the coordinates of point C is $(\frac{4}{\sqrt{3} } ,0)$

Required to find, the equation of AB or AC

Equation of line A $(\frac{4}{\sqrt{3} } ,0)$, C(0,4) is

$\frac{x-\frac{4}{\sqrt{3} } }{\frac{4}{\sqrt{3} }-0 } = \frac{y-0}{0-4}$

$\frac{x-\frac{4}{\sqrt{3} } }{\frac{4}{\sqrt{3} } } = \frac{y}{-4}$

$-4({x-\frac{4}{\sqrt{3} }) = y (\frac{4}{\sqrt{3} } )$

$\frac{-4(\sqrt{3}x - 4) }{\sqrt{3} } = \frac{4y}{\sqrt{3} }$

$-(\sqrt{3}x - 4) = y$

√3x+y-4 = 0

The equation of the one side other than the base is √3x+y-4 = 0

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