Question

ABC is an equilateral triangle of coordinates of vertices of B and C are(3,0)and (-3,0) find coordinates of vertices of A which on positive y -Axis​

Answer 1

Step-by-step explanation:

Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(-3,0)(-3,0)(3,0)\qbezier(-3,0)(-3,0)(0,5.2)\qbezier(3,0)(3,0)(0,5.2)\put(-3.2,-0.4){\sf C(-3,0)}\put(3.2,-0.4){\sf B(3,0)}\put(0.4,5.3){\sf A(0,5.2)}\thinlines\put(0,0){\vector(0,1){6}}\put(0,0){\vector(0,-1){2}}\put(0,0){\vector(1,0){5}}\put(0,0){\vector(-1,0){5}}\end{picture}$

Concept:-

As ∆ABC is an Equilateral triangle ,then The distance from each vertices to another will be same.

Given:-

• ∆ABC is an Equilateral triangle
• B(3,0)
• C(-3,0)

To find:-

• A(x,y)

Formula Used:

Distance Formula

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

Solution:-

As ∆ABC is Equilateral thus

• AB=AC=BC
• AB=AC

$\tt \longmapsto \: \sqrt{(3 - x) {}^{2} + (0 - y) {}^{2} } = \sqrt{( x + 3) {}^{2} + (y - 0) {}^{2} }\\ \\ \tt \longmapsto \: \sqrt{ {3 }^{2} - 2 \times 3 \times x + {x}^{2} + {( - y)}^{2} } = \sqrt{ {x}^{2} + 2 \times 3 \times x + {3}^{2} + {y}^{2} } \\ \\ \tt \longmapsto \: \sqrt{9 - 6x + {x}^{2} + {y}^{2} } = \sqrt{{x}^{2} + 6x + 9 + {y}^{2} } \\ \\ \tt \longmapsto \: {x}^{2} + {y}^{2} - 6x + 9 = {x}^{2} + {y}^{2} + 6x + 9 \\ \\ \tt \longmapsto \: {x}^{2} - {x}^{2} + {y}^{2} - {y}^{2} + 9 - 9 = 6x + 6x \\ \tt \longmapsto \: 12x = 0 \\ \\ \tt \longmapsto \: x = 0$

Now

• A=(0,y)

• Compare it with BC
• AB=BC

$\\ \tt \longmapsto \: \sqrt{(3-0)^2+(0-y)^2}=\sqrt{(3+3)^2+(0-0)^2}$

$\\ \tt \longmapsto \: \sqrt{3^2+y^2}=\sqrt{6^2+0}$

$\\ \tt \longmapsto \: 9+y^2=6^2$

$\\ \tt \longmapsto \: y^2+9=36$

$\\ \tt \longmapsto \: y^2=36-9$

$\\ \tt \longmapsto \: y^2=27$

$\\ \tt \longmapsto \: y=\sqrt{27}$

$\\ \tt \longmapsto \: y=5.2$

$\\ \tt \longmapsto \: y=5(Approx)$

$\\ \\ \therefore \bf A(0,5)$

Learn more:-

Values of x and y in each Quadrant:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\vector(0,1){6}}\put(0,0){\vector(0,-1){2}}\put(0,0){\vector(1,0){5}}\put(0,0){\vector(-1,0){5}}\put(3,3){\bf (x> 0 ,y>0) }\put(-3,3){\bf (x<0,y>0) }\put(-3,-3){\bf (x<0,y<0) }\put(3,-3){\bf (x> 0,y<0) }\put(3.3,3.5){\boxed{\sf Q_1}}\put(-3.3,3.5){\boxed{\sf Q_2}}\put(-3.3,-3.5){\boxed{\sf Q_3}}\put(3.3,-3.5){\boxed{\sf Q_4}}\end{picture}$

Answer 2

Given :- ABC is an equilateral triangle of coordinates of vertices of B and C are(3,0)and (-3,0) find coordinates of vertices of A which on positive y -Axis .

Answer :-

since ∆ABC is an equilateral triangle , length of all sides will be equal .

so,

• AB = BC = AC .

we know that, distance between two points P(x1, y1) and Q(x2, y2) is given by,

• PQ = √[(x2 - x1)² + (y2 - y1)²]

given that,

• A = (0 , y) { since at y - axis x will be zero. }
• B = (3 , 0)
• C = (-3 , 0)

then,

→ AB = BC

→ √[(3 - 0)² + (0 - y)²] = √[(-3 - 3)² + (0 - 0)²]

squaring both sides,

→ 3² + (-y)² = (-6)² + 0²

→ 9 + y² = 36 + 0

→ y² = 36 - 9

→ y² = 27

square root both sides,

→ y = √(27)

→ y = ± 3√3

since , it is given that, vertices of A which on positive y axis .

therefore, the coordinates of A will be (0 , 3√3).

Learn more :-

prove that the mid point of the line joining points (-5,12) and (-1,12) is a point of trisection of the line joining the...

https://brainly.in/question/15118408

in what ratio is tge line joining the points (9, 2) and (-3, - 2 ) divided by the y axis? also find the coordinate of th...

https://brainly.in/question/15022879

find the equation of the straight line which passes through the point 4,3 and parallel to line 3x +4y=7

https://brainly.in/question/15527182