How can I find the third vertex of various triangles?
For first part assume the third vertex to be (a,b), now use the fact that two line segments in a right isoceles triangle are perpendicular(product of slope =-1) and then use the distance formula to equate the lengths of two equal sides.Hint for 2nd part-there are many methids , you can just use distance formula to get two equations
Find the cosines of the angles of the triangle with vertices (1,2,0), (0,2,3) and (4,0,1)?
first, find the lengths of each side, using the distance formula in 3D A(1,2,0) B(0,2,3) C(4,0,1) AB = sqrt[1^2 + 0^2 + 3^2] = sqrt(10) AC = sqrt[3^2 + 2^2 + 1^2] = sqrt(14) BC = sqrt[4^2 + 2^2 + 2^2] = sqrt(24) hey, it's a right triangle (since AB^2 + AC^2 = BC^2) angle A is the right angle, and cos 90 = 0 at this point, make sure to sketch it out use the basic definition that cos = adj / hyp cos B = AB / BC = sqrt(10) / sqrt(24) cos C = AC / BC = sqrt(14) / sqrt(24) these could also be rationalized if desired ex: cos B = sqrt(240) / 24 = 4sqrt(15)/ 24 = sqrt(15) / 6 cos C = sqrt(336) / 24 = 4sqrt(21)/24 = sqrt(21) / 6 note: this simplified because this is a right triangle if it were an oblique (non-right) triangle, then you would have to use the law of cosines: cos A = [b^2 + c^2 - a^2] / (2bc) cos B = [a^2 + c^2 - b^2] / (2ac) cos C = [a^2 + b^2 - c^2] / (2ab) here, a = BC, b = AC, and c = AB by convention, side a is opposite angle A side b is opposite angle B side c is opposite angle C
How do I find the type of triangle with the help of 3 coordinates?
If the coordinates of a triangle are given, we can find the type of triangle by measuring the distance between the points..For eg,Let A(x1,y1) , B(x2,y2) , and C(x3,y3) be the 3 coordinates of a triangle…Now, To find the type of triangle we now find the distance between the points as follows…Distance between the points AB,Distance between the points AC,And, distance between the points BC…Now, ifAB=AC=BC , then the triangle is an equilateral triangle..AB=AC≠BC / AB=BC≠AC / AC=BC≠AB , then the triangle is an isosceles triangle..AB≠AC≠BC, then the triangle is a scalene triangle…This is how we find the type of triangle using the 3 given coordinates…P.S.- The formula for finding the distance between 2 points is as follows..Let us consider 2 points, A(x1,y1) and B(x2,y2)..Thus, the distance between AB is given by the formulaAB=√{(x1-x2)²+(y1-y2)²}
How do you find the volume of a right triangle?
Actually only 3D figures has volume. Volume stands for the capacity of any solid which is possible only in 3D figures not in 2D figures.So,therefore, a right angle triangle do not have volume because it doesn't have any capacity and can only be drawn on a plane.
How do I find the third vertex of a 3D equilateral triangle?
3DTrianglePick one.Edit:If you're asking for the locus, it will be a circle with the center at the midpoint of those two points and having a radius ✓3/2 times the distance between the two points, in a plane with a normal of the plane having the same DCs as the line joining the two given points. Hope that answers your question.
What is the coordinates of the third vertex of an equilateral triangle whose two vertices are at (3,4) and (-2,3)?
GIVEN: Equilateral triangle ABC , A(x,y),B(3,4), C(-2,3)TO FIND: Coordinates of A, x & yBy distance formula, we calculate AB, BC & AC. Then we equate them to get x & y.Distance between 2 points= √[( x1-x2)² + ( y1-y2)²], where( x1, y1) &( x2,y2) are coordinates of 2 given pointsSo, in the given triangle ABC,BC= √(5²+1²) = √26 ………….(1)AB = √[(x-3)² +(y-4)²] …………(2)AC = √[(x+2)² +(y-3)² …………..(3)Let's equate (1) & (3) AC = BC => AC² =BC²(x+2)² + (y-3)² =26=> x² +4x +4 + y² -6y +9 = 26=> x² + y² +4x -6y -13 =0 ……………..(4)Now, we equate(2) & (3) AB = AC => AB² = AC²=> (x-3)² +(y-4)² = (x+2)² +(y-3)²=> x² + 9 - 6x + y² + 16 -8y = x² +4 + 4x + y² +9-6y=> 10x +2y -12 =0=> 5x + y -6 =0=> 5x +y = 6=> y = 6–5x ……………..(5)Now, put the value of y in equation (4)we get, x² + (6–5x)² + 4x -6( 6–5x) -13 =0=> x² +36 +25x² - 60x +4x -36+30x -13 =0=> 26x² -26x -13 = 0=> 2x² - 2x -1 =0= x = (2 +,- √12)/4 ( by quadratic formula)= > x = (2+,- 2√3)/4=> x = (1+ ,- √3)/2 ……………..(6)Now, get Y by eq (5)y = 6 - 5{( 1+, - √3)/2}y= 6 - ( 5+,-5√3 )/2= y= (7 +,- 5√3)/2 …………..(7)Now, after verifying we get the coordinates of the third vertex AIf x= (1+√3)/2 then y= (7–5√3)/2 ……….ANS★&if x = (1-√3)/2 then y = (7+5√3)/2………..ANS★●VERIFICATION●:AC = √[ {( 1+√3/2) + 2}² + {(7- 5√3)/2) -3)}²]= √[{(5+√3)/2}² + {(1–5√3)/2}²]= √[( 25+ 3 +10√3 +1 +75 -16√3) /4]=√104/4=√26Similarly AB =√[{(1-√3)/2+ 2}² + {(7+5√3)/2–3}²]= √[{(5-√3)/2}² +{ (1+5√3)/2}²]=√(25+3 -10√3 +1 +75 +10√3) /4= √104/4=√26And BC = √26 ( already calculated)So, AB= BC = AC[ Verified]
How do I find the length of a triangle on a 3D plane?
Oh, it's pretty much the same, with any number of dimensions, because each additional dimension, is at right angles to all of the earlier ones. ∆x² + ∆y² + ∆z² = D² with each delta meaning the difference between the same coordinate for the two points, and D the distance between the two points. It's the Pythagorean theorem, applied over again. Imagine it all, how the x's and y's give the diagonal in the x-y- plane; and that diagonal squared (which is the x's squared and the y's squared), added to the z-difference squared, gives the longer diagonal within the three dimensions. (You can rightly extend this to any number of dimensions...!) Ok, so PQ² = (5-3)² +(3-1)² + (5-1)² = 4 + 4 + 16 = 24, so PQ = √24 = 2√6. And QR² = (3-3)² + (1-7)² + (4-1)² = 36 + 9 = 45, so QR = √45 = 3√5. And RP = (3-5)² + (7-3)² + (1-5)² = 4 + 16 + 16 = 36, so RP = √36 = 6. And so we find the three sides measure: √24, √45, and 6 = √36 Clearly it's not isosceles (isosceles means two sides measure the same). If it were a right triangle, the squares of the short sides should add to the square of the long side, 24 + 36 = 45, but that's clearly not so either, so it's not a right triangle. Tip: if you had a hundred dimensions, and two points, same formula works, as extended: take the same coordinate from each point, subtract them, and square the difference; do it for each coordinate, and add all the results: the square root of the sum is the distance between the two points. By the way, it's not usual to call it "on a 3D plane" but rather "in a 3D space." It's just custom, no particular reason for it. And the proof for that formula would be inductive, it works in one and two dimentions, and if it works in n dimentions (in a n-dimensional space), then it works in n+1 dimensions as well, for the one in n-dimensions is a diagonal, and the n+1st difference will be at right angles to it... etc.
In coordinate geometry, is one of all vertex of equilateral triangles equal or not?
A vertex in geometry is the intersection of two lines forming an angle.Vertices is the plural of vertex. A triangle has 3 vertices.An equilateral triangle by definition has 3 equal sides which means that all the angles are going to be 60 degrees. All equilateral triangles have the same vertices, 60 degrees.