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# for the traingle with the vertex A (5,1) B(3,-5) & c(-3,7) find the equation of the altitude from B

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## Research, Knowledge and Information :

### Find the area of a triangle with vertices A(1, 3, 5), B(-2 ...

draw a sketch of the triangle b. draw the altitude from vertex A c. find the ... 5), B(-2, -3, -4) and C(0, 3, -1). Trig Find the ... (7,3,4)(1,0,6)(4,5,-2 ...

### Triangle - Wikipedia

A triangle with vertices A, B, and C is denoted ... (one for each vertex) of any triangle is 360 degrees. ... and the altitude of the triangle from the base of length ...

### the vertices of a triangle are A(-1,7), B(-2,-2) and C(5,1 ...

... (5,1) the altitudes AD and BE meet at H. find the equation ... any altitude drawn from any vertex must be ... to find the equation for altitude ...

### Triangles - UH

b) 7.5 in, 8.3 in, and 4.2 in 2. The ... c) Hold an angle at its vertex and fold so that the sides meet along a line that ... 3. The triangles have the same size and ...

### Triangle XYZ with vertices X(3,-1) Y(7,-3)Z(3,-5) undergoes ...

Triangle XYZ with vertices X(3,-1) Y(7 ... angle in the traingle b) ... triangle b. draw the altitude from vertex A c. find the slope of side BC which is ...

### /&1 .I·l. I I, I - JMAP

(3) (2) (4) 2 An equation of a line perpendicular to the line represented by the ... vertex? (1) no and C'(l,2) (3) yes and A'(6,2) (2) ... (1) 5 (3) 10 I (2) 7 @20

### A triangle ABC with vertex \$C(4,3)\$. The bisector and the ...

A triangle \$\triangle ABC\$ one of his vertex is the point \$C(4,3) ... x+2y-5=0\$ and the median line equation is \$4x+3y ... B and the altitude drawn from C meet ...

### (688,#23) Find the lengths of the medians of the triangle ...

(688,#23) Find the lengths of the medians of the triangle with vertices A(1,2,3), B(-2,0,5) and C(4,1,5). Solution: Determine the midpoint of each line segment:

### Geometry Regents Review - White Plains Public Schools / Overview

1 Geometry Regents Review ... altitude CD? A) 6 B) 6 5 C) 3 D) 3 5 7. ... 1 B) 2 C) 3 D) 0 14. The equation of a circle is (x −2)2 +(y +5)2 =32.

### Altitude (triangle) - Wikipedia

In geometry, an altitude of a triangle is a line segment through a vertex and perpendicular to (i.e., forming a right angle with) a line containing the base (the side ...

## Suggested Questions And Answer :

### for the traingle with the vertex A (5,1) B(3,-5) & c(-3,7) find the equation of the altitude from B

?????????????????? "altitide" ?????????? ???? yu gonna fli plane inside thr triangel ??????? me assume yu want the HITE, but kant duit yer wae Hite go from a SIDE, not from a point The base=longest side=from (-3,7) tu (+3,-5) deltax=6, deltay=12, leng=sqrt(6^2+12^2)=sqrt(36+144)=sqrt 180=13.41640786 side2...from (-3,7)  tu (5,1): deltax=8, deltay=6...leng=sqrt(8^2+6^2)=sqrt(64+36)=sqrt(100)=10 side3...from (3,-5) tu (5,1); deltax=2, deltay=6...leng=sqrt(4+36)=sqrt(40)=6.32455532 now, kan make resunabel triangel from 3 side lenths Hite=4.47236 triangel area=30.00074, rimleng=29.74063 top=(4.47192, 4.47236

### if each side of a rhombus is 8 cm and is 36 cm square.find its altitude

Here the side length a=8(cm) and the area of rhombus S=36(cm^2) are known values, but the height of altitude h is the unknown which we are to find relating to a and S. Label each vertex of the rhombus A,B,C and D, respectively, counterclockwise from an acute vertex. Draw a line connecting B and D, and an altitude from D to the foot H on side AB. The altitude DH intersects side AB perpendicularly at H. Therfore, the area of triangle ABD is s=1/2(side ABx altitude BD)=ah/2. While triangle ABD is congruent to triangle BCD since each side of a rhombus is congruent to each other, and both triangles share side BD. (S.S.S) Therfore, S=2xs=2xah/2=ah. This equation S=ah expesses: The area of a rhombus=(the side)x(the altitude). Plug S=36(cm^2) and a=8(cm) into the equation obtained above. 36=8xh, h=36/8=4.5 CK: 4.5(cm)x8(cm)=36(cm^2) The altitude of the rhombus is 4.5cm.

### Find the equation of the altitude of the triangle

A(1,-5),B(2,2) and C(-2,4) are the vertices of triangle ABC . Find the equation of the altitude of the triangle through B. The answer given at the back of my book is 3y=x+4. But how is it coming. The altitude of the point B is a line perpinficular to the base line AC, and passing through the point B(2,2). Find the slope of the base line AC Hence find the slope of any line perpindicular to it. This line will now pass through the point B(2,2) You have a line with a slope that passes through a fixed point. Hence you can find the equation of this line.

### area under y=3(x-2)^2+1

The first step is to sketch the curve on a graph. This curve is a parabola, a U-shaped curve in which the arms spread out gradually. The key points of a parabola are intercepts, one at least, and the vertex, which is where the curve changes direction. The vertical line through the vertex is a line of symmetry acting like a mirror reflecting the two halves of the curve. If we rewrite the equation: y-1=3(x-2)^2 we can find the origin of the parabola. When y=1, x=2 which tells us the origin is (2,1), which is the vertex and the line x=2 is the line of symmetry. When x=0, y=13, so the curve cuts the y axis at (0,13). When y=0 (the x axis) the curve has no real value so it doesn't touch or cut the x axis. This is enough to sketch the curve and actually see the area you need to find.  The next step is to enclose the area by establishing limits. The normal limits will trap an area between the curve and one or both axes. These will become the integration limits so that a definite integration can take place.  Now we build the integral by dividing the area into thin rectangular strips. The length of a vertical rectangle is y and its width is dx. The area is ydx and the sum of the areas of the rectangles gives the area under the curve so we have integral(ydx)=integral((3(x-2)^2-1)dx) and the limits for integration will be a to b, a Read More: ...

### Finding the vertex of a parabola?

The formula for x is -b/2a so your problem is x^2 +8 because x*x is x^2 and 4*2=8. The formula for a quadratic equation is ax^2+bx+c and you don't have a b so your b is 0. -0/2(1) =0. Your axis of symmetry equation is x=0. If you want to graph as well, then you plug it into a graph... my teacher has us put three x-intercepts on either side so you would have -3,-2,-1,0,1,2,3 as your x-intercepts. Then you plug each one back into the original equation and those are your y's. They should be the same numbers on either side of the vertex- so if one coordinate is (-3,2), your other coordinate should be (3,2.) Hope this helps!

### In a triangle ABC, line joining orthocenter and the circumscenter is parallel to AC, then find tanA *tanC.

Draw the base of the triangle AC and mark its midpoint, D. Draw a perpendicular from D. Somewhere along the perpendicular will be the circumcentre, that is, the centre of the circle that circumscribes the triangle ABC.  The position of vertex B has to be such that the perpendicular bisector of AB meets the perpendicular bisector of AC at X, and the perpendicular from vertex C on to AB meets the perpendicular from vertex B on to AC at Y, such that XY is parallel to AC. This is another way of framing the question. No other construction lines are required because the missing perpendiculars are superfluous to solving the problem. Represent the problem graphically. Plot the points A(0,0), B(p,q), C(t,0) where p and q are to be determined and t is an arbitrary constant t=AC. Midpoint of AC is N(t/2,0); midpoint of AB is M(p/2,q/2). AB is a segment of the line y=qx/p. The equation of the bisector of AB: -p/q is its gradient, so y=-px/q+c, where c is found by plugging in M: q/2=-p^2/2q+c, c=q/2+p^2/2q and the perpendicular bisector is a segment of y=q/2+p^2/2q-px/q. Therefore, the coords of X are where this line meets the perpendicular bisector of AC, which is a segment of the line x=t/2. The intersection is X(t/2,q/2+p^2/2q-pt/2q). Now we need to find Y, which must lie on the line x=p, because this is the perpendicular from vertex B on to AC. The perpendicular from vertex C on to AB is parallel to the perpendicular bisector of AB so has the same gradient: -p/q. The equation of the perpendicular from C is y=-px/q+k, where k is found by plugging in C(t,0): 0=-pt/q+k, k=pt/q and y=-px/q+pt/q=p(t-x)/q. The point Y is therefore Y(p,p(t-p)/q). XY is parallel to AC, which means their y coord is the same: q/2+p^2/2q-pt/2q=p(t-p)/q; q^2+p^2-pt=2p(t-p); q^2+3p^2=3pt; q^2=3p(t-p). Also, tanA=q/p and tanC=q/(t-p), so tanA*tanC=q^2/(pt-p^2)=3p(t-p)/(p(t-p))=3. Note that when tanA=tanB=sqrt(3), B is (1/2,sqrt(3)/2), the triangle is equilateral and XY=0 because the circumcentre and orthocentre coincide.

### Finding the vertex of a quadratic function?

To find the y-intercept, set x equal to 0. 0^2= 0 2*0 = 0 and you are left with y=-3 (0,-3) is the y-intercept. Yo find the Vertex: x coordinate of vertex is -b/2a = -2 / 2 = -1 y coordinate of the vertex = (-1)^2+2(-1)-3=1-2-3 = -4 vertex is (-1,-4) --------------- You can use the quadratic equation with the following values to find the zeros: a=1, b=2 and c=-3

### find the vertex of a parabola of f(x)=3x^2+18x+10 quadratic function

The x-coordinate of the vertex will be = -b/(2a) when you're given an equation of the form y = ax^2 + bx + c. So here a = 3, b = 18, and so the x-coord is -18/(2*3) = -3. Now just plug that into the equation to find the y-coord of the vertex: 3(-3)^2 + 18(-3) + 10 = -17 So the vertex is (-3,-17)

### Mechanics projectile M2 question

The trajectory of the particle is a parabola because of gravity. The acceleration due to gravity is 9.81m/s^2 and is, of course, a negative influence on the particle's motion. If we draw a graph of the parabola and make it symmetrical, the point A is on the y axis at y=8, representing 8m. On the x axis at O(-10,0) and P(10,0) the parabola intersects the x axis. The point A(0,8) is the vertex at which the slope is horizontal. Let y=ax^2+bx+c. A is the y intercept so c=8. When y=0, x=10 and -10, so 0=100a+10b+8=100a-10b+8, making b=0, and 100a=-8, so a=-0.08, making the equation of the parabola y=8-0.08x^2. We can relate x and t and y and t, where t is time. The general equation is s=ut-(1/2)gt^2, where s is distance, u an initial speed, and g is the acceleration of gravity which in this case acts against the particle's movement because it is initially projected against gravity. This is a scalar equation, but we need to apply it in a vector situation. Now that we have the equation of the parabola, we need to find out the gradient at O(-10,0) to give us the projection angle. So differentiate y=8-0.08x^2 and we get y'=-0.16x. At x=-10 y'=1.6. The angle of projection z is given by tanz=1.6. Velocity is a vector and the initial projection velocity can be split into a horizontal and vertical component. Gravity acts only on the vertical component. We can apply the motion equation by replacing s with y and u with vsinz, where v is the magnitude of the projection velocity. Therefore y=v(sinz)t-(1/2)gt^2. For x the motion equation is simply x=v(cosz)t.  We can use t to represent the time taken to go from O(-10,0) to P(10,0). The vertical displacement (y) then becomes 0, because the particle starts on the ground and ends on the ground. The horizontal displacement is 20m. By eliminating t between equations we can find the magnitude of v. Putting y=0 in the motion equation: v(sinz)t=(1/2)gt^2, so t=2v(sinz)/g and x=v(cosz)t=2v^2(sinz)(cosz)/g. The horizontal displacement (x) is 20m so 2v^2(sinz)(cosz)/g=20 and v=sqrt(10g/(sinzcosz)). We can calculate sinz and cosz from tanz. sinz=1.6/sqrt(3.56) and cosz=1/sqrt(3.56), so sinzcosz=1.6/3.56. Therefore v=sqrt(98.1*3.56/1.6)=14.774m/s. z, the projection angle=tan^-1(1.6)=58 degrees.

### x2 − 6x + 12y + 57 = 0

12y+x^2-6x+57=0; 12y+x^2-6x+9+48=0;12y+48=-(x-3)^2; y+4=-(1/12)(x-3)^2. The equation of the parabola in this form shows that the vertex V(h,k) is at (3,-4) (y+4=0, y=-4; x-3=0, x=3). The axis of symmetry is x=3. The directrix is the line y=d (=2k-f) and the focus is at F(3,f), where d and f are to be calculated.  To find the focus, we take the coefficient of (x-3)^2 which is -1/12, which is 1/4p where p is the distance of the curve from both the focus and directrix (see section in square brackets below). So 4p=-12, making p=-3. At the vertex, the curve is equidistant from the focus and the directrix. Therefore, f=p-4=-3-4=-7 and the focus is at (3,-7). The directrix is the line y=d=-4-(-3)=-1. The parabola is an upturned, rather fat U. To illustrate the use of focus and directrix, put x=9 in the equation of the parabola, y+4=-(1/12)36=-3, so y=-7. The point (9,-7) is 9-3=6 away from the focus and -1-(-7)=6 away from the directrix. This is as fundamental property of all parabolas: all points are equidistant from the focus and directrix line. The vertex is 3 away from the focus and directrix line. [Derivation of focus and directrix from first principles and geometry: all points on the parabola are equidistant from the focus and directrix line: Equation of parabola: y-k=a(x-h)^2 (y+4=-(1/12)(x-3)^2, a=-1/12, h=3, k=-4); focus: F(h,f); directrix: y=k-(f-k)=2k-f; P(x,y); V(h,k); FV=f-k=VM, where VM is perpendicular to the directrix line. PF=PN, where PN is perpendicular to the directrix line. By Pythagoras: PF=(y-f)^2+(x-h)^2=(y-(2k-f))^2=PN (y-f)^2+(y-k)/a=(y-2k+f)^2=(y+f-2k)^2 y^2-2fy+f^2+(y-k)/a=y^2+2fy+f^2-4k(y+f)+4k^2 -4fy+(y-k)/a+4k(y+f)-4k^2=0 (y-k)/a+4(-fy+ky+kf-k^2)=0=(y-k)/a+4(k(y-k)-f(y-k))=(y-k)(1/a+4(k-f)) 1/a+4(k-f)=0; 1/4a+k-f=0, because y in general is not equal to k (only at the vertex) f-k=1/4a=p (see paragraph 3 above) p=1/4a=-3; f+4=-3, f=-7, so focus is F(3,-7). Equation of directrix: y=-8+7=-1.]