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# 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.

## Research, Knowledge and Information :

### Find the Equation of an Altitude in Triangle - YouTube

Dec 15, 2010 · Find the Equation of an Altitude in Triangle Dan Nechodom. ... find the equation and length of altitude from the vertex A In the triangle ABC with ...

### Equation of the Altitudes of a Triangle | eMathZone

To find the equation of the altitude of a triangle, we examine the following example: ... This is the required equation of the altitude from to .

### Altitude of Triangles Tutorial, Theorem, Formula, Definition

Altitude of a Triangle Tutorial. ... In the above right triangle, BC is the altitude ... Learn How To Find Discriminant Value Of Quadratic Equation ;

### Equilateral Triangle Equations Formulas Calculator - Altitude ...

Geometry calculator for solving the altitude of an equilateral triangle ... Triangle Equations ... These equations apply to any type of triangle. Reduced equations ...

### BBC - Higher Bitesize Maths - The straight line part 2 ...

Applying your knowledge. ... Find the equation of the median through A in triangle ABC. ... Find the equation of the altitude through B in Triangle ABC.

### How do you find the equation of a triangles altitude using ...

how do you find the equation of a triangles altitude using point slope ... An altitude of a triangle is a line segment from a vertex perpendicular to the opposite ...

### How to Write Equations of the Altitudes of Triangles | Sciencing

How to Write Equations ... Altitude = (2 * Area)/Base. The final equation is Altitude ... To solve for the height of a scalene triangle using a single equation, ...

### Finding the Altitude of a Triangle - dummies

Finding the Altitude of a Triangle; Finding the Altitude of a Triangle. Related Book. Geometry For Dummies, 2nd Edition. By Mark Ryan . The altitude of a triangle is ...

## Suggested Questions And Answer :

### 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.

### 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.

### how do you find the altitude of a triangle

?????????????????????? "altitude" ????????????????? weerz the plane???? & tell us how hi it fli !!!!!

### line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them

line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them There are two lines that satisfy the given parameters, one above and to the left of line d, and one below and to the right of line d. To find a point 6 units away from line d, we need a line that is perpendicular to d. We get the equation of that line by using the negative reciprocal of the m component, 2: that gives us -1/2. The equation of this line is y = -1/2 x + 4, using the same y intercept. A segment of that line that is 6 units represents the hypotenuse of a right triangle. If we are looking at the segment that extends to the left of the given y intercept, we consider the distance to be negative (-6). If we are looking at the segment that extends to the right of the given y intercept, we consider the distance to be positive (+6). To find the x and y co-ordinates, we use the sine and the cosine of the angle (we'll call it a) formed by this new line and the x axis. The slope m of a line is the y distance divided by the x distance. That is also the definition of the tangent of the angle. Using the inverse tangent, we determine that the angle is tan^-1 (-1/2) =  -26.565 degrees. We need to keep all the signs straight in order to get the correct values. The sine of -26.565 degrees is -0.4472. The cosine of -26.565 degrees is 0.8944. y1 / -6 = sin a y1 = -6 sin a = -6 * -0.4472 = 2.6832 This is measured from a horizontal line through the y intercept, because we are constructing a right-triangle with one corner at the y intercept, so the point's y co-ordinate is actually y = 2.6832 + 4 = 6.6832 x / -6 = cos a x = -6 * 0.8944 = -5.3664 Making a quick check using x^2 + y^2 = r^2: (-5.3664)^2 + 2.6832^2 = 28.79825 + 7.19956 = 35.99781   -6^2 = 36 Close enough considering the rounding in the calculations. We have co-ordinate (-5.3664, 6.6832) lying 6 units from line d. Substituting those values into y = 2x + c we have 6.6832 = 2 * (-5.3664) + c 6.6832 = -10.7328 + c 6.6832 + 10.7328 = c = 17.416 The equation of a line 6 units away from d and parallel to it, located above it, is y = 2x + 17.416    <<<<<<<< Working in the other direction, on the line below and to the right of line d, y2 / 6 = sin a y2 = 6 * sin a = 6 * -0.4472 = -2.6832 (that is, 2.6832 below d's y intercept) y = 4 - 2.6832 = 1.3168 x / 6 = cos a x = 6 * cos a = 6 * 0.8944 = 5.3664 The co-ordinates for this point, (5.3664, 1.3168), when substituted into the equation y = 2x + f (f is a new y intercept) gives 1.3168 = 2 * 5.3664 + f 1.3168 - 10.7328 = f = -9.416 The equation of a line 6 units away from d and parallel to it, located below it, is y = 2x - 9.416    <<<<<<<<

### equations of the lines that contain the three altitudes of triangle (0,0), (12,6), and (18,0)

Problem: equations of the lines that contain the three altitudes of triangle (0,0), (12,6), and (18,0) Need help on special segments Apparently you mean the three corners of the triangle. Let's start with the easy one. We have two points that lie on the x-axis: (0, 0) and (18, 0). The slope of that line is zero, and we can see that the y-intercept is zero. Therefore, the equation is y = 0. The next one to consider is the line with points at (0, 0) and (12, 6). Obviously, the y-intercept is 0.The slope is the change in y divided by the change in x. The slope has to be 6/12, which is 1/2. Why? As the line moves 12 spaces to the right, it moves 6 spaces up. So, the equation is y = 1/2 x. Likewise, we can calculate the slope of the third line. As it moves backwards from (18, 0) to (12, 6), the x value changes by -6, while the y value changes by 6. The slope is 6/-6, which is -1. All we need now is the y-intercept of this last line. From the equation in slope-intercept form, y = mx + b, we can get b = y - mx. We get x and y from the point that was given to us in the problem statement. b = 6 - (-1)12 b = 6 + 12 b = 18 The equation for this line is y = -x + 18. Our three equations are: y = 0 y = 1/2 x y = -x + 18

### find the area of a triangle that the sides are 12cm, 8cm and 6cm.

find the area of a triangle that the sides are 12cm, 8cm and 6cm. Layout the triangle with point A at the origin, point B 8cm to the right, and point C somewhere above, 6cm from B and 12cm from the origin. If this were a right-triangle, the hypotenuse would be 10cm, but in this triangle, what would be the hypotenuse is longer than that. We need another point, D, directly below C, forming two additional triangles, both of which will be right-triangles. They will be triangle ADC (which encloses triangle ABC) and triangle BDC (which is outside of triangle ABC). We will then find the area of triangle ADC and subtract the area of triangle BDC, leaving the area of triangle ABC. The endpoints are layed out similar to this:                  C A          B   D Call side AB a. Call side BC b. Call side AC c. Call angle ABC theta. The length of c is given by the general equation a^2 + b^2 - 2ab(cos theta) = c^2 We know a, b and c, so all we need to do is solve for the cosine of angle theta. Subtract a^2 from both sides, subtract b^2 from both sides, and divide both sides by -2ab. cos theta = (c^2 - a^2 - b^2) / (-2ab) cos theta = (12^2 - 8^2 - 6^2) / (-2 * 8 * 6) cos theta = (144 - 64 - 36) / (-96) cos theta = 44 / -96 = -0.458333.... Using the inverse cosine, we find that angle theta is 117.2796 degrees. That's the angle inside the given triangle, facing the 12cm side. We want the supplement, the angle CBD, facing away from the origin. 180 - 117.2796 = 62.72 degrees With that, we can find the length of CD. Call that d. d / 6cm = sin 62.72 d = 6cm * (sin 62.72) = 6cm * 0.88878 = 5.3327cm We need the length of side BD. Call it e. e / 6cm = cos 62.72 e = 6cm * (cos 62.72) = 6cm * 0.4583 = 2.75cm Those two lengths can be used to find the area of this small triangle: area1 = 1/2 * d * e area1 = 1/2 * 5.3327cm * 2.75cm = 7.3325cm^2 Area2, the area of triangle ADC is found by first adding the length of BD to the length of AB. Call that f. f = a + e = 8cm + 2.75cm = 10.75cm area2 = 1/2 * d * f area2 = 1/2 * 5.3327cm * 10.75cm = 28.66327cm^2 Area3, the area of the original triangle, is found by subtracting the area of the smaller triangle from the area of the larger triangle. area3 = 28.66327cm^2 - 7.3325cm^2 = 20.33cm^2

### (-5,3) (5,-3) (6,6) ARE THE MID PIONTS OF A TRIANGLE THEN FIND THE EQUATIONS OF THE SIDES?

(-5,3) (5,-3) (6,6) ARE THE MID PIONTS OF A TRIANGLE THEN FIND THE EQUATIONS OF THE SIDES? Let the vertices of the triangle be A, B and C. The coordinates of the vertices are: A(ax, ay), B(bx, by), C(cx, cy). Mid-point of AB is: ((ax + bx)/2, (ay + by)/2) = (-5, 3), say Mid-point of AC is: ((ax + cx)/2, (ay + cy)/2) = (5, -3), say Mid-point of BC is: ((bx + cx)/2. (by + cy)/2) = (6, 6), say Comparing these expressions with the point coordinates given, ax + bx = -10 ay + by = 6 ax + cx = 10 ay + cy = -6 bx + cx = 12 by + cy = 12 substituting for ax = -bx – 10 and ay = -by + 6 into the bottom 4 eqns, -bx – 10 + cx = 10  -> -bx + cx = 20 -by + 6 + cy = -6     -> -by + cy = -12 bx + cx = 12 by + cy = 12 Adding together the 1st and 3rd eqns just above and the 2nd and 4th eqns just above gives us, 2cx = 32 2cy = 0 i.e. cx = 16, cy = 0. So the point C is C(16,0) Substituting for cx = 16 and cy = 0 gives -bx + cx = 20 -> -bx + 16 = 20 -> bx = -4 -by + cy = -12 -> -by + 0 = -12 -> by = 12 i.e. bx = -4, by = 12. So the point B is B(-4,12)   Substituting for cx = 16, cy = 0, bx = -4 and by = 12 gives ax = -bx – 10 -> ax = 4 – 10 -> ax = -6 ay = -by + 6 -> ay = -12 + 6 -> ay = -6 i.e. ax = -6, ay = -6. So the point A is A(-6, -6) The vertices of the triangle are: A(-6, -6), B(-4, 12), C(16, 0)   The equations of the sides Side AB A = (-6, -6), B = (-4, 12) (y – 12)/(x – (-4)) = (-6 – 12)/(-6 – (-4)) (y – 12)/(x + 4) = -18/(-2) = 9 y – 12 = 9x + 36 y = 9x + 48 Side AC A = (-6, -6), C = (16, 0) (y – 0)/(x – 16) = (-6 – 0)/(-6 – 16) y /(x – 16) = -6/(-22) = 3/11 11y = 3x – 48 Side BC B = (-4, 12), C = (16, 0) (y – 0)/(x – 16) = (12 – 0)/(-4 – 16) y/(x – 16) = 12/(-20) = -3/5 5y = -3x + 48

### X is midpoint side AB in triangle ABC.y is the midpoint of CX.let by cut AC at Z.show that AZ=2ZC

To tackle this problem I decided to try using a graph. Point A of the triangle is at the origin (0,0); point B is at (2X,0) and point X at (X,0), so that X is the midpoint of AB. The third point of the triangle at C is (a,b), an arbitrary point. We know that Y is halfway along CX so that means Y is ((X+a)/2,b/2), because its x coord along base AB is halfway between X (X,0) and C (a,b) and its y coord is halfway between B and C. We can write an equation for AC: y=bx/a. And we can write an equation for BZ because it's a continuation of BY, and we know the coords of B and Y. The slope of the line is (b/2)/((X+a)/2-2X)=b/(a-3X), the difference of the y values divided by the difference of the x values of the two points. The equation for the line becomes y=bx/(a-3X)+k, where k is the y intercept we have to find. Since point B (2X,0) is on the line we can substitute the coords to find k: 0=2Xb/(a-3X)+k, so k=-2Xb/(a-3X) and the equation of the line BZ is y=bx/(a-3X)-2Xb/(a-3X)=(b/(a-3X))(x-2X). The two lines AC and BZ intersect at bx/a=(b/(a-3X))(x-2X). We can divide through by b to get: x/a=(x-2X)(a-3X). Multiply through by a(a-3X): x(a-3X)=a(x-2X), so -3xX=-2aX, and x=2a/3. The y value  can be found by substitution: y=bx/a=2b/3 because the a's cancel out. The point C is (a,b). Therefore the point Z is (2a/3,2b/3) and this implies that AZ lies 2/3 the way along AC and AZ=2ZC. QED.

### Question related to trigonometry.... Scroll down

C is equidistant from the points A and B. Since A = (2a,0) and B = (0, 2b), then the coords for the point C is (a,b). Hence distance of the point C from O is √(a^2 + b^2). OAB is a right-angled triangle with height = 2b and length = 2a. By Pythagoras' Theorem, the length of AB is √[(2b)^2 + (2a)^2] = 2√(a^2 + b^2) i.e. OC is half the length of AB. Hence C is equidistant form A, B and O. Equation of line For a line passing through the points (x1, y1), and (x2, y2), the equation of that straight line is given by (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1) SInce we have the points (0,2b) and (2a,0), then the eqn of the line is, (y - 2b)/(x - 0) = (0 - 2b)/(2a - 0) (y - 2b)/x = -b/a ay + bx = 2ab Area of triangle Area = (1/2) base times height A = (1/2)(2a * 2b) A = 2ab