How to find the equation of a perpendicular in a triangle?
829618
1.find the midpoint of BC usingthe midpoint formula\n(x1+x2/2,y1+y2/2) say (x,y)\n2.find the slope os BC using theformula \ny2/y1/x2-x1 say m\n3.the slope of the perpendicular bisector will be the negative inverse of this -1/m\n4. now use the point (x,y) and slope (-1/m) and find the equation of the perpendicular bisector
551883
The answer is C\nwell, one side's length of a triangle should be shorter than the other 2's plus (2 sides plus should be longer than one side)\nthen: 3x+4+8x+2>6x-1\n      8x+2+6x-1>3x-4\n      6x-1+3x+4>8x+2\n--> x>-7/5\n    x>3/11\n    x>-1\nThus, the answer should be x>3/11\n:)
1045403
First put the line in slope-intercept form so you can get the slope.\n\n-4y = -3x + 18\ny = 0.75x - 4.5\n\nSo the slope of the line is 0.75 or 3/4.\n\nSo the slope of a line that is perpendicular to this line is -1/0.75 = -4/3 (The slope of a perpendicular is -1/m)\n\nSo we know the slope for the perpendicular, and a point on the perpendicular is (3,4), so we can get the intercept.\n\ny = mx + b\n4 = (-4/3)(3) + b\n4 = -4 + b\nb = 8\ny = (-4/3)x + 8\n\nSo now we can find the point where the two lines intersect by solving the system of equations.\n\ny = (3/4)x - 4.5\ny = (-4/3)x + 8\nSo (3/4)x - 4.5 = (-4/3)x + 8.\nSo (3/4 + 4/3)x = 8 +4.5.\nSo (25/12)x = 12.5.\nSo x = 12.5(12/25) = 6.\nSo y = (3/4)(6) - 4.5.\nSo y = 0.\n\nSo now we can find the distance between the point (3,4) and the point that is on the line and on the perpendicular (6,0)\n\nsqrt( (3-6)^2 + (4 - 0)^2) = sqrt(25) = 5\n\nSo the distance is 5 units.
962974
the reflection of point (3,4) will be on a line perpendicular to 3x+4y=50 and passing by the given point, and at equal dist from the line to that of the point(3,4) and the line.\n\nanyway I'll try to show u how to solve that\n\nfirst we have to write the equation of the line passing thru (3,4) and perpendicular of y= -3/4x + 50/3\nm1 is the slope of the equation y=-3/4x + 50/3 so m1=-3/4\nm2 is the slope of the equation perpendicular to the above one\nm1 * m2 = -1\nso m2=4/3\n\nthen we will have y=4/3x + b\nreplacing x and y in the above equation we get\n4 = 4/3 *3 +b so b = 0\nthen the equation passing thru point A(3,4) and perpendicular to the equation 3x+4y=50 will be:\ny = 4/3 x\n\nNow to find the distance between point A(3,4) and the equation\n y=-3/4x +50/4 we should get the intersection point of the two perpendicular lines\nto do that we set the two y coordinates equal\n4/3x = -3/4x + 50/4\n(4/3 + 3/4)x = 50/4\n25/12 x = 50/4\nx = 6\nand replacing x in either equation we get y=4/3 * 6 = 8\nthen the intersection point will be I(6,8)\nLet point B(xb,yb) be the reflection of A\nso I midpoint of AB\nthe xi = (xb + xa)/2 => 2xi - xa = xb = 2*6 - 3 = 9\nand yi = (yb + ya)/2=> yb = 2yi -ya = 2*8 - 4 = 12\n\nso the reflection of point A(3,4) is point B(9,12)
174825
(-1,4), perpendicular to 2x+3y=8\n\n2x+3y=8\n3y = -2x+8\ny = (-2/3)x + (8/3)\nThen slope is: (-2/3)\nA rect with slope m perpendicular to the first:\nm(-2/3) = -1\nthen m = 3/2\nand the equation is \n(y-4) = (3/2)(x+1)\n2y-8 = 3x+3\n2y + 3x = 11
920764
Perpendicular lines have slopes whose product is -1.\n\nIf the original line has slope 3, the line that is perpendicular has slope -(1/3)\n\nNow use the point slope form of the line.\n\ny - y1 = m(x - x1)\n\nIf (x1,y1) = (2,2) - your point, then:\n\ny - 2 = (-1/3)(x - 2)\n\nWe can convert this into slope-intercept form.\n\ny - 2 = (-1/3)x + (2/3)\ny = (-1/3)x +(8/3)\n\nBoth equations represent the same line.
227708
There are two ways to do it...\n\n1.> If you know the base and the height of the triangle:\n\n(Base*Height)/2\n\nso if one of the side was 12 cm., and the perpendicular on it from the point where the other two sides meet is 6 cm., the area is = (12*6)/2 = 36 sq.cm.\n\n2.> If you know all the sides:\n\nHero's formula:\n\nFirst find s (semi-perimeter) as:\n\ns=0.5*(side1+side2+side3)\n\nnow the area= root (s * (s-side1) * (s-side2) * (s-side3))\n\nso if a triangle's sides are 3,4 and 5 cm.:\n\nsemi-perimeter = (3+4+5)/2 = 6\n\narea= root (6(6-3)*(6-4)*(6-5)) = root (6*3*2*1) = root (36) = 6 sq. cm.
1184743
Heron's formula for triangles:\na = √s(s-a)(s-b)(s-c)\ns is the semiperimeter or 1/2 (4+4+4)\ns = 6\na = √6(2)(2)(2)\na = √3×2²×2²\na = 2×2×√3\na = 4×1.7\na = 6.8\n\n\nJose, you calculated the area of the right triangle, not the entire equilateral triangle you started with.
1102848
Slope of k = (2 - 4) / (4 - 1) = -2 / 3\n\nSlope of perpendicular = -1 / (-2 / 3) = 3 / 2\n\n(To find the slope of any line perpendicular to any other line you divide -1 by the slope of the first line).
380640
F.G is the component of F laying on G.\nl=F.G/g=(10*0.1-6*0.2+5*0.3)/g\nl=(1-1.2+1.5)/sqrt(0.1^2+0.2^2+0.3^2)\nl=(1.3)/sqrt(0.14)=3.5\nNow the component of F perpendicular to G is:\np=sqrt(f^2-l^2)=sqrt(100+36+25-12.25)=13.2\n\nThe answer is 13.2\n\nthe direction is parallel to: (G*F)*G\nwhere * is cross product.
333253
You can solve this using x, y, and r (the lengths of the 3 sides of the triangle made by the angle).  The trigonometric functions are originally defined as:\n\nsin = y/r\ncos = x/r\ntan = y/x\nsec = r/x\ncsc = r/y\ncot = x/y\n\nYou need to find x, y, and r for angles a and b.  Do this using the given sin and cot values and the equation\n\nx^2 + y^2 = r^2\n\nSo for a:\n\nsin a = 3/5\ny=3, r=5\n\nx^2 + 9 = 25\nx^2 = 16\nx=-4\n(this is negative because the angle is in the 2nd quadrant)\n\nFor b:\ncot b = -12/5\nx=-12, y=5\n(x is the negative one because this is in the 2nd quadrant)\n\n144 + 25 = r^2\n169 = r^2\nr=13\n\nNow you need to turn tan (a-b) into a form where you can plug in your x, y, and r values, and then plug and solve.  Remember that tan = y/x.\n\ntan (a-b) = (tan a + tan b) / (1 + tan a * tan b)\n=(-3/4 + -5/12) / (1 + (-3/4)(-5/12)\n=(-14/12)/(1 + 15/48)\n=(-7/6)/(63/48)\n=-8/9\n\nTherefore, your final answer is -8/9.\n\nHope this helps!
897842
The first thing you need to be able to figure out is what the question is asking for.  Slope-intercept form (which I prefer to standard form for any and all linear equations) is y = mx + b, where m is the slope and b is the y-intercept (the place where x=0).  To be perpendicular to a line, another line must have a slope which is the negative reciprocal of the slope of the first (e.g. 2 & -1/2).\n\nFirst, figure out the slope of the given line.  In standard form, which is ax + by = c, the slope is -a/b, so the slope of this line must be 5/2.  Take the negative reciprocal of this to find that your second line must have a slope of -2/5.\n\nNow set the equation up in slope-intercept form:\ny = -2/5x + b\nPlug in your point:\n-5 = (-2/5)(3) + b\nSolve for b\n-5 = -6/5 + b\n-19/5 = b\n\ny = -2/5x -19/5
350722
The steps I followed were:  pick a vector in that subspace of R3, then find another vector perpendicular to that one that's still in R3.  These two should form a basis.  So, I picked (a,0,-a)to start from.  A vector perpendicular to that one that's still in the subspace is (a,2a,a).  \n\nA more general way to  find a basis (not necessarily normal) would be to pick a vector in that subspace, and then pick another one.  If the only solution of the equation A(vector 1)+B(vector 2)=0 is A=B=0, the two vectors are linearly independent, and form a basis for the subspace.
831376
First, put both equations into slope-intercept form (y = mx + b).\n\nL1:\nx - 4y = 12\nx - 12 = 4y\ny = (1/4)x - 3\n\nL2:\n4x + y = 4\ny = -4x + 4\n\nWhen two lines are parallel, they will have the same slope. When two lines are perpendicular, the slopes will be the opposite reciprocals of each other.\n\n-4 and 1/4 are the opposite reciprocals of each other. Therefore, the two lines are perpendicular.
247800
it is basically:\na2 + b2 = c2   -->  (A squared)+(B squared)=(C squared)\n  \n\nif you have a right triangle.you have three sides.the two legs.and the hypotenus(sp?).a and b are the two legs and c is the hypotenus.\n\nin an equation.if you are given a and b.you can square each of them and then add the two values together.the will give you c squared.to find c you then need to take the square root of the value(a squared + b squared).\n\nif you have c and a leg(example:a).you square c.and then subtract a squared from that value(c squared).that will give you b squared.to get b you just get the square root of c squared - a squared.\n\n\nhope that helps.
1175454
y=-3x\nand\ny=3x\n\nare perpendicular, but only if you change the scale into 1/3 of the y-scale(meaning, you divide the y-scale by 3). Otherwise y = -3x and y = 3x are not perpendicular.\n\n^_^
385116
2x + y = 7\ny = -2x + 7\n\nThe slope of this is -2, so the slope of the perpendicular is 1/2.  Using the point-slope formula, you can find the equation of the other line:\n\ny - y1 = m(x - x1)\ny - 4 = (1/2)(x - 4)\ny - 4 = x/2 - 2\ny = x/2 + 2