six.1 and 6.step 3 Test
Explanation: SV = VU 2x + eleven = 8x – 1 8x – 2x = 11 + 1 6x = 12 x = dos Ultraviolet = 8(2) – 1 = fifteen
Explanation: Keep in mind your circumcentre of good triangle are equidistant regarding the vertices out-of a beneficial triangle. Assist Good(- cuatro, 2), B(- 4, – 4), C(0, – 4) end up being the vertices of your own provided triangle and assist P(x,y) be the circumcentre for the triangle. Following PA = PB = Pc PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + 16 + y? – 4y + 4 = x? + 8x + sixteen + y? + 8y + 16 12y = -twelve y = -step 1 PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + sixteen + y? + 8y + 16 = x? + y? + 8y + 16 8x = -16 x = -dos The fresh circumcenter was (-2, -1)
Explanation: Recall your circumcentre away from a great triangle are equidistant on vertices out of an excellent triangle. Assist D(3, 5), E(seven, 9), F(11, 5) be the vertices of one’s given triangle and you may help P(x,y) become circumcentre associated with triangle. Upcoming PD = PE = PF PD? = PE? = PF? PD? = PE? (x – 3)? + (y – 5)? = (x – 7)? + (y – 9)? x? – 6x + 9 + y? – 10y + twenty five = x? – 14x + 44 + y? – 18y + 81 -6x + 14x – 10y + 18y = 130 – 34 8x + 8y = 96 x + y = 12 – (i) PE? = PF? (x – 7)? + (y – 9)? = (x – 11)? + (y – 5)? x? – 14x + forty two + y? – 18y + 81 = x? – 22x + 121 + y? – 10y + 25 -14x Schwul Dating + 22x – 18y + 10y = 146 – 130 8x – 8y = 16 x – y = 2 – (ii) Add (i) (ii) x + y + x – y = twelve + dos 2x = 14 x = eight Lay x = 7 from inside the (i) seven + y = several y = 5 The latest circumcenter try (eight, 5)
Explanation: NQ = NR = NS 2x + 1 = 4x – 9 4x – 2x = ten 2x = ten x = 5 NQ = ten + 1 = eleven NS = eleven
Explanation: NU = NV = NT -3x + six = -5x -3x + 5x = -six 2x = -6 x = -3 NT = -5(-3) = 15
Explanation: NZ = Ny = NW 4x – 10 = 3x – step one x = nine NZ = 4(9) – ten = thirty six – 10 = 26 NW = twenty six
Explanation: 5x – 4 = 4x + 3 times = eight ?JGK = 4(7) + 3 = 31 yards?GJK = 180 – (30 + 90) = 180 – 121 = 59
Get the coordinates of one’s centroid of one’s triangle wilt brand new considering vertices. Concern nine. J(- step one, 2), K(5, 6), L(5, – 2)
Explanation: The slope of TU = \(\frac < 1> < 0>\) = -2 The slope of the perpendicular line is \(\frac < 1> < 2>\) The perpendicular line is y – 5 = \(\frac < 1> < 2>\)(x – 2) 2y – 10 = x – 2 x – 2y + 8 = 0 The slope of UV = \(\frac < 5> < 2>\) = 2 The slope of the perpendicular line is \(\frac < -1> < 2>\) The perpendicular line is y – 5 = \(\frac < -1> < 2>\)(x + 2) 2y – 10 = -x – 2 x + 2y – 8 = 0 equate both equations x – 2y + 8 = x + 2y – 8 -4y = -16 y = 4 x – 2(4) + 8 = 0 x = 0 So, the orthocenter is (0, 4) The orthocenter lies inside the triangle TUV