six.1 and you may six.step 3 Test
Explanation: SV = VU 2x + 11 = 8x – step one 8x – 2x = 11 + 1 6x = 12 x = 2 Ultraviolet = 8(2) – step 1 = 15
Explanation: 5x – 4 = 4x + three times = eight ?JGK = 4(7) + 3 = 30 yards?GJK = 180 – (31 + 90) = 180 – 121 = 59
Explanation: Remember the circumcentre off a beneficial triangle is equidistant in the vertices out of an effective triangle. After that PA = PB = Desktop computer PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + sixteen + y? – 4y + 4 = x? + 8x + sixteen + y? + 8y + 16 12y = -several y = -1 PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + sixteen + y? + 8y + sixteen = x? + y? + 8y + 16 8x = -sixteen x = -dos The circumcenter was (-2, -1)
Explanation: Recall your circumcentre off good triangle are equidistant in the vertices out-of a great triangle. Help D(3, 5), E(eight, 9), F(eleven, 5) function as the vertices of your considering triangle and you may assist P(x,y) end up being the circumcentre in the triangle. Upcoming PD = PE = PF PD? = PE? = PF? PD? = PE? (x – 3)? + (y – 5)? = (x – 7)? + (y – 9)? x? – 6x + nine + y? – 10y + twenty five = x? – 14x + 49 + y? – 18y + 81 -6x + 14x – 10y + 18y = 130 – 34 8x + 8y = 96 x + y = a dozen – (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 + 22x – 18y + 10y = 146 – 130 8x – 8y = 16 x – y = dos – (ii) Incorporate (i) (ii) x + y + x – y = a dozen + dos 2x = 14 x = 7 Lay x = seven inside the (i) 7 + y = 12 y = 5 The brand new circumcenter are (7, 5)
Explanation: NQ = NR = NS 2x + step one = 4x – nine 4x – 2x = ten 2x = 10 x = 5 NQ = ten + step one = 11 NS = eleven
Explanation: NU = NV = NT -3x + six = -5x -3x + 5x = -6 2x = -six x = -3 NT = -5(-3) = 15
Explanation: NZ = Nyc = NW 4x – 10 = 3x – step one x = nine NZ = 4(9) – ten = thirty six – ten = twenty-six NW = 26
Find the coordinates of your own centroid of the triangle wilt new given vertices. Question nine. J(- step 1, 2), K(5, 6), L(5, – 2)
Help Good(- cuatro, 2), B(- cuatro, – 4), C(0, – 4) function as vertices of your own provided triangle and you will let P(x,y) become circumcentre associated with triangle
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 ÑasualDates Birine Nasıl Mesaj -4y = -16 y = 4 x – 2(4) + 8 = 0 x = 0 So, the orthocenter is (0, 4) The orthocenter lies inside the triangle TUV