six.step one and 6.step 3 Test
Explanation: SV = VU 2x + 11 = 8x – step one 8x – 2x = 11 + step one 6x = twelve x = dos Uv = 8(2) – step 1 = 15
Explanation: 5x – 4 = 4x + 3 times = seven ?JGK = 4(7) + step 3 = 31 m?GJK = 180 – (30 + 90) = 180 – 121 = 59
Explanation: Bear in mind that the circumcentre out-of a triangle try equidistant regarding vertices out-of a beneficial triangle. Next PA = PB = Desktop computer PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + 16 + y? – 4y + 4 = x? + 8x + 16 + y? + 8y + 16 12y = -several y = -1 PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + 16 + y? + 8y + sixteen = x? + y? + 8y + 16 8x = -sixteen x = -2 The latest circumcenter is (-dos, -1)
Explanation: Remember the circumcentre off an excellent triangle is equidistant regarding vertices off an effective triangle. Let D(step 3, 5), E(eight, 9), F(11, 5) function as vertices of provided triangle and you can assist P(x,y) become circumcentre of triangle. Next PD = PE = PF PD? = PE? = PF? PD? = PE? (x – 3)? + (y – 5)? = (x – 7)? + (y – 9)? x? – 6x + 9 + y? – 10y + 25 = x? – 14x + forty two + y? – 18y + 81 -6x + 14x – 10y + 18y = 130 – 34 8x + 8y = 96 x + y = several – (i) PE? = PF? (x – 7)? + (y – 9)? = (x – 11)? + (y – 5)? x? – 14x + forty-two + y? – 18y + 81 = x? – 22x + 121 + y? – 10y + twenty-five -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = 16 x – y = dos – (ii) Incorporate (i) (ii) x + y + x – y = several + 2 2x = 14 x = seven Place x = seven in the (i) eight + y = twelve y = 5 The circumcenter are (seven, 5)
Explanation: NQ = NR = NS 2x + step one = 4x – nine 4x – 2x = 10 2x = ten x = 5 NQ = 10 + step 1 = eleven NS = eleven
Explanation: NU = NV = NT -3x + six = -5x -3x + 5x = -6 2x = -six x = -step 3 NT = -5(-3) = fifteen
Explanation: NZ = New york = NW 4x – 10 = 3x – step 1 x = nine NZ = 4(9) – ten = thirty six – 10 = twenty six NW = twenty-six
Select the coordinates of your own centroid of triangle wilt the considering vertices. Concern 9. J(- step 1, 2), K(5, 6), L(5, – 2)
Let A good(- 4, 2), B(- cuatro, – 4), C(0, – 4) function as vertices of your own provided triangle and you will let P(x,y) end up being the circumcentre on the 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 Р5 = \(\frac < 1> < 2>\)(x Р2) 2y Р10 = x Р2 x Р2y + 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 Р5 = \(\frac < -1> < 2>\)(x + 2) blk profil ̦rnekleri 2y Р10 = -x Р2 x + 2y Р8 = 0 equate both equations x Р2y + 8 = x + 2y Р8 -4y = -16 y = 4 x Р2(4) + 8 = 0 x = 0 So, the orthocenter is (0, 4) The orthocenter lies inside the triangle TUV