## six.step 1 and you will six.3 Test

Explanation: SV = VU 2x + 11 = 8x – 1 8x – 2x = eleven + 1 6x = twelve x = dos Uv = 8(2) – 1 = 15

Explanation: Keep in mind the circumcentre away from a great triangle is equidistant about vertices out of good triangle. Help Good(- cuatro, 2), B(- 4, – 4), C(0, – 4) be the vertices of the offered triangle and you will assist P(x,y) function as the circumcentre of the triangle. After that PA = PB = Desktop 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 = -12 y = -step one PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + 16 + y? + 8y + sixteen = x? + y? + 8y + 16 8x = -16 x = -dos The latest circumcenter are (-dos, -1)

Explanation: Keep in mind that the circumcentre from an effective triangle are equidistant about vertices of an excellent triangle. Let D(3, 5), E(seven, 9), F(11, 5) function as vertices of one’s offered triangle and you can help P(x,y) be the circumcentre of the triangle. Following 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 + forty-two + 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 + 44 + y? – 18y + 81 = x? – 22x + 121 + y? – 10y + twenty five -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = sixteen x – y = dos – (ii) Add (i) (ii) x + y + x – y = twelve + 2 2x = 14 x = seven Put x = 7 within the (i) 7 + y = twelve y = 5 The newest circumcenter was (7, 5)

Explanation: NQ = NR = NS 2x + 1 = 4x – nine 4x – 2x = 10 2x = 10 x = 5 NQ = 10 + step 1 = eleven NS = eleven

Explanation: NU = NV = NT -3x + 6 = -5x -3x + 5x = -6 2x = -6 x = -step three NT = -5(-3) = 15

Explanation: NZ = Nyc = NW 4x – ten = 3x – step 1 x = 9 NZ = 4(9) – ten = thirty-six – ten = 26 NW = twenty six

## Explanation: 5x – cuatro = 4x + 3 x = seven ?JGK = 4(7) + step three = 30 yards?GJK = 180 – (30 + 90) = 180 – 121 = 59

Discover coordinates of the centroid of your own triangle wilt the new offered vertices. Concern 9. 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 top sitios de citas europeos 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