six.1 and 6.step three Quiz
Explanation: SV = VU 2x + 11 = 8x – step 1 8x – 2x = eleven + 1 6x = a dozen x = 2 Uv = 8(2) – step one = 15
Explanation: 5x – cuatro = 4x + three times = seven ?JGK = 4(7) + step 3 = 29 yards?GJK = 180 – (30 + 90) = 180 – 121 = 59
Explanation: Keep in mind the circumcentre from a good triangle was equidistant regarding the vertices out of good triangle. After that PA = PB = Pc PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + 16 + y? – 4y + cuatro = x? + 8x + 16 + y? + 8y + sixteen 12y = -several y = -step one https://www.datingranking.net/tr/blackdatingforfree-inceleme/ PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + sixteen + y? + 8y + sixteen = x? + y? + 8y + sixteen 8x = -16 x = -2 The fresh new circumcenter was (-2, -1)
Explanation: Bear in mind that circumcentre out of an effective triangle was equidistant on the vertices regarding a great triangle. Assist D(step three, 5), E(seven, 9), F(11, 5) function as the vertices of one’s offered triangle and you can help P(x,y) function as circumcentre of the triangle. Next 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 = several – (i) PE? = PF? (x – 7)? + (y – 9)? = (x – 11)? + (y – 5)? x? – 14x + 49 + y? – 18y + 81 = x? – 22x + 121 + y? – 10y + twenty-five -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = sixteen x – y = 2 – (ii) Incorporate (i) (ii) x + y + x – y = twelve + 2 2x = fourteen x = seven Place x = 7 inside (i) 7 + y = 12 y = 5 This new circumcenter was (seven, 5)
Explanation: NQ = NR = NS 2x + 1 = 4x – nine 4x – 2x = 10 2x = ten x = 5 NQ = ten + step 1 = eleven NS = 11
Explanation: NU = NV = NT -3x + 6 = -5x -3x + 5x = -6 2x = -6 x = -step three NT = -5(-3) = fifteen
Explanation: NZ = Nyc = NW 4x – 10 = 3x – step 1 x = 9 NZ = 4(9) – ten = thirty six – ten = twenty six NW = 26
Discover coordinates of your own centroid of one’s triangle wilt the newest offered vertices. Matter 9. J(- 1, 2), K(5, 6), L(5, – 2)
Let Good(- cuatro, 2), B(- cuatro, – 4), C(0, – 4) become vertices of one’s offered triangle and you will help P(x,y) end up being the circumcentre of 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 -4y = -16 y = 4 x – 2(4) + 8 = 0 x = 0 So, the orthocenter is (0, 4) The orthocenter lies inside the triangle TUV