Explanation: QP = QR 6x = 3x + nine 3x = 9 x = 3 QP = 6(3) = 18

6.step 1 and you can six.step 3 Quiz

Explanation: SV = VU 2x + 11 = 8x – step one 8x – 2x = eleven + step one 6x = a dozen x = dos Ultraviolet = 8(2) – step 1 = 15

Explanation: 5x – 4 = 4x + three times = 7 ?JGK = 4(7) + 3 = 30 m?GJK = 180 – (31 + 90) = 180 – 121 = 59

Explanation: Remember the circumcentre out of a triangle is equidistant about vertices of good triangle. Up coming PA = PB = Pc 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 = -a dozen y = -step one PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + sixteen + y? + 8y + 16 = x? + y? + 8y + sixteen 8x = -16 x = -dos This new circumcenter is (-dos, -1)

Explanation: Keep in mind your circumcentre off a good triangle try equidistant about vertices away from an excellent triangle. Assist D(step three, 5), E(seven, 9), F(11, 5) function as vertices of your own given triangle and let P(x,y) become 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 = 12 – (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 = 2 – (ii) Create https://datingranking.net/pl/whatsyourprice-recenzja/ (i) (ii) x + y + x – y = a dozen + dos 2x = 14 x = seven Set x = seven inside (i) seven + y = 12 y = 5 The new circumcenter try (7, 5)

Explanation: NQ = NR = NS 2x + step one = 4x – 9 4x – 2x = 10 2x = 10 x = 5 NQ = ten + 1 = 11 NS = 11

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

Explanation: NZ = Nyc = NW 4x – ten = 3x – 1 x = 9 NZ = 4(9) – ten = 36 – 10 = twenty-six NW = twenty six

Find the coordinates of your centroid of your own triangle wilt the offered vertices. Concern nine. J(- step one, 2), K(5, 6), L(5, – 2)

Assist An effective(- 4, 2), B(- 4, – 4), C(0, – 4) end up being the vertices of considering triangle and you may let P(x,y) become circumcentre for 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 – 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