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

## six.step one and you can six.step 3 Quiz

Explanation: SV = VU 2x + eleven = 8x – step one 8x – 2x = eleven + step 1 6x = several x = 2 Ultraviolet = 8(2) – step one = fifteen

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

Explanation: Recall your circumcentre out-of a triangle are equidistant about vertices away from a beneficial triangle. Then PA = PB = Pc PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + 16 + y? – 4y + cuatro = x? + 8x + sixteen + 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 + sixteen 8x = -16 x = -2 This new circumcenter was (-dos, -1)

Explanation: Recall that circumcentre of an excellent triangle was equidistant about vertices regarding a beneficial triangle. Assist D(3, 5), E(7, 9), F(eleven, 5) be the vertices of your own offered triangle and you can help P(x,y) function as circumcentre from the 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 = 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 + twenty five -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = sixteen x – y = dos – (ii) https://datingranking.net/tr/bookofmatches-inceleme/ Include (i) (ii) x + y + x – y = 12 + dos 2x = 14 x = eight Put x = 7 when you look at the (i) 7 + y = a dozen y = 5 The brand new circumcenter is actually (eight, 5)

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

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

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

Select the coordinates of the centroid of your triangle wilt the fresh new offered vertices. Question nine. J(- 1, 2), K(5, 6), L(5, – 2)

## Help A great(- cuatro, 2), B(- cuatro, – 4), C(0, – 4) function as the vertices of your own considering triangle and you will let P(x,y) end up being the circumcentre associated with the triangle

Explanation: The slope of TU = \(\frac \) = -2 The slope of the perpendicular line is \(\frac \) The perpendicular line is y – 5 = \(\frac \)(x – 2) 2y – 10 = x – 2 x – 2y + 8 = 0 The slope of UV = \(\frac \) = 2 The slope of the perpendicular line is \(\frac \) The perpendicular line is y – 5 = \(\frac \)(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

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