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

6.1 and you may 6.step three Quiz

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

Explanation: 5x – 4 = 4x + 3 times = seven ?JGK = 4(7) + step 3 = 29 yards?GJK = 180 – (30 + 90) = 180 – 121 = 59

Explanation: Keep in mind that circumcentre off a good triangle was equidistant on the vertices out-of a great triangle. After that PA = PB = Desktop PA? = PB? = PC? PA? bilgisayara airg indir = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + sixteen + y? – 4y + cuatro = x? + 8x + sixteen + y? + 8y + sixteen 12y = -12 y = -step 1 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 was (-2, -1)

Explanation: Recall that the circumcentre out-of good triangle try equidistant from the vertices out-of a great triangle. Help D(step 3, 5), E(7, 9), F(eleven, 5) function as vertices of the provided triangle and you can help P(x,y) become circumcentre of triangle. Upcoming 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 + 44 + 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 + 49 + y? – 18y + 81 = x? – 22x + 121 + y? – 10y + twenty five -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = 16 x – y = 2 – (ii) Put (i) (ii) x + y + x – y = several + 2 2x = fourteen x = eight Set x = seven in (i) seven + y = a dozen y = 5 The circumcenter are (seven, 5)

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

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

Explanation: NZ = Ny = NW 4x – 10 = 3x – 1 x = nine NZ = 4(9) – ten = thirty six – 10 = twenty six NW = 26

Discover the coordinates of the centroid of triangle wilt the newest considering vertices. Concern nine. J(- 1, 2), K(5, 6), L(5, – 2)

Let A good(- 4, 2), B(- 4, – 4), C(0, – 4) end up being the vertices of given triangle and you can let P(x,y) end up being the circumcentre associated with 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

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