## six.1 and 6.step 3 Quiz

Explanation: SV = VU 2x + eleven = 8x – step 1 8x – 2x = eleven + step 1 6x = several x = 2 Uv = 8(2) – step 1 = 15

Explanation: 5x – 4 = 4x + 3 x = 7 ?JGK = 4(7) + step 3 = 29 yards?GJK = 180 – (31 + 90) = 180 – 121 = 59

Explanation: Bear in mind that circumcentre out-of good triangle was equidistant on vertices from a beneficial triangle. Then PA = PB = Desktop computer PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + 16 + y? – 4y + cuatro = x? + 8x + sixteen + y? + 8y + sixteen 12y = -twelve y = -step one PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + sixteen + y? + 8y + 16 = x? + y? + 8y + 16 8x = -16 x = -2 This new circumcenter is (-2, -1)

Explanation: Keep in mind that circumcentre out-of an excellent triangle was equidistant from the vertices out of good triangle. Help D(step 3, 5), E(seven, 9), F(eleven, 5) function as vertices of the provided triangle and you may let P(x,y) function as the circumcentre from the triangle. Up coming 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 + bicupid ilk mesaj y? – 18y + 81 = x? – 22x + 121 + y? – 10y + twenty five -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = 16 x – y = 2 – (ii) Add (i) (ii) x + y + x – y = several + dos 2x = fourteen x = 7 Lay x = seven inside the (i) eight + y = several y = 5 The new circumcenter is (seven, 5)

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

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

Explanation: NZ = New york = NW 4x – ten = 3x – step one x = 9 NZ = 4(9) – ten = thirty-six – 10 = twenty six NW = 26

Discover the coordinates of your centroid of your triangle wilt the latest given vertices. Question 9. J(- 1, 2), K(5, 6), L(5, – 2)

## Help A great(- cuatro, 2), B(- cuatro, – 4), C(0, – 4) become vertices of your provided triangle and you may let P(x,y) function as the 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

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