6.step one and you can six.step 3 Test

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

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

Explanation: Remember your circumcentre from an effective triangle are equidistant in the vertices out-of an effective triangle. Then 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 FatFlirt? + 8y + sixteen 12y = -several y = -step one PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + 16 + y? + 8y + sixteen = x? + y? + 8y + 16 8x = -sixteen x = -dos The fresh circumcenter is actually (-2, -1)

Explanation: Recall that circumcentre away from a great triangle are equidistant on vertices regarding an excellent triangle. Assist D(step three, 5), E(7, 9), F(eleven, 5) end up being the vertices of your considering triangle and assist P(x,y) end up being the circumcentre of this triangle. Following 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 + 44 + 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 + 25 -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = sixteen x – y = dos – (ii) Put (i) (ii) x + y + x – y = a dozen + dos 2x = fourteen x = eight Place x = seven in the (i) eight + y = twelve y = 5 The brand new circumcenter are (eight, 5)

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

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

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

Discover the coordinates of one’s centroid of triangle wilt the fresh new given vertices. Question nine. J(- step 1, 2), K(5, 6), L(5, – 2)

Let A(- 4, 2), B(- cuatro, – 4), C(0, – 4) become vertices of offered triangle and you will assist P(x,y) end up being the circumcentre associated with 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