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

six.step 1 and you can 6.3 Test

Explanation: SV = VU 2x + eleven = 8x – 1 8x – 2x = eleven + step 1 6x = 12 x = 2 Ultraviolet = 8(2) – 1 = fifteen

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

Explanation: Remember that circumcentre away from an effective triangle are equidistant on vertices from good triangle. Up coming PA = PB = Desktop PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + sixteen + y? – 4y + cuatro = x? + 8x + sixteen + y? + 8y + sixteen 12y = -a dozen y = -1 PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + 16 + y? + 8y + sixteen = x? + y? + 8y + sixteen 8x = -16 x = -dos The new circumcenter was (-dos, -1)

Explanation: Recall the circumcentre out-of a triangle is actually equidistant in the vertices of a great triangle. Let D(step 3, 5), E(eight, 9), F(eleven, 5) be the vertices of your given triangle and you may let P(x,y) function as circumcentre from the triangle. Then PD = PE = PF PD? = PE? = PF? PD? = PE? (x – 3)? + (y – 5) farmers dating site gibi uygulamalar? = (x – 7)? + (y – 9)? x? – 6x + nine + 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 = 16 x – y = 2 – (ii) Create (i) (ii) x + y + x – y = several + 2 2x = 14 x = eight Lay x = eight during the (i) eight + y = twelve y = 5 The newest circumcenter try (7, 5)

Explanation: NQ = NR = NS 2x + step one = 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 three NT = -5(-3) = 15

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

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

Help A great(- cuatro, 2), B(- 4, – 4), C(0, – 4) be the vertices of one’s considering triangle and you will assist P(x,y) be the circumcentre on 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