A(-6,p), B(10,6) and C(30,q) and 3 collinear points. also 4q+5p=k. find the value of k.​

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1. Given :-

   Following three points are collinear,

   • A (- 6, p)

   • B(10, 6)

   • C(30, q).

   and

   • 4q + 5p = k

   To Find :-

   • The value of k

   Solution :-

   Given that

   Following three points are collinear,

   • A (- 6, p)

   • B(10, 6)

   • C(30, q).

   We know,

   Condition for 3 points to be collinear is given by

   [tex]\rm :\longmapsto\:x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0[/tex]

   Here,

   [tex] \rm :\longmapsto\:x_1 = – 6[/tex]

   [tex] \rm :\longmapsto\:y_1 = p[/tex]

   [tex] \rm :\longmapsto\:x_2 = 10[/tex]

   [tex] \rm :\longmapsto\:y_2 = 6[/tex]

   [tex] \rm :\longmapsto\:x_3 = 30[/tex]

   [tex] \rm :\longmapsto\:y_3 = q[/tex]

   Thus,

   On substituting the values, we get

   [tex]\rm :\longmapsto\: – 6(6 – q) + 10(q – p) + 30(p – 6) = 0[/tex]

   [tex]\rm :\longmapsto\: – 36 + 6q+ 10q -10 p+ 30p – 180= 0[/tex]

   [tex]\rm :\longmapsto\:20p + 16q – 216 = 0[/tex]

   [tex]\rm :\longmapsto\:4(5p + 4q – 54) = 0[/tex]

   [tex]\rm :\longmapsto\:5p + 4q – 54= 0[/tex]

   [tex]\rm :\longmapsto\:k – 54= 0 \: \: \: \: \: \: \: \: \: \: \{ \because \: 5p + 4q = k \}[/tex]

   [tex]\bf\implies \:k = 54[/tex]

   Additional Information :-

   Section Formula :-

   [tex]\rm :\longmapsto\: \:( x, y) = \bigg(\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)[/tex]

   Midpoint Formula :-

   [tex]\rm :\longmapsto\: \: ( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg)[/tex]

   Area of triangle :-

   [tex]\rm :\longmapsto\:\ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)][/tex]

   Distance Formula :-

   [tex]\rm :\longmapsto\: \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }[/tex]

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