Question

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

Answer 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

$\rm :\longmapsto\:x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0$

Here,

$\rm :\longmapsto\:x_1 = - 6$

$\rm :\longmapsto\:y_1 = p$

$\rm :\longmapsto\:x_2 = 10$

$\rm :\longmapsto\:y_2 = 6$

$\rm :\longmapsto\:x_3 = 30$

$\rm :\longmapsto\:y_3 = q$

Thus,

On substituting the values, we get

$\rm :\longmapsto\: - 6(6 - q) + 10(q - p) + 30(p - 6) = 0$

$\rm :\longmapsto\: - 36 + 6q+ 10q -10 p+ 30p - 180= 0$

$\rm :\longmapsto\:20p + 16q - 216 = 0$

$\rm :\longmapsto\:4(5p + 4q - 54) = 0$

$\rm :\longmapsto\:5p + 4q - 54= 0$

$\rm :\longmapsto\:k - 54= 0 \: \: \: \: \: \: \: \: \: \: \{ \because \: 5p + 4q = k \}$

$\bf\implies \:k = 54$

Additional Information :-

Section Formula :-

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

Midpoint Formula :-

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

Area of triangle :-

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

Distance Formula :-

$\rm :\longmapsto\: \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$