Question

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Answer 1

Step-by-step explanation:

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Answer 2

Solution:-

For solving this question we have two methods

1. Triangle area method
2. Determinant method

Triangle area

In this method if area of triangle comes zero than point are collinear

$\mathtt{A = \frac{1}{2} \times x_1(y_2 - y_3) + x_2(y_3 - y_1) + x3(y_1 - y_2)}$

Putting values of variables

$\mathtt{A = \frac{1}{2} \times 3(p - ( - 5)) + 5( - 5- 1) + 7(1- p)}$

$\mathtt{0 = \frac{1}{2} \times 3(p + 5) + 5( - 6) + 7(1- p)}$

$\mathtt{0 = 3p + 15 - 30+ 7- 7p}$

$\mathtt{p = \frac{ -8 }{ \: \: \:4} = - 2}$

Determinant method:-

Here also area should be zero For collinear

$\mathtt{\frac{1}{2}\left|\begin{array}{cc}X_1 & Y_1&1 \\X2 &Y_2&1\\X_3&Y_3&1\end{array}\right| = A}$

$\left|\begin{array}{cc}3& 1&1 \\5 & p&1\\7& - 5&1\end{array}\right| = 0$

Solving determinant we get

$\mathtt{3p + 15 + 2 - 25 - 7p = 0}$

From here

$\large\mathtt{p = \frac{ - 8}{ \: \: 4} = - 2}$