Find the equation of the line which passes through the points (-1, 1) and (2, -4).

Question

If the components of \( \vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \) along and perpendicular to \( \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \) respectively are \( \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) \) and \( \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k}) \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:

Answer 1

It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is \(y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\)
Therefore, the equation of the line passing through the points (-1, 1) and (2, -4) is
\(y-1=\frac{-4-1}{2+1}\left(x+1\right)\)

\(y – 1 =\frac{ -5}{3} (x + 1)\)

\(3 (y – 1) = (-5) (x + 1)\)
\(3y – 3 = -5x – 5\)
\(3y – 3 + 5x + 5 = 0\)
\(i.e,5x + 3y + 2 = 0\)

Find the equation of the line which passes through the points (-1, 1) and (2, -4).

Solution and Explanation

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