Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

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

Given:

The required line:

Passes through the point (2, 2)
Cuts off intercepts on the x- and y-axes whose sum is 9

Step 1: Use intercept form of a straight line

If a line cuts intercepts a and b on the x- and y-axes respectively, its equation is:

x/a + y/b = 1

Given:

a + b = 9 … (1)

Step 2: Use the condition that the line passes through (2, 2)

Substitute x = 2, y = 2 in the intercept form:

2/a + 2/b = 1

Divide throughout by 2:

1/a + 1/b = 1/2 … (2)

Step 3: Solve equations (1) and (2)

From (2):

( a + b ) / (ab) = 1/2

Substitute a + b = 9:

9 / (ab) = 1/2

ab = 18

So we have:

a + b = 9, ab = 18

These are satisfied by:

a = 3, b = 6 or a = 6, b = 3

Step 4: Write the equations of the lines

For a = 3, b = 6:

x/3 + y/6 = 1

For a = 6, b = 3:

x/6 + y/3 = 1

Final Answer:

The equations of the required line(s) are:

x/3 + y/6 = 1 and x/6 + y/3 = 1

Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Solution and Explanation

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