The co-ordinates of two points are A(4, -1) and B(7, t) - Leaving Cert Mathematics - Question 3 - 2015

The distance from a point to a line can be calculated using the formula:

$d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$

For line l₁: $3x - 4y - 12 = 0$, we have:

• A = 3
• B = -4
• C = -12

Substituting P(10, k) into the distance formula:

$d = \frac{|3(10) - 4(k) - 12|}{\sqrt{3^2 + (-4)^2}}$ $= \frac{|30 - 4k - 12|}{\sqrt{9 + 16}}$ $= \frac{|18 - 4k|}{5}$

To find the possible values of k, we need the conditions of being on the angle bisector. The slopes of line l₂ (from the equation $5x + 12y - 20 = 0$) can be computed as follows:

Rearranging gives:

$12y = -5x + 20 \Rightarrow y = -\frac{5}{12}x + \frac{20}{12}$

Thus, the slope of l₂ is $-\frac{5}{12}$. Using the angle bisector formula:

$\frac{y_1 - y_2}{x_1 - x_2} = \frac{m_1 - m_2}{1 + m_1m_2}$

Substituting the values leads to the solutions for k:

From the calculations:

• The process results in two possible values of k: $k = 4$ or $k = -48$.

If $k > 0$, we will take the positive value found:

Let’s use $k = 4$.

Then the distance from P(10, 4) to l₁ can be calculated:

Using the distance formula: $d = \frac{|3(10) - 4(4) - 12|}{\sqrt{3^2 + (-4)^2}}$ $= \frac{|30 - 16 - 12|}{5} = \frac{|2|}{5} = \frac{2}{5}$

Thus, the distance from point P to line l₁ is:

$3$