The line p makes equal intercepts on the axes at A and at B, as shown - Leaving Cert Mathematics - Question 2 - 2015

To find the equation of line p that passes through the point (1, 5) with a slope of 1, we can use the point-slope form of a line:

$y - y_1 = m(x - x_1)$

Substituting in the values:

$y - 5 = 1(x - 1)$

Simplifying this gives:

$y - 5 = x - 1$ $y - x + 4 = 0$

This can be rearranged into the form $ax + by + c = 0$.

The slope of line p is 1, therefore the slope of line q, which is perpendicular to p, will be the negative reciprocal:

$m_q = -1$

Using the point-slope formula with point O(0, 0):

$y - 0 = -1(x - 0)$

This simplifies to:

$y + x = 0$.

In triangles OCA and OBC:

• |OA| = |OB|, which are equal intercepts.
• |OC| is a common line segment.
• ∠OCA = ∠OBC, both angles are right angles.

By the criteria of R.H.S (Right angle-Hypotenuse-Side), the triangles OCA and OBC are congruent.