Hi Hazel,
You have two lines that intersect on the y axis at (0,5). My first approach is to imagine that I translate the diagram downwards 5 units so that the line intersect at the origin, and the rotate the diagram so that the line y = 5 + 2x lies along the x axis. This will not change the shape of the locus, so if I can fond the shape in this position then I will know the shape in the original position.
In the new position the line y = 5  2x become a line y = mx. I chose a point that satisfies the required property and called it P. I let d be the distance from P to the line y = mx. Below is my diagram.
Angle PQT is a right angle so angle QTP plus angle TPQ is 90 degrees. Also angle TPQ plus angle QPS is 90 degrees so angle QTP is congruent to angle QPS. Also angle PSQ is congruent to angle TPQ so triangles QTP and SQP are similar. Thus
^{SP}/_{TP }= ^{SQ}/_{QP}
But t ^{SP}/_{TP } is m, the slope of the line so
SQ = m d
The important fact here is that SQ is a constant times d.
Now apply Pythagoras Theorem to triangle SQP and you will see that SP is also a constant times d. Hence SR is a constant times d. But SR is the ycoordinate of S, and the ycoordinate of S satisfies = y mx. Thus the xcoordinate is a constant times d. But the xcoordinate of P is the same as the xcoordinate of S and thus the coordinates of P are
(a constant d, 2 d)
Hence the ycoordinate of P divided by the xcoordinate of P is a constant. If you do this carefully you will see that this constant only depends on m. I want to call this constant k.
Hence every point on the locus satisfies y = k x. Hence the locus is a line through the origin. Now move the diagram back to the original position and the required locus is a line through the intersection point of the two given lines.
Actually I biased the argument by placing P below the line y = mx. If P is above the line y = mx you get another line and hence the locus is two lines that intersect at the point of intersection of the two given lines.
Penny
