Draw a line #l# and two points A and B not lying on l. Make sure that the
line #bar(AB)# is not perpendicular to #l#. Find the point C on #l# such that AC = BC?

Question

Sadie Allen · Accepted Answer

See below.

Given a line

#L_1->p=p_1+lambda_1 vec v_1#

and two points #p_a# and #p_b# not pertaining to #L_1# and not perpendicular to #L_1#

for #p in L_1# we have

#norm(p-p_a)=norm(p-p_b)# or

#norm(p_1+lambda_1 vec v_1-p_a)^2=norm(p_1+lambda_1 vec v_1-p_b)^2# or

#norm(p_1-p_a)^2+2lambda_1 << p_1-p_a, vec v_1 >> +lambda_1^2 norm(vec v_1)^2=norm(p_1-p_b)^2+2lambda_1 << p_1-p_b, vec v_1 >> +lambda_1^2 norm(vec v_1)^2# so

#norm(p_1-p_a)^2+2lambda_1 << p_1-p_a, vec v_1 >> =norm(p_1-p_b)^2+2lambda_1 << p_1-p_b, vec v_1 >> # so

#lambda_1^@ = (norm(p_1-p_b)^2-norm(p_1-p_a)^2)/(2<< p_b-p_a, vec v_1 >>)#

The contact point in #L_1# is

#p_c=p_1+lambda_1^@ vec v_1#

Attached an example with

#p_a = (1,3)#
#p_b = (4,3)#
#p_1 = (1,1)# and
#vec v_1 = (1,1)#

giving

#p_c = (5/2,5/2)#

Claire Akin · Answer

undefined To find the point $C$ on line $l$ such that $AC = BC$, where $A$ and $B$ are two points not lying on line $l$, follow these steps:

1. Draw line $l$.
2. Place points $A$ and $B$ anywhere not on line $l$.
3. Draw segment $\overline{AB}$ connecting points $A$ and $B$.
4. Bisect segment $\overline{AB}$ to find the midpoint $M$.
5. Draw a perpendicular line from midpoint $M$ to line $l$. Label the point of intersection with line $l$ as $C$.
6. $C$ is the point on line $l$ such that $AC = BC$.

Draw a line #l# and two points A and B not lying on l. Make sure that the line #bar(AB)# is not perpendicular to #l#. Find the point C on #l# such that AC = BC?