consider points A(3,4) and B(7,13). If p be a point on line y=x such that PA + PB is minimum, then coordinates of P are
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let the point be (x,x)
PA = √((x-3)²+(x-4)²)
PB = √((x-7)²+(x-13)²)
consider the following diagram,
the mirror of A on line y=x be it A' is (4,3)
PA+PB is minimum when the A'B intersects line y=x at P (x,x)
equation of line → y=mx+c
where m=(13-3)/(7-4)=10/3
c=-31
solving for P = (31/7,31/7)
PA = √((x-3)²+(x-4)²)
PB = √((x-7)²+(x-13)²)
consider the following diagram,
the mirror of A on line y=x be it A' is (4,3)
PA+PB is minimum when the A'B intersects line y=x at P (x,x)
equation of line → y=mx+c
where m=(13-3)/(7-4)=10/3
c=-31
solving for P = (31/7,31/7)
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Answer:
31/7,31/7
Step-by-step explanation:
step 1 find the image of A wrt y=x line
step 2 find the equation of A'B
step 3 solve equation y=x and A'B and answer is 31/7,31/7
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