Question

Consider the point $A=(3,4)$ $B=(7,13)$

If $P$ be a point of the line $y=x$ such that $PA+PB$ is minimum then coordinates of $P$ is

• A. $(\frac{13}{7},\frac{73}{7})$
• B. $(\frac{23}{7}\cdot \frac{23}{7})$
• C. $(\frac{31}{7},\frac{31}{7})$
• D. $(\frac{33}{7},\frac{33}{7})$

Answer 1

The correct option is C $(\frac{31}{7},\frac{31}{7})$

$PointA(3,4)andB(7,13),liney=xletpoint{A}^{\prime }theimageofAiny=xthen,PA=P{A}^{\prime }ifP{A}^{\prime }+PBisminimum,i.e,ifP,A^{\prime },Barecollinearslopeofliney=xis1e{q}^{n}ofline{\perp }^{r}toy=xandpassingthrough(3,4)willbe\left(\frac{y-4}{x-3}\right)=-1ory=-x+7forMy=-x+7orx=-x+7orx=\left(\frac{7}{2}\right)then\left(\frac{a+3}{2}\right)=\left(\frac{7}{2}\right)ora=4\left(\frac{b+4}{2}\right)=\left(\frac{7}{2}\right)orb=3\therefore A^{\prime }is(4,3)nowco-ordinatesofPis(x,x)asP,A^{\prime },Barecollinearhenceslopewillbesameor\left(\frac{13-x}{7-x}\right)=\left(\frac{13-3}{7-4}\right)or(13-x)\times 3=10(7-x)or39-3x=70-10xor7x=31\therefore x=\left(\frac{31}{7}\right)henceP=\left(\right(\frac{31}{7}),(\frac{31}{7}\left)\right)$