Abscissa and Ordinate....

Geometry Level 4

A straight line L with negative slope which passes through point (8,2) and cuts the positive coordinate axes at points P and Q. Find the absolute minimum value of OP + OQ, as L varies, where O is the origin.

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The answer is 18.

Harshvardhan Mehta by Harshvardhan Mehta , 23, India

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1 solution

展豪 張
Mar 6, 2016

Let the equation of L be y − 2 x − 8 = m \frac{y-2}{x-8} = m where m m is the slope.
Solving for the intercepts, we have O P , O Q = − 8 m + 2 , 8 − 2 m OP, OQ = -8m+2, 8-\frac 2m .
O P + O Q = − 8 m + 10 − 2 m OP+OQ = -8m+10-\frac 2m
d ( O P + O Q ) d m = − 8 + 2 m 2 \frac {d(OP+OQ)}{dm} = -8 + \frac 2 {m^2}
it equals 0 0 when m = − 0.5 m = -0.5
putting back the value of m m , O P + O Q = 18 OP+OQ = 18


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