Question

The product of perpendicular drawn from the point (1,2) to the pair of lines x^2+ 4xy + y^2=0 is

Answer 1

Answer:

product of the slopes of the line x+y =0 and the perpendicular upon it from (1,2) will be = -1

Answer 2

Given : point = (1,2) and line = x² +4xy+y²=0

Solution :

The product of perpendicular drawn from the point$(x_1,y_1)$  to the line is $ax^2+2hxy+by^2=0$ is

$\left | \frac{ax_1^2+2hx_1y_1+by_1^2}{\sqrt{(a-b)^2+4h^2}} \right |$

here , The point is (1,2) and line = x² +4xy+y²=0

thus

$\left | \frac{1.1^2+2\times 2\times 1\times 2+1\times 2^2}{\sqrt{(1-1)^2+4\times 2^2}} \right |= \frac{13}{4}$

hence ., The product of perpendicular drawn from the point (1,2) and line = x² +4xy+y²=0 is $\frac{13}{4}$

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