A line passing through the point P(1,2) meets the line x+y=7 at the distance of 3 units from P. Then theslope of this line satisfies the equation :

A line passing through the point P(1,2) meets the line x+y=7 at the di

1.	A line passing through the point P(1,2) meets the line x+y=7 at the distance of 3 units from P. Then theslope of this line satisfies the equation :
Answer» `8x^(2)-9x+1=0` `7x^(2)-18x+7=0` `16x^(2)-39x+16=0` `7x^(2)-6x-7=0` Solution :The equation of a line through P(1,2) is `(x-1)/(cos theta)=(y-2)/(sin theta)` The COORDINATES of point of this line at a distance of 3 units from P(1,2) are given by `(x-1)/(cos theta)=(y-2)/(sin theta)=PM3` . Letthe coordinates of the points be `(1 pm 3 cos theta, 2 pm sin theta)` . Thesepoints lie on `x+y=7`. `(1 pm 3 cos theta)+(2 pm 3 sintheta)=7` `IMPLIES pm3(cos theta + sin theta)=4` `implies 9(1+sin 2theta)=16` `implies (18tan theta)/(1+tan^(2)theta)=7` `implies 7 tan^(2)theta-18 tan theta +7=0` `implies tan theta ` is a root of `7x^(2)-18x+7=0`