A line passing through the point P(1,2) meets the line `x+y=7` at the distance of 3 units from P. Then the slope of this line satisfies the equation :A. `8x^(2)-9x+1=0`B. `7x^(2)-18x+7=0`C. `16x^(2)-39x+16=0`D. `7x^(2)-6x-7=0`

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

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 the slope of this line satisfies the equation :A. `8x^(2)-9x+1=0`B. `7x^(2)-18x+7=0`C. `16x^(2)-39x+16=0`D. `7x^(2)-6x-7=0`
Answer» Correct Answer - B 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)` . These points 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`

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A line passing through the point P(1,2) meets the line `x+y=7` at the distance of 3 units from P. Then the slope of this line satisfies the equation :A. `8x^(2)-9x+1=0`B. `7x^(2)-18x+7=0`C. `16x^(2)-39x+16=0`D. `7x^(2)-6x-7=0`

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