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Question
Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
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Solution
Let the slope of the required line PQ be m.
The equation of the line PQ, which passes through the point P(−1, 2) and has slope m, is
y – y1 = m(x – x1)
y – 2 = m(x + 1)
or mx – y + m + 2 = 0 ....…(i)
equation of line AB x+ y = 4
∴ y = 4 – x
Putting the value of y in equation (1),
mx – (4 – x) + m + 2 = 0
or (m + 1) x + m – 2 = 0
∴ x = `- ("m" - 2)/("m" + 1)`
Now y = 4 − x
= `4 + ("m" - 2)/("m" + 1)`
= `(4"m" + 4 + "m" - 2)/("m" + 1) = (5"m" + 2)/("m" + 1)`
Given: PQ = 3 or PQ2 = 9
∴ `(- ("m" - 2)/("m" + 1) + 1)^2 + ((5"m" + 2)/("m" + 1) - 2)^2 = 9`
or `((-"m" + 2 + "m" + 1)/("m" + 1))^2 + ((5"m" + 2 - 2"m" - 2)/("m" + 1))^2 = 9`
or `9/("m" + 1)^2 + ((3"m")/("m"+ 1))^2 = 9`
or `(9 + 9"m"^2)/("m" + 1)^2 = 9`
or 1 + m2 = (1 + m)2
∴ 1 + m2 = 1 + 2m + m2
or 2m = 0
or m = 0
Hence, the slope of line PQ is 0 i.e., the line is parallel to the x-axis.
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