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Question
Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
Solution
Let the equation of line AB be, 3x – 4y – 16 = 0 ....…(i)
or y = `3/4"x" - 4`
Slope of line AB = `3/4`
The perpendicular drawn from point C(−1, 3) to AB is CD.
∴ AB ⊥ CD
∴ Slope of CD = `(-1)/(("Slope of line AB")"`
= `(-1)/(3/4)`
= `(-4)/3`
Hence, the equation of line CD,
y – y1 = m(x – x1)
y – 3 = `(-4)/3 ("x" + 1)`
or 3y – 9 = –4x – 4
or 4x + 3y – 5 = 0 ....…(ii)
Multiplying equation (i) by 3 and equation (ii) by 4,
9x – 12y = 48
16x + 12y = 20
on adding these
25x = 68 or x = `68/25`
Putting the value of x in (i),
`3 xx 68/25 - 4"y" = 16`
∴ `4"y" = 204/25 - 16`
= `(204 - 400)/25`
∴ y = `-196/25 xx 1/4 = -49/25`
Hence, the coordinates of perpendicular foot D are `(68/25, -49/25)`.
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