The coordinates of a point on the line y=x where perpendicular from the line 3x+4y=12 is 4 units, are

(\frac{3}{7}, \frac{5}{7})

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(\frac{3}{2}, \frac{3}{2})

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(- \frac{8}{7}, - \frac{8}{7})

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(\frac{32}{7}, - \frac{32}{7})

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Solution

The correct option is A $(\frac{3}{7}, \frac{5}{7})$
Let any point on line $y = x$
from point of P(a,a),
Distance from point,
$\begin{matrix} ∣ ∣ ∣ \frac{3 a + 4 a - 12}{\sqrt{9 + 16}} ∣ ∣ ∣ = 4 ∣ ∣ ∣ \frac{7 a - 12}{5} ∣ ∣ ∣ = 4, \frac{7 a - 12}{5} = \pm 4 w e h a v e t w o c o n d i t i o n s, 7 a - 12 = 20 - - - - - (1) a = \frac{32}{7} 7 a - 12 = - 20 - - - - - - (2) a = \frac{- 8}{7} C o o r d i n a t e o f p o i n t i s a, t h e n, P (\frac{32}{7}, \frac{32}{7}) a n d P (\frac{- 8}{7}, \frac{- 8}{7}) S o, t h e c o r r e c t o p t i o n i s C . \end{matrix}$

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Similar questions

Q. The coordinates of the foot of the perpendicular from the points

(1, - 2, 1)

on the plane containing the lines,

\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}

and

\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}

is:

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Q. Find the value of

A

2 + 3 = 8

3 + 7 = 27

4 + 5 = 32

5 + 8 = 60

6 + 7 = 72

7 + 8 = A

Q. The slope intercept form of the line

3 x + 7 y + 8 = 0

Q. The length (in units) of the perpendicular from the point

(1, 2, 3)

to the line

\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}

is :

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