Points on the line $y = x$ whose perpendicular distance from the line $3 x + 4 y = 12$ is $4$ have the coordinates

(- \frac{8}{7}, - \frac{8}{7}), (- \frac{32}{7}, - \frac{32}{7})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(\frac{8}{7}, \frac{8}{7}), (\frac{32}{7}, \frac{32}{7})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- \frac{8}{7}, - \frac{8}{7}), (\frac{32}{7}, \frac{32}{7})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(- \frac{8}{7}, - \frac{8}{7}), (\frac{32}{7}, \frac{32}{7})$
Let the required point be $(x_{1}, x_{1})$
Now, $d = \frac{| 3 x_{1} + 4 x_{1} - 12 |}{5} = 4$
$\Rightarrow 7 x_{1} - 12 = \pm 20$
$\Rightarrow x_{1} = - \frac{8}{7}, \frac{32}{7}$
$\Rightarrow y_{1} = - \frac{8}{7}, \frac{32}{7}$
Hence, the points are $(- \frac{8}{7}, - \frac{8}{7}), (\frac{32}{7}, \frac{32}{7})$

Suggest Corrections

Similar questions

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. The slope intercept form of the line

3 x + 7 y + 8 = 0

Q. Find the value of

A

2 + 3 = 8

3 + 7 = 27

4 + 5 = 32

5 + 8 = 60

6 + 7 = 72

7 + 8 = A

Find:

(a) (− 7) − 8 − (− 25)

(b) (− 13) + 32 − 8 − 1

(d) 50 − (− 40) − (− 2)

7 + 7 + 7 + 7 + 8 + 8 + 8 = 7 \times 4 + 8 \times 2

Join BYJU'S Learning Program

Points on the line y=x whose perpendicular distance from the line 3x+4y=12 is 4 have the coordinates

The correct option is C (−87,−87),(327,327)Let the required point be (x1,x1)Now, d=|3x1+4x1−12|5=4⇒7x1−12=±20⇒x1=−87,327⇒y1=−87,327Hence, the points are (−87,−87),(327,327)

Points on the line $y = x$ whose perpendicular distance from the line $3 x + 4 y = 12$ is $4$ have the coordinates