You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Condition for Concurrency of Three Lines

The line 4-3 ...

Question

The line 4 – 3y + 12 = 0 meets the y-axis at A. Write down the coordinates of A.

Determine the equation of the line passing through A and perpendicular to 4 – 3y + 12 = 0.

A(0 , 4); 3 + 4y = 16

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

A(0 , 4); + 4y = 9

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

A(0 , -4); 3 + y = -9

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

A(0 , -4); - 4y = 10

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

A(0 , 2); - 4y = 6

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

Open in App

Solution

The correct option is A
A(0 , 4); 3 + 4y = 16

Suggest Corrections

Similar questions

The line 4 – 3 + 12 = 0 meets the -axis at A. Write down the co-ordinates of A. Determine the equation of the line passing through A and perpendicular to 4 – 3 + 12 = 0.

The line 4x - 3y + 12 = 0 meets x-axis at A. Write the co-ordinates of A.

Determine the equation of the line through A and perpendicular to 4x - 3y + 12 = 0.

Q. The equation of the straight line passing through the intersection of the lines

x - y - 1 = 0

and

2 x - 3 y + 1 = 0

and parallel to

3 x + 4 y - 14 = 0

Q. A line passing through

A (- 5, - 4)

meets the line

x + 3 y + 2 = 0, 2 x + y + 4 = 0

and

x - 5 - y = 0

B, C

and

D

respectively. If

{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2}

, then the equation of line is

Q. A line through A (-5, -4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0, 2x + y + 4 = 0 and x – y – 5 = 0 at B, C and D respectively. If

{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2},

, then the equation of the line is

Join BYJU'S Learning Program

The line 4 – 3y + 12 = 0 meets the y-axis at A. Write down the coordinates of A. Determine the equation of the line passing through A and perpendicular to 4 – 3y + 12 = 0.

The correct option is A A(0 , 4); 3 + 4y = 16

The line 4 – 3y + 12 = 0 meets the y-axis at A. Write down the coordinates of A.

Determine the equation of the line passing through A and perpendicular to 4 – 3y + 12 = 0.

The correct option is A
A(0 , 4); 3 + 4y = 16